Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Q.uestion 1.
Find the co-ordinates of the point which divides the join (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let required point be P (x, y) which divides the join of given points A (- 1, 7)
and B (4, – 3) in the ratio of 2 : 3.
(-1, 7) (x, y) (4, – 3)

∴ x = \(\frac{2 \times 4+3 \times-1}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)

and y = \(\frac{2 \times-3+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Hence, required point be (1, 3).

Question 2.
Find the co-ordinates of the points of trisection of the line segment joining (4, – 1) and (2, – 3).
Solution:
Let P (x₁, y₁) and Q (x₂, y₂) be the required points which trisect the line segment joining A (4, – 1)and B (- 2, – 3) i.e., P(x₁, y₁) divides AB in ratio 1: 2 and Q divides AB in ratio 2 : 1.

∴ x₁ = \(\frac{1 \times-2+2 \times 4}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y₁ = \(\frac{1 \times-3+2 \times-1}{1+2}=\frac{-3-2}{3}=-\frac{5}{3}\)
∴ P(x₁, y₁) be (2, \(-\frac{5}{3}\))

Now, x₂ = \(\frac{2 \times-2+1 \times 4}{2+1}\)
= \(\frac{-4+4}{3}\) = 0

y₂ = \(\frac{2 \times-3+1 \times-1}{2+1}=\frac{-6-1}{3}=-\frac{7}{3}\)

∴ Q(x₂, y₂) be (0, \(-\frac{7}{3}\))
Hence, required points be (2, \(-\frac{5}{3}\)) and (0, \(-\frac{7}{3}\)).

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs \(\frac{1}{4}\) th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) th the distance AD on the eighth line and posts a red flag. What is the distance betweenboth the flags? If Rashmi has to post a blue flag exactly half way between the line (segment) joining the two flags, where should she post her flag?

Solution:
In the given figure, we take A as origin. Taking x-axis along AB and y-axis along AD.
Position of green flag = distance covered by Niharika
= Niharika runs \(\frac{1}{4}\)th distance AD on the 2nd line
= \(\frac{1}{4}\) × 100 = 25 m
∴ Co-ordinates of the green flag are (2, 25)
Now, position of red flag = distance covered by Preet = Preet runs \(\frac{1}{5}\)th the distance AD on the 8th line
= \(\frac{1}{5}\) × 100 = 20 m.
Co-ordinates of red flag are (8, 20)
∴ distance between Green and Red flags = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)
= \(\sqrt{36+25}=\sqrt{61}\) m.
Position of blue flag = mid point of green flag and red flag
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
= (5, 22.5).
Hence, blue flag is in the 5th line and at a distance of 22.5 m along AD.

Question 4.
Find the ratio in which (he segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let point P (- 1, 6) divides the line segment joining the points A (- 3, 10) and B (6, – 8) the ratio K : 1.

∴ -1 = \(\frac{6 \times \mathrm{K}-3 \times 1}{\mathrm{~K}+1}\)
or – K – 1 = 6K – 3
or – K – 6K = – 3 + 1
or – 7K = – 2
K : 1 = \(\frac{2}{7}\) : 1 = 2 : 7
Hence, required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the co
ordinates of the point of division.
Solution:
Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1, – 5) and B (- 4, 5) in the
ratio K : 1.

Consider, y co-ordinates of P ¡s:
0 = \(\frac{5 \times \mathrm{K}+(-5) \times 1}{\mathrm{~K}+1}\)

or 0 = \(\frac{5 \mathrm{~K}-5}{\mathrm{~K}+1}\)
or 5K – 5 = 0
or 5K = 5
or K = \(\frac{5}{5}\) = 1
∴ Required ratio is K : 1 = 1 : 1.
Now, x co-ordinate of P is:
x = \(\frac{-4 \times K+1 \times 1}{K+1}\)
Putting the value of K = 1, we get:
x = \(\frac{-4 \times 1+1 \times 1}{1+1}=\frac{-4+1}{2}\)
x = \(-\frac{3}{2}\)
Hence, required point be (\(-\frac{3}{2}\), 0).

Question 6.
If (1, 2); (4, y); (x, 6) and (3, 5)are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let points of parallelogram ABCD are A (1, 2) (4, y) ; C (x, 6) and D (3, 5)
But diagonals of a || gm bisect each other.
Case I. When E is the mid point of A (1, 2) and C (x, 6)
∴ Co-ordinates of E are:
E = \(\left(\frac{x+1}{2}, \frac{6+2}{2}\right)\)
E = (\(\frac{x+1}{2}\), 4) …………..(1)

Case II. When E is the mid point B (4, y) and D (3, 5)
∴ Co-ordinates of E are:

E = \(\left(\frac{3+4}{2}, \frac{5+y}{2}\right)\)

E = \(\left(\frac{7}{2}, \frac{5+y}{2}\right)\) …………….(2)
But values of E in (1) and (2) are same, so comparing the coordinates, we get
\(\frac{x+1}{2}=\frac{7}{2}\)
or x + 1 = 7
or x = 6.

and 4 = \(\frac{5+y}{2}\)
or 8 = 5 + y
or y = 3
Hence, values of x and y are 6 and 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let, coordinates of A be (x, y). But, centre is the’ niij ioint of the vertices of the diameter.

∴ O is the mid point of A(x, y) and B(1, 4)
∴ \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\) = (2, -3)
On comparing, we get
\(\frac{x+1}{2}\)
or x + 1 = 4
or x = 3

and \(\frac{y+4}{2}\) = – 3
or y + 4 = – 6
or y = – 10
Hence, required point A be (3, – 10).

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = AB and P lies ¡n the line segment AB.
Solution:
Let required point P be (x, y)
Also AP = \(\frac{3}{7}\) AB …(Given)
But, PB = AB – AP
= AB – \(\frac{3}{7}\) AB = \(\frac{7-3}{7}\) AB
PB = \(\frac{4}{7}\) AB
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\frac{3}{7} \mathrm{AB}}{\frac{4}{7} \mathrm{AB}}=\frac{3}{4}\).

∴ P divides given points A and B in ratio 3 : 4.
Now,
x = \(\frac{3 \times 2+4 \times-2}{3+4}\)

or x = \(\frac{6-8}{7}=-\frac{2}{7}\)

and y = \(\frac{3 \times-4+4 \times-2}{3+4}\)
= \(\frac{-12-8}{7}=-\frac{20}{7}\)

Hence, coordinates of P be (\(-\frac{2}{7}\), \(-\frac{20}{7}\)).

Question 9.
Find the coordinates of the points which divides the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Solution:
Let required points are C, D and E which divide the line segment joming the points A (- 2, 2) and B (2, 8) into four equal parts. Then D is mid point of A and B ; C is the mid point of A and D ; E is the mid point of D and B such that
AC = CD = DE = EB

Now, mid point of A and B (i.e., Coordinates of D)
= \(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\) = (0, 5)

Mid point of A and D (i.e., Coordinates of C)
= \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Mid point of D and B (i.e., Coordinates of E)
= \(\left(\frac{2+0}{2}, \frac{8+5}{2}\right)=\left(1, \frac{13}{2}\right)\)

Hence, requned points be (0, 5), (-1, \(\frac{7}{2}\)), (1, \(\frac{13}{2}\)).

Question 10.
Find the area of a rhombus if the vertices are (3, 0); (4, 5); (- 1, 4) and(- 2, – 1) taken in order.
[Hint: Areas of a rhombus = \(\frac{1}{2}\) (Product of its diagonals)]
Solution:
Let coordinates of rhombus ABCD are A (3, 0); B(4, 5); C(-1, 4) and D(- 2, – 1).
Diagonal, AC = \(\sqrt{(-1-3)^{2}+(4-0)^{2}}\)
= \(\sqrt{16+16}=\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

and diagonal BD
BD = \(\sqrt{(-2-4)^{2}+(-1-5)^{2}}\)
= \(\sqrt{36+36}=\sqrt{72}=\sqrt{36 \times 2}\) = 6√2.

∴ Area of rhombus ABCD = \(\frac{1}{2}\) × AC × BD
ABCD = [\(\frac{1}{2}\) × 4√2 × 6√2] sq. units
(\(\frac{1}{2}\) × 24 × 2) sq. units
= 24 sq. units
Hence, area of rhombus is 24 sq. units.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:
(i) Given points are: (2, 3); (4, 1)
Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
\(\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}\)
= 2√2.

(ii) Given points are: (-5, 7); (-1, 3)
Required distance = \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
\(\sqrt{16+16}=\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= 4√2.

(iii) Given points are: (a, b); (-a, -b)
Required distance = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= √4 \(\sqrt{a^{2}+b^{2}}\)
= \(2 \sqrt{a^{2}+b^{2}}\)

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:
Given points are: A (0, 0) and B (36, 15)
Distance, AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)
\(\sqrt{1296+225}=\sqrt{1521}\) = 39.
According to Section 7.2
Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Draw BC ⊥ on X-axis.
Now, In rt. ∠d ∆ACB,
AB = \(\sqrt{\mathrm{AC}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}=\sqrt{1521}\)
= 39.
Hence, required distance between points is 39.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Given point are : A (1. 5); B (2.3) and C (- 2, – 11).
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
= \(\sqrt{1+4}=\sqrt{5}\)

BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
= \(\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1+2)^{2}+(5+11)^{2}}\)
= \(\sqrt{9+256}=\sqrt{265}\)
From above distances, it is clear that sum of any two is not equal to third one.
Hence, given points are not collinear

Question 4.
Check whether (5, – 2); (6, 4) and (7, – 2) are the Vertices of an isosceles triangle.
Solution:
Given points be A (5, – 2); B (6, 4) and C (7, – 2).
AB = \(\sqrt{(5-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

BC = \(\sqrt{(6-7)^{2}+(4+2)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

CA = \(\sqrt{(7-5)^{2}+(-2+2)^{2}}\)
= \(\sqrt{4+0}=2\)
From above discussion, it is clear that AB = BC = √37.
Given points are vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

DA=\(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\)
= \(\sqrt{36+0}=6\)

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\)
= \(\sqrt{0+36}\) = 6
From above discussion, it is clear that
AB = BC = CD = DA = √18 and AC = BD = 6.
ABCD formed a square and Champa is correct about her thinking.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:
(i) Given points be A (- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0).
AB = \(\sqrt{(1+1)^{2}+(0+2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(-3+1)^{2}+(0-2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(-1+3)^{2}+(-2+0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+16}=4\)

BD = \(\sqrt{(-3-1)^{2}+(0-0)^{2}}\)
= \(\sqrt{16+0}=4\)

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

(ii) Given points be A (- 3, 5); B (3, 1); C (0, 3) and D (- 1,- 4)
AB = \(\sqrt{(-3-3)^{2}+(5-1)^{2}}\)
= \(\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times 13}\)
= 2√13

BC = \(\sqrt{(3-0)^{2}+(1-3)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)

CA = \(\sqrt{(0+3)^{2}+(3-5)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)
Now, BC + CA = \(\sqrt{13}+\sqrt{13}\) = 2√13 = AB
∴A, B and C are collinear then A, B, C and D do not form any quadrilateral.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)
AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

CD = \(\sqrt{(1-4)^{2}+(2-3)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+4}\) = 2

BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\)

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)² = (PB)²
or (2 – x)² + (- 5- 0)² = (- 2 – x)² + (9 – 0)²
or 4 + x² – 4x + 25 = 4 + x²+ 4x + 81
-8x = 56
x = \(\frac{4}{4}\) = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given points are P (2, – 3) and Q (10, y)
PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
= \(\sqrt{64+y^{2}+9+6 y}\)
= \(\sqrt{y^{2}+6 y+73}\)
According to question,
PQ = 10
or \(\sqrt{y^{2}+6 y+73}\) = 10
Squaring
or y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 9y – 3y – 27 = 0
S = 6 P = – 27
or y (y + 9) – 3 (y + 9) = 0
or (y + 9) (y – 3) = 0
Either y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence, y = – 9 and 3.

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given points Q (0, 1); P (5, – 3) and R (x, 6)
QP = \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)
= \(\sqrt{25+16}=\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{x^{2}+25}\)

According to question,
QP = QR
or \(\sqrt{41}=\sqrt{x^{2}+25}\)
Squaring
or 41 = x² + 25
or x² = 16
or x = ± √16 = ± √4.

When x = 4 then R (4, 6).
QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{1+81}=\sqrt{82}\)

When x = – 4 then R (- 4, 6).
QR = \(\sqrt{(-4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(-4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{81+81}=\sqrt{162}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let required points be P (x, y) and given points are A (3, 6) and B (- 3, 4)
PA = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
= \(\sqrt{9+x^{2}-6 x+36+y^{2}-12 y}\)
= \(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\)

and PB = \(\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
= \(\sqrt{9+x^{2}+6 x+16+y^{2}-8 y}\)
= \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)

According to question,
PA = PB
\(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\) = \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)
sq,. both sides, we have,
or x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y – 25
or -12x – 4y + 20 = 0
or 3x + y – 5 = 0 is the required relation.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is 15 for the first km and 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs 150 for the first metre and rises by 50 for each msubsequent metre.

(iv) The amount of money in the account every year when 10000 is deposited at compound interest at 8% per annum.

Solution:
(i) Let T_n denotes the taxi fare in n^th km.
According to question,
T₁ = 15 km;
T₂ = 15 + 8 = 23;
T₃ = 23 + 8 = 31
Now, T₃ – T₂ = 31 – 23 = 8
T₂ – T₁ = 23 – 15 = 8
Here, T₃ – T₂ = T₂ – T₁ = 8
∴ given situation form an AP.

(ii) Let T_n denotes amount of air present in a cylinder.
According to question,
T₁ = x;
T₂ = x – \(\frac{1}{4}\)x
= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x
T₃ = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)

= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on
Now, T₃ – T₂ = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x
= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x

T₂ – T₁ = \(\frac{3}{4}\)x – x
= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation donot form an AP.

(iii) Let T_n denotes cost of digging a well for the nth metre,
According to question,
T₁ = ₹ 150; T₂ = (150 + 50) = ₹ 200;
T₃ = ₹ (200 + 5o) = 250 and so on
Now, T₃ – T₂ = ₹ (250 – 200) = 50
T₂ – T₁ = ₹ (200 – 150) = 50
Here, T₃ – T₂ = T₂– T₁ = 50
∴ given situation form an A.P.

(iv) Let T_n denotes amount of money in the nth year.
According to question
T₁ = ₹ 10,000
T₂ = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + ₹ 800 = ₹ 10,800
T₃ = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + ₹ 864
= ₹ 11,640 and so on.
Now, T₃ – T₂ = ₹ (11,640 – 10,800) = ₹ 840
T₂ – T₁ = ₹ (10,800 – 10,000) = ₹ 800
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation do not form an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(1) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac{1}{2}\)
(w) a = -1.25, d = -0.25
Solution:
(i) Given that first term = a = 10
and common difference = d = 10
∴ T₁ = a = 10;
T₂ = a + d = 10 + 10 = 20;
T₃ = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T₄ = a + 3d = 10 + 3 × 10
= 10 + 30 = 40
Hence, first four terms of an A.P. are 10, 20, 30, 40………….

(ii) Given that first term = a = -2
and common iifference = d = 0
∴ T₁ = a = -2;
T₂ = a + d = -2 + 0 = -2
T₃ = a + 2d = -2 + 2 × 0 = -2
T₄ = a + 3d = -2 + 3 × 0 = -2
Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….

(iii) Given that first term = a = 4
and common difference = d = -3
∴ T₁ = a = 4;
T₂= a + d = 4 – 3 = 1
T₃ = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T₄ = a + 3d = 4 + 3(-3) = 4 – 9 = -5
Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….

(iv) Given that first term = a = -1
and common difference = d = \(\frac{1}{2}\)
∴ T₁ = a = -1; T₂ = a + d
= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)
T₃ = a + 2d = -1 + 2(\(\frac{1}{2}\))
= -1 + 1 = 0
T₄ = a + 3d = -1 + 3(\(\frac{1}{2}\))
= \(\frac{-2+3}{2}=\frac{1}{2}\)
Hence, first four terms of an AP are -1, –\(\frac{1}{2}\), 0, \(\frac{1}{2}\), …………..

(v) Given that first term = a = – 1.25
and common difference = d = – 0.25
∴ T₁ = a = – 1.25;
T₂ = a + d = – 1.25 – 0.25 = -1.50
T₃ = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T₄ = a + 3d = – 1.25 + 3(- 0.25)
= – 1.25 – 0.75 = – 2
Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..

Question 3.
For the following APs, wilte the first term and the common difference:
(i) 3, 1, -1, -3, …………
(ii) 5, -1, 3, 7, ………….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\), …………..
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) Given A.P., is 3, 1, -1, -3, ………
Here T₁ = 3, T₂ = 1,
T₃ = -1, T₄ = -3
First term = T₁ = 3
Now, T₂ – T₁ = 1 – 3 = – 2
T₃ – T₂ = – 1 – 1 = -2
T₄ – T₃ = -3 + 1 = -2
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
Hence, common difference = – 2 and first term = 3.

(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T₁ = – 5, T₂ = – 1,
T₃ = 3, T₄ = 7
First term T₁ = -5
Now, T₂ – T₁ = -1 + 5 = 4
T₃– T₂ = 3 + 1 = 4
T₄ – T₃ = 7 – 3 = 4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4
Hence, common difference = 4 and first term = – 5.

(iii) Given AP. is:
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
Here T₁ = \(\frac{1}{3}\), T₂ = \(\frac{5}{3}\),
T₃ = \(\frac{9}{3}\), T₄ = \(\frac{13}{3}\)
First term = T₁ = \(\frac{1}{3}\)
Now, T₂ – T₁ = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)
T₃ – T₂ = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)
T₄ – T₃ = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{4}{3}\)

Hence, common difference = \(\frac{4}{3}\) and first term = \(\frac{1}{3}\).

(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…
Here, T₁ = 0.6, T₂ = 1.7, T₃ = 2.8, T₄ = 3.9
First term = T₁ = 0.6
Now, T₂ – T₁ = 1.7 – 0.6 = 1.1
T₃ – T₂ = 2.8 – 1.7 = 1.1
T₄ – T₃ = 3.9 – 2.8 = 1.1
Hence, common difference = 1.1 and first term = 0.6.

Question 4.
WhIch of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ………….
(iv) – 10, – 6, – 2, 2, ………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ……….
(vi) 0.2, 0.22, 0.222, 0.2222, ………….
(vii) 0, -4, -8, -12, …………..
(viii) \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ………..
(ix) 1, 3, 9, 27 …………….
(x) a, 2a, 3a, 4a, ………………
(xi) a, a², a³, a⁴, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 1², 3², 5², 7², ………..
(xv) 1², 5², 7², 73, ………….

Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T₁ = 2, T₂ = 4, T₃ = 8, T₄ = 16
T₂ – T₁ = 4 – 2 = 2
T₃ – T₂ = 8 – 4 = 4
∵ T₂ – T₁ ≠ T₃ – T₂
Hence, given terms do not form an A.P.

(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
Here T₁ = 2, T₂ = 4, T₃ = 3, T₄ = 16
T₂ – T₁ = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)
T₃ – T₂ = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)
T₄ – T₃ = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{1}{2}\)
∴ Common difference = d = \(\frac{1}{2}\)
Now, T₅ = a + 4d = 2 + 4\(\frac{1}{2}\) = 4

T₆ = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)

T₇ = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.

(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T₁ = – 1.2, T₂ = – 3.2,
T₃ = – 5.2, T₄ = – 7.2
T₂ – T₁ = – 3.2 + 1.2 = – 2
T₃ – T₂ = – 5.2 + 3.2 = – 2
T ₄ – T₃ = – 7.2 + 5.2 = – 2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
∴ Common difference = d = – 2
Now, T₅ = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T₆ = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T₇ = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2

(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T₁ = – 10,T₂ = – 6
T₃ = – 2, T₄=2 .
T₂ – T₁ = – 6 + 10 = 4
T₃ – T₂ = – 2 + 6 =4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁=T₃ – T₂ = T₄ – T₃ = 4 .
∴ Common difference = d = 4
Now, T₅ = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T₆ = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T₇ = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.

(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T₁ = 3, T₂ = 3 + √2,
T₃ = 3 + 2√2, T₄= 3 + 3√2
T₂ – T₁ = 3 + √2 – 3 = √2
T₃ – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T₄ – T₃ = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T₂ -T₁ = T₃ – T₂ = T₄ – T₃ = √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = 3 + 4(√2) = 3 + 4√2
T₆ = a + 5d = 3 + 5√2
T₇ = a + 6d = 3 + 6√2

(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T₁ = 0.2, T₂ = 0.22,
T₃ = 0.222, T₄ = 0.2222.
T₂ – T₁ = 0.22 – 0.2 = 0.02
T₃ – T₂ = 0.222 – 0.22 = 0.002
∵ T₂ – T₁ ≠ T₃ – T₂
∴ given terms do not form an A.P.

(vii) Given terms are 0, -4, -8, -12
Here T₁ = 0, T₂ = -4,
T₃ = -8, T₄ = -12
T₂ – T₁ = – 4 – 0 = -4
T₃ – T₂= – 8 + 4 = -4
T₄ – T₃= – 12 + 8 = -4.
T₂ – T₁ = T₃ – T₂ = T₄ – T₃
∴ Common difference = d = -4
Now, T₅= a + 4d = 0 + 4(-4) = -16
T₆ = a + 5d = 0 + 5(-4) = -20
T₇ = a + 6d = 0 + 6(-4) = -24.

(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….
Here T₁ = \(-\frac{1}{2}\), T₂ = –\(\frac{1}{2}\)
T₃ = \(-\frac{1}{2}\), T₄ = \(-\frac{1}{2}\)
T₂ – T₁ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
T₃ – T₂ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
∵ T₂ – T₁ = T₃ – T₂ = 0
∴ Common difference = d = 0
Now, T₅ = T₆ = T₇ = –\(\frac{1}{2}\)
[∵ a = –\(\frac{1}{2}\), d = 0]

(ix) Given terms are 1, 3, 9, 27
T₁ = 1, T₂ = 3, T₃ = 9, T₄ = 27
T₂ – T₁ = 3 1 = 2
T₃ – T₂ = 9 – 3 = 6.
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(x) Given terms are a, 2a, 3a, 4a, …
T₁ = a, T₂ = 2a, T₃ = 3a, T4 = 4a
T₂ – T₁ = 2a – a = a
T₃ – T₂ = 3a – 2a = a
T₄ – T₃ = 4a – 3a = a
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = a
∴ Common difference = d = a
Now T₅ = a + 4d = a + 4(a) = a + 4a = 5a
T₆ = a + 5d = a + 5a = 6a
T₇ = a + 6d = a + 6a = 7a

(xi) Given terms are a, a², a³, a⁴, …………
T₁ = a, T₂ = a², T3 = a³, T₄ = a⁴
T₂ – T₁ = a² – a
T₃ – T₂ = a³ – a²
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xii) Given terms are √2, √8, √18, √32, …………
T₁ = √2, T₂ = √8, T₃ = √18, T₄ = √32
or T₁ = √2, T₂ = 2√2 T₃ = 3√2, T₄ = 4√2
T₂ – T₁ = 2√2 – √2 = √2
T₃ – T = 3√2 – 2√2 = √2
T₄ – T₃ = 4√2 – 3√2 = √2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃= √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = √2 + 4√2 = 5√2
T₆ = a + 5d = √2 + 5√2 = 6√2
T₇ = a + 6d = √2 + 6√2 = 7√2

(xiii) Given terms are √3, √6, √9, √12, ……………..
T₁ = √3, T₂= √6, T₃= √9, T₄= √12
or T₁ = √3, T₂ = √6, T₃ = 3, T₄ = 2√3
T₄ – T₁ = √6 – √3
T₃ – T₂ = 3 – √6
∵ T₂ – T₁ ≠ T₃ – T₂
∴Given terms do not form an A.P.

(xiv) Given terms are 1², 3², 5², 7², ………..
T₁ = 1², T₂ = 3², T₃ = 5², T₄ = 7²
or T₁ = 1, T₂ = 9, T₃ = 25, T₄ = 49
T₄ – T₁ = 9 – 1 = 8
T₃ – T₂ = 25 – 9 = 16
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xv) Given terms are 12, 52, 72, 73
T₁ = 12, T₂ = 52, T₃ = 72, T₄ = 73
or T₁ = 1, T₂ = 25, T₃ = 49, T₄ = 73
T₂ – T₁ = 25 – 1 = 24
T₃ – T₂ =49 – 24= 24
T₄ – T₃ = 73 – 49 = 24
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 24
∴ Common difference = d = 24
T₅ = a + 4d = 1 + 4(24) = 1 + 96 = 97
T₆ = a + 5d = 1 + 5(24) = 1 + 120 = 121
T₇ = a + 6d = 1 +6(24) = 1 + 144 = 145

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 1
Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the
AP:

Solution:
(i) Here a = 7, d = 3, n = 8
∵ a_n = a + (n – 1)d
∴ a₈ = 7 + (8 – 1)3
= 7 + 21 = 28.

(ii) Here a = – 18, n = 10, a_n = 0
∵ a_n = a + (n – 1)d
∴ a₁₀ = – 18 + (10 – 1)d
or 0 = – 18 + 9d .
or 9d = 18
d = \(\frac{18}{2}\) = 2.

(iii) Here d = – 3, n = 18, a_n = – 5
∵ a_n = a + (n – 1)d
∴ a₁₈ = a + (18 – 1)(-3)
or -5 = a – 51
or a = – 5 + 51 = 46.

(iv) Here a = – 18.9, d = 2.5 a_n = 3.6
∵ a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1) 2.5
or 3.6 + 18.9 = (n – 1) 2.5
or (n – 1) 2.5 = 22.5
or n – 1 = \(\frac{22.5}{2.5}\)
or n = 9 + 1 = 10.

(v) Here a = 3.5, d = 0, n = 105
∵ a_n = a + (n – 1) d
∴ a_n = 3.5 + (105 – 1) 0
a_n = 3.5 + 0 = 3.5.

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …………….. is
(A) 97 (B) 77 (C) – 77 (D) – 87

(ii) 11th term of the AP: – 3, – \(\frac{1}{2}\), 2, ………. is
(A) 28 (B) 22 (C) – 38 (D) – 48\(\frac{1}{2}\)

Solution:
(i) Given A.P. is 10, 7, 4 ……………
T₁ = 10, T₂ = 7, T₃ = 4
T₂ – T₁ = 7 – 10 = – 3
T₃ – T₂ = 4 – 7 = – 3
∵ T₂ – T₁ = T₃ – T₂ = – 3 = d(say)
∵ T_n = a + (n – 1) d
Now, T₃₀ = 10 + (30 – 1)(-3)
= 10 – 87 = – 77
∴ Correct choice is (C).

(ii) Given A.P. is – 3, –\(\frac{1}{2}\), 2, ……….
T₁ = – 3 T₂ = –\(\frac{1}{2}\), T₃ = 2, …………..
T₂ – T₁ = –\(\frac{1}{2}\) + 3 = \(\frac{-1+6}{2}=\frac{5}{2}\)
T₃ – T₂ = 2 + \(\frac{1}{2}\) = \(\frac{4+1}{2}=\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{5}{2}\) = d(say)
∵ T_n = a + (n – 1) d
Now, T₁₁ = -3 + (11 – 1) \(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\) = – 3 + 25 = 22
∴ Correct choice is (B).

Question 3.
In the following APs, find the missing terms in the boxes:
(i) 2, , 26
(ii), 13, , 3
(iii) 5,, , 9
(iv) – 4, , , , , 6
(v) , 38, , , , – 22
Solution:
Let a be the first term and ‘d’ be the common difference of given A.P.
(i) Here T₁ = a = 2
and T₃ = a + 2d = 26
or 2 + 2d = 26
or 2d = 26 – 2 = 24
or d = 12
∴ Missing term = T₂ = a + d = 2 + 12 = 14.

(ii) Here, T₂ = a + d = 13 ……………(1)
and T₄ = a + 3d = 3 …………….(3)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a – 5 = 13
a = 13 + 5 = 18.
∴ T₁ = a = 18
T₃ = a + 2d = 18 + 2(-5)
= 18 – 10 = 8.

(iii)Here T₁ = a = 5
and T₄ = a + 3d = 9
or a + 3d = \(\frac{19}{2}\)
or 5 + 3d = \(\frac{19}{2}\)
or 3d = \(\frac{19}{2}\) – 5
or 3d = \(\frac{19-10}{2}=\frac{9}{2}\)
or d = \(\frac{9}{2} \times \frac{1}{3}=\frac{3}{2}\)
T₂ = a + d = 5 + \(\frac{3}{2}\)
= \(\frac{10+3}{2}=\frac{13}{2}\)
T₃ = a + 2d = 5 + 2(\(\frac{3}{2}\)) = 5 + 3 = 8.

(iv) Here T₁ = a = —
T₆ = a + 5d = 6
or -4 + 5d = 6
or 5d = 6 + 4
or 5d = 10
or d = \(\frac{10}{2}\) = 2
Now, T₂ = a + d = -4 + 2 = -2
T₃ = a + 2d = – 4 + 2(2)
= – 4 + 4 = 0
T₄ = a + 3d = – 4 + 3(2)
= – 4 + 6 = 2
T₅ = a + 4d = – 4 + 4(2)
= – 4 + 8 = 4

(v) Here T₂ = a + d = 38 ………….(1)
and T₆ = a + 5d = -22 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + (-15) = 38
a = 38 + 15 = 53
∴ T₁ = a = 53
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23.
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = – 7.

Question 4
Which term of the A.P. 3, 8, 13, 18, …………… is 78?
Solution:
Given A.P. is 3, 8, 13, 18, ………….
T₁ = 3, T₂ = 8, T₃ = 13, T₄ = 18
T₂ – T₁ =8 – 3=5
T₃ – T₂= 13 – 8=5
T₂ – T₁ = T₃ – T,= 5 = d (say)
Using, T_n = a + (n – I) d
or 78 = 3 + (n – 1) 5
or 5(n – 1) = 78 – 3 = 75
or n – 1 = 15
or n = 15 + 1 = 16
Hence, 16th term of given AP. is 78.

Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19,…, 205
(ii) 18, 15\(\frac{1}{2}\), 13, ………….., – 47
Solution:
(i) Given A.P. is 7, 13, 19, …………..
T₁ = 7, T₂ = 13, T₃ = 19
T₂ – T₁ = 13 – 7 = 6
T₃ – T₂ = 19 – 13 = 6
T₂ – T₁ = T₃ – T₂ = 6 = d(say)
Using formula, T_n = a + (n – 1) d
205 = 7 + (n – 1) 6
or (n – 1) 6 = 205 – 7 = 198
or (n – 1) = \(\frac{198}{6}\)
or n – 1 = 33
n = 33 + 1 = 34
Hence, 34th term of an AP. is 205.

(ii) Given A P. is 18, 15\(\frac{1}{2}\), 13, …………..
T₁ = 18, T₂ = 15\(\frac{1}{2}\) = \(\frac{31}{2}\), T₃ = 13
T₂ – T₁ = \(\frac{31}{2}\) – 18 = \(\frac{31-36}{2}=-\frac{5}{2}\)
T₃ – T₂ = 13 – \(\frac{31}{2}\) = \(\frac{26-31}{2}=-\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{-5}{2}\) = d (say)
Using formula. T_n = a + (n – 1) d
– 47 = 18 + (n – 1) \(\frac{-5}{2}\)
or (n – 1) (\(\frac{-5}{2}\)) = – 47 – 18
or (n – 1) (\(\frac{-5}{2}\)) = – 65
or n – 1 = – 65 × – \(\frac{2}{5}\)
or n – 1 = 26
or n = 26 + 1 = 27
Hence, 27th term of an A.P. is – 47.

Question 6.
Is – 150 a term of 11, 8, 5, 2….? why?
Solution:
Given sequence is 11, 8, 5, 2, ………..
T₁ = 11, T₂ = 8, T₃ = 5, T₄ = 2
T₂ – T₁ = 8 – 11 = – 3
T₃ – T₂ = 5 – 8 = – 3
T₄ – T₃ = 2 – 5 = – 3
T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 3 = d (say).
Let – 150 be any term of given A.P.
then T_n = – 150
a+(n – 1)d = – 150
or 11 +(n – 1)(- 3) = – 150
or (n – 1)( – 3) = – 150 – 11 = – 161
or n – 1 = \(\frac{161}{3}\)
or n = \(\frac{161}{3}\) + 1 = \(\frac{161+3}{3}\)
n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\),
which is not a natural number.
Hence, – 150 cannot be a term of given A.P.

Question 7.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:
Let ‘a’ and 4d’ be the first term and common difference of given A.P.
Given that T₁₁ = 38
a +(11 – 1) d = 38
[∵ T_n = a + (n – 1) d]
a + 10 d = 38
and T₁₆ = 73
a + (16 – 1) d = 73
[∵ T_n = a + (n – 1) d]
a + 15 d = 73
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 10 (7) = 38
or a + 70 = 38
or a = 38 – 70 = – 32
Now, T₃₁ = a + (31 – 1) d
= – 32 + 30 (7) = – 32 + 210 = 178.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 291h term.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that, T₃ = 12
a + (3 – 1) d = 12
∵ T_n = a + (n – 1) d
or a + 2d = 12 ………………(1)
and Last term = T₅₀ = 106
a + (50 – 1) d = 106
∵ T_n = a + (n – 1) d
a + 49 d = 106 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(2) = 12
or a + 4 = 12
or a + 12 – 4 = 8
Now, T₂₉ = a + (29 – 1) d
= 8 + 28 (2) = 8 + 56 = 64.

Question 9.
If the 3rd and 9th tenus of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given AP.
Given that: T₃ = 4
a + (3 – 1) d = 4
∵ T_n = a + (n – 1) d
a + 2d = 4 …………..(1)
and T₉ = – 8
a + (9 – 1) d = 8
and T₉ = – 8
a + (9 – 1)d = 8
∵ T_n = a + (n – 1) d
or a + 8d = – 8
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(- 2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, T_n = 0 (Given)
a + (n – 1) d = 0
or 8 + (n – 1)(- 2)=0
or -2 (n – 1) = – 8
or n – 1 = 4
or n = 4 + 1 = 5
Hence, 5th term of an AP. is zero.

Question 10.
The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Now, T₁₇ = a + (17 – 1) d = a + 16 d
and T₁₀ = a + (10 – 1) d = a + 9 d
According to question
T₁₇ – T₁₀ = 7
(a + 16 d) – (a + 9 d) = 7
or a + 16 d – a – 9 d = 7
7 d = 7
or d = \(\frac{7}{7}\) = 1
Hence, common difference is 1.

Question 11.
Which term of the A.P. 3, 15, 27, 39, …………. will be 132 more than its 54th term?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given A.P. is 3, 15, 27, 39, …
T₁ = 3, T₂ = 15, T₃ = 27, T₄ = 39
T₂ – T₁ = 15 – 3 = 12
T₃ – T₂ = 27 – 15 = 12
:. d=T₂ – T₁ = T₃ – T₂ =12
Now, T₅₄ = a + (54 – 1) d
= 3 + 53 (12) = 3 + 636 = 639
According to question
T = T₅₄ + 132
a + (n – 1)d = 639 + 132
3 + (n – 1)(12) = 771
(n – 1) 12 = 771 – 3 = 768
or n – 1 = \(\frac{768}{12}\) = 64
or n = 64 + 1 = 65
Hence, 65th term of A.P. is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of first AP.
Also, ‘A’ and ‘d’ be the first term and common difference of second A.P.
According to question
[T₁₀₀ of second A.P.] – [T₁₀₀ of first A.P.] = 100
or[A +(100 – 1)d] – [a +(100 – 1)d] = 100
or A + 99d – a – 99d = 100
or A – a = 100
Now, [T₁₀₀₀ of second A.P.] – [T₁₀₀₀ of first A.P.]
= [A + (1000 – 1) d) – (a + (1000 – 1) d]
= A + 999 d – a – 999 d
= A – a = 100 [Using (I)].

Question 13.
How many three-digits numbers are divisible by 7?
Solution:
Three digits numbers which divisible by 7 are 105, 112, 119 , 994
Here a = T₁ = 105, T₂ = 112, T₃ = 119 and T_n = 994
T₂ – T₁ = 112 – 105=7
T₃ – T₂ = 119 – 112=7
∴ d = T₂ – T₁ = T₃ – T₂ = 7
Given that, T_n = 994
a + (n – 1) d = 994
or 105 + (n – 1) 7 = 994
or (n – 1) 7 = 994 – 105
or (n – 1) 7 = 889
or n – 1 = \(\frac{889}{7}\) = 127
or n = 127 + 1 = 128.
Hence, 128 terms of three digit number are divisible by 7.

Question 14.
How many multiples of 411e between 10 and 250?
Solution:
Multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, … 248
Here a = T₁ = 12, T₂ = 16, T₃ = 20 and T_n = 248
T₂ – T₁ = 16 – 12 = 4
T₃ – T₂ = 20 – 16 = 4
∴ d = T₂ – T₁ = T₃ – T₂ = 4
Given that, T_n = 248
a + (n – 1) d = 248
or 12 + (n – 1)4 = 248
or 4(n – 1) = 248 – 12 = 236
or n – 1 = \(\frac{236}{4}\) = 59
or n = 59 + 1 = 60
Hence, there are 60 terms which are multiples of 4 lies between 10 and 250.

Question 15.
For what value of n, are the n terms of two A.P.s 63, 65, 67, …………. and 3, 10, 17, …………….. equal?
Solution:
Given A.P. is 63, 65, 67, ……………..
Here a = T₁ = 63, T₂ = 65, T₃ = 67
T₂ – T1 = 65 – 63 = 2
T₃ – T₂ = 67 – 65 = 2
∴ d = T₂ – T₁ = T₃ – T₂ = 2
and second A.P. is 3, 10, 17, …
Here a = T₁ = 3, T₂ = 10, T₃ = 17
T₂ – T₁ = 10 – 3 = 7
T₃ – T₂ = 17 – 10 = 7
According to question.
[nth term of first A.P.] = [nth term of second A.P.]
63 + (n – 1)2 = 3 + (n – 1) 7
or 63 + 2n – 2 = 3 + 7n – 7
or 61 + 2n = 7n – 4
or 2n – 7n = – 4 – 61
– 5n = – 65
n = \(\frac{65}{5}\) = 13.

Question 16.
Determine the AP. whose third term is 16 and 7 term exceeds the by 12.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that T₃ = 16
a + (3 – 1) d = 16
a + 2d = 16
According to question
T₇ – T₅ = 12
[a + (7 – 1) d] – [a + (5 -1) d] = 12
a + 6 d – a – 44 = 12
2d = 12
d = \(\frac{12}{2}\) = 6
Substitute this value of d in (1), we get
a + 2(6) = 16
a = 16 – 12 = 4 .
Hence, given A.P. are 4, 10, 16, 22, 28, ………….

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ………., 253.
Solution:
Given A.P. is 3, 8, 13, …………., 253
Here, a = T₁ = 3, T₂ = 8, T₃ =13 and T_n = 253
T₂ – T₁ = 8 – 3 = 5
T₃ – T2 = 13 – 8 = 5
∴ d = T₂ – T₁ = T₃ – T₁ = 5
Now, T_n = 253
3 + (n – 1)5 = 253
∵ T_n = a + (n – 1) d
(n – 1) 5 = 250
n-1 = \(\frac{250}{5}\) = 50
n = 50 + 1 = 51
20th term from the end of AP = (Total number of terms) – 20 + 1
= 51 – 20 + 1 = 32nd term
∴ 20th term from the end of AP
= 32nd term from the starting
= 3 + (32 – 1)5
∵ Tn = a + (n – 1)d
= 3 + 31 × 5
= 3 + 155 = 158.

Question 18.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6’ and 10th terms is 44. FInd the first three terms of the A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
T₄ + T₈ = 24
a + (4 – 1) d + a + (8 – 1) d = 24
∵ T_n = a + (n – 1) d
or 2a + 3d + 7d = 24
2a + 10d = 24
a + 5d = 12 …………(1)
According to 2nd condition
T₆ + T₁₀ = 44
a + (6 – 1) d + a +(10 – 1) d = 44
∵ T_n = a + (n – 1) d
2a + 5d + 9d = 44
2a + 14d = 44
a + 7d = 22
Now (2) – (1) gives

Substitute this value of d in (I). we get
a + 5(5) = 12
a + 25 = 12
a = 12 – 25 = -13
T₁ = a = -13
T₂ = a + d = 13 + 5 = -8
T₂ = a + 2d = – 13 + 2(5) = – 13 + 10 = -3
Hence, given A.P. is -13, -8, -3, ……………

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Subba Rao’s starting salary = ₹ 5000
Annual increment = ₹ 200
Let ‘n’ denotes number of years.
∴ first term = a = ₹ 5000
Common diflerence = d = ₹ 200
and T_n = ₹ 7000
5000 + (n – 1) 200 = 7000
∵ T_n = a + (n – 1) d
(n – 1) 200 = 7000 – 5000
or (n – 1) 200 = 2000
or n – 1 = \(\frac{2000}{200}\) = 10
or n = 10 + 1 = 11
Now, in case of year the sequence are 1995. 1996, 1997, 1998, ……………
Here a = 1995, d = 1 and n = 11
Let T_n denotes the required year.
∴ T_n = 1995 + (11 – 1) 1
= 1995 + 10 = 2005
Hence, in 2005, Subba Rao’s salary becomes 7000.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving becomes ₹ 20.75, find n.
Solution:
Amount saved in first week = ₹ 5
Increment in saving every week = ₹ 1.75
It is clear that, it form an A.P. whose terms are
T₁ = 5, d = 1.75
∴ T₂ = 5 + 1.75 = 6.75
T₃ = 6.75 + 1.75 = 8.50
Also. T_n = 20. 75 (Given)
5 + (n – 1) 1.75 = 20.75
∵ T_n = a + (n – 1) d
or (n – 1) 1.75 = 20.75 – 5
or (n – 1) 1.75 = 15.75
or (n – 1) = \(\frac{1575}{100} \times \frac{100}{175}\)
or n – 1 = 9
or n = 9 + 1 = 10
Hence, in 10th week, Ramkali’s saving becomes ₹ 20.75.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of it.
Solution:
Radius of cone = Radius of hemisphere = 1 cm

R = 1 cm
Height of cone (H) = 1 cm
Volume of solid = volume of cone + volume of hemisphere
= \(\frac{1}{3}\) πR²H + \(\frac{2}{3}\) πR³
= \(\frac{1}{3}\) πR²2 [H + 2R]
= \(\frac{1}{3}\) π × 1 × 1 [1 + 2 × 1]
= \(\frac{1}{3}\) π × 3 = π cm³
Hence, Volume of solid = π cm³.

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:

Radius of cone = Radius of cylinder (R) = \(\frac{3}{2}\) cm
∴ R = 1.5 cm
Height of eah cone (h) = 2 cm
∴ Height of cylinder = 12 – 2 – 2 = 8 cm
Volume of air in cylinder = volume of cylinder + 2 (volume of cone)

Volume of air in cylinder = \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}\)
= 22 × 3 = 66 cm³

Hence, Volume of air in cylinder =66 cm³.

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:
Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm

Radius of cylinder = Radius of hemisphere (R)
= \(\frac{28.2}{2}\) = 1.4 cm
R = 1.4 cm
Height of cylindrical part = 5 – 1.4 – 1.4
= (5 – 2.8) cm = 2.2 cm.
Volume of one gulab Jamun = Volume of cylinder + 2 [Volume of hemisphere]
= πR²H + 2 \(\frac{2}{3}\) πR³

= πR² H + \(\frac{4}{3}\) R

= \(\frac{4}{4}\) × 1.4 × 1.4 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{14}{10}\) × \(\frac{14}{10}\) 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{196}{100}\) [2.2 + 1.86]

= \(\frac{22 \times 28}{100}\) [4.06]

Volume of one gulab Jamun = 25.05 cm³
Now volume of 45 gulab Jamuns = 45 × 25.05 cm³ = 1127.25 cm³
Volume of sugar syrup = 30% volume of 45 gulab Jamuns
= \(\frac{30 \times 1127.25}{100}\) = 338.175 cm3
Hence, Approximately sugar syrup = 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution. Length of cuboid (L) = 15 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Radius of conical cavity (r) = 0.5 cm
Height of conical cavity (h) = 1.4 cm
Volume of wood in Pen stand = volume of cuboid – 4 [volume of cone]
= LBH – 4 \(\frac{1}{3}\) πr²h

= 15 × 10 × 3.5 – \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4

= \(\frac{15 \times 10 \times 35}{10}-\frac{4}{3} \times \frac{22}{7} \times \frac{5}{10} \times \frac{5}{10} \times \frac{14}{10}\)

= \(15 \times 35-\frac{22}{3 \times 5}\)
= 525 – 1.466 = 523.534 cm³
Hence, Volume of wood in Pen stand = 523.53 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height ¡s 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = \(\frac{1}{4}\) Volume of water in cone

= 10 × 10 = 100
Hence, Number of lead shots = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cfi and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of Iron has approximately 8 g mass. (Use n = 3.14)
Solution:
Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of Lower cylinder + volume of upper cylinder
= πR²H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6

Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8 = 892262.4 gm
= \(\frac{892262.4}{1000}\) kg = 892.2624 kg
Hence, Mass of Pole = 892.2624 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of cone = Radius of hemisphere = Radius of cylinder

Height of cone (h) = 120 cm
Height of cylinder (H) = 180 cm
Volume of cylindrical vessel = πR²H
= \(\frac{22}{7}\) × 60 × 60 × 180 = 2036571.4 cm³
Volume of solid inserted in cylinder = Volume of hemisphere + Volume of cone
= \(\frac{2}{3}\) πR³ + \(\frac{1}{3}\) πR²h

= \(\frac{1}{3}\) πR² [2R + h]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 60 × 60 [2 × 60 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 [120 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 × 240 = 905142.86 cm³
Volume of water flows out = 90514186 cm³
∴ Volume of water left in cylinder = Volume of cylinder – Volume of solid inserted in th’e vessel
= (2036571.4 – 905142.86) cm³ = 1131428.5 cm³
= \(\frac{1131428.5}{100 \times 100 \times 100}\) m³ = 1.131 m³
Hence, Volume of water left in cylinder = 1.131 m³.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Diameter of neck (cylindrical Portion) = 2 cm

Radius of neck (r) = 1 cm
Height of cylindrical portion (H) = 8 cm
Diameter of spherical portion = 8.5 cm
Radius of spherical portion (R) = \(\frac{8.5}{2}\) cm = 4.25 cm
Volume of water in vessel = Volume of sphere + Volume of cylinder
= \(\frac{4}{3}\) πR³ + πR²h
= \(\frac{4}{3}\) × 3.14 × 4.25 × 4.25 × 4.25 × 3.14 × 1 × 1 × 8
= 321.39 + 25.12 = 346.51 cm³
Hence, Volume of water in vessel = 346.51 cm³ and She is wrong.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
Find the surface area of the solid.
Solution:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 294 + 38.5 = 332.5 cm²

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 2πRH + πRL
= πR [2H + L]
= \(\frac{22}{7}\) × 2[2(2.1) + 2.8]
= \(\frac{22}{7}\) × 2[4.2 + 2.8]
= \(\frac{22}{7}\) × 2 × 7
= 44 m²
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in Fig., If PQ = 24 cm, PR =7 cm and O is the centre of the circle.

Solution:
PQ = 24 cm
PR = 7 cm
RQ is diameter of circle
∠RPQ = 90° Angle in semi circle
In ∆PQR,
QR² = RP² + PQ²
QR = \(\sqrt{(7)^{2}+(24)^{2}}=\sqrt{49+576}\)
= \(\sqrt{625}\)
QR = 25 cm
∴ Diameter of circle (QR) = 25 cm
Radius of circle (R) = \(\frac{25}{2}\) cm
Area of shaded region = Area of the semicircle – Area of ∆RPQ
= \(\frac{1}{2} \pi \mathrm{R}^{2}-\frac{1}{2} \mathrm{RP} \times \mathrm{PQ}\)

= \(\left[\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 7 \times 24\right]\) cm²

= \(\left[\frac{6875}{28}-84\right]\)
= 245.53 – 84 = 161.53 cm²
∴ Area of shaded region = 161.53 cm².

Question 2.
Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Radius of smaller circle (r) = 7 cm
Radius of bigger circle (R) = 14 cm
Central angle ∠AOC (θ) = 40°
Area of shaded region = Area of bigger sector OAC – Area of smaller sector OBD
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{\pi r^{2} \theta}{360^{\circ}}\)

= \(\frac{\pi \theta}{360^{\circ}}\) [R² – r²]

= \(\frac{22}{7} \times \frac{40}{360}\) × [14² – 7²]

= \(\frac{22}{63}\) [196 – 49]

= \(\frac{22}{63}\) × 147 = 51.33 cm²
∴ Shaded Region = 51.33 cm².

Question 3.
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semi circles.

Solution:
Side of square = 14 cm
Diameter of semicircle (AB = BC) = 14 cm
Radius of semi circle (R) = 7 cm
Area of square = (Side)²
= 14 × 14 = 196cm²
Area of a semi circles = \(\frac{1}{2}\) πR²
= \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 77 cm²
Area of two semi circle = 2(77) = 154 cm²
Area of shaded region = Area of square ABCD – Area of two semi circles
= (196 – 154) = 42 cm²
Area of shaded region = 42 cm².

Question 4.
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

[Each angle of equilateral triangle is 60°]
Area of major sector of circle = Area of circle – Area of sector
= πR² – \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7}\) × 6 × 6 – \(\frac{22}{7}\) × 6 × 6 × \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 6 × 6 1 – \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 36 1 – \(\frac{1}{6}\)

= \(\frac{22}{7}\) × 36 × \(\frac{5}{6}\)
= 94.28 cm²
∴ Area of major sector of circle = 94.28 cm²
Area of equilateral triangle OAB = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{1.73}{4}\) × 12 × 12
= 1.73 × 36 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle
= 62.28 + 94.28 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle
= 62.28 + 94.28 = 156.56
Shaded Area = 156.56 cm².

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.
Solution:
Side of square = 4 cm
Radius of each semi circle cut out (r) = 1 cm

Diameter of circle (R) = 2 cm
.. Radius of circle (R) = 1 cm
Area of square = (Side)²
= (4)² = 16 cm²
Area of 4 quadrants = 4\(\left[\frac{\pi^{2} \theta}{360^{\circ}}\right]\)

= \(\frac{4 \times 90}{360} \times \frac{22}{7} 1 \times 1\)

= 1 × \(\frac{22}{7}\) × 1 × 1 = 3.14 cm²
Area of circle = πR²
= \(\frac{22}{7}\) × 1 × 1
Area of circle = 3.14 cm²
Required area = Area of square – Area of 4 quadrants – Area of circle
= (16 – 3.14 – 3.14) cm² = 9.72 cm²
Required Area = 9.72 cm².

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as
shown in fig. Find the area of the design (shaded region).

Solution:
Radius of table cover (R) = 32 cm
OA = OB = OC = 32 cm

∆ABC is equilateral triangle with AB = AC = BC
∠AOB = ∠BOC = ∠COA = 120°
Now, in ∆BOC,
From O draw, angle bisector of ∠BOC as well as perpendicular bisector 0M of BC.
∴ BM = MC = \(\frac{1}{2}\) BC
Also, OB = OC [radii of the circle]
∴ ∠B = ∠C
∴ ∠O + ∠B + ∠C = 180°
120° + 2∠B = 180°
∠B = 30°
and ∠B = ∠C = 30°
Also, ∠BOM = ∠COM = 60°
∆OMB ≅ ∆OMC [RHS Cong.]
∴ In ∠OMB,
∠OBM = 30° [∠O = 60° and ∠M = 90°]
∴ \(\frac{\mathrm{BM}}{\mathrm{OB}}\) = cos 30°

\(\frac{\mathrm{BM}}{32}=\frac{\sqrt{3}}{2}\)

BM = 16√3 cm.

∴ BC = 2 MB = 32√3 cm
Area of circle = πR² = \(\frac{22}{7}\) × (32)²
= \(\frac{22}{7}\) × 32 × 32 = 3218.28 cm²

Area of ∆ABC = 4 (side)²
= \(\frac{1.73}{4}\) × 32√3 × 32√3 = 1328.64 cm²

∴ Required Area = Area of circle – Area of ∆ABC
= 3218.28 – 1328.64 = 1889.64 cm²

Question 7.
In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Side of square ABCD = 14 cm
Radius of circle (R) = 7 cm
Sector angle (θ) = 90° [Each angle of square 90°]
Area of square = (side)²
= 14 × 14 = 196 cm²
Area of four quadrants = 4 \(\left[\frac{\pi R^{2} \theta}{360}\right]\)
= 4 × \(\frac{22}{7} \times \frac{7 \times 7 \times 90}{360}\)
= 22 × 7 = 154 cm²
∴ Required shaded area = Area of square – Area of 4 quadrants
= 196 – 154 = 42 cm².

 

Question 8.
Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:

(i) Here AB = DC = 106 m
AF = BE = CG = HD = 10m
Diameter of inner semicircle (APD and BRC) =60 m
∴ Radius of inner semicircle (APD (r) = 30 m
Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m
Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA
= 2 AB + 2 [circumference of semi circle BRC]
= 2 (106) + 2(\(\left(\frac{2 \pi r}{2}\right)\))
= 212 + 2πr
= 212 + 2 × \(\frac{22}{7}\) × 30
= 212 + \(\frac{60 \times 22}{7}\)
= 212 + 188.57 = 400.57 m.
∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.
= 2 Area of rectangle ABCD + 2 Area of region (II)
= 2 (AB × AF) + 2
[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]
= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]
= 2 × 1060 + \(\frac{2 \pi}{2}\) [R² – r²]
= 2120 + \(\frac{22}{7}\) (40² – 30²)
= 2120 + \(\frac{22}{7}\) [1600 – 900]
= 2120 + \(\frac{22}{7}\) [700]
= 2120 + 2200 = 4320 m²
Area of track = 4320 m²

Question 9.
In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Diameter of circle = 14 cm
Radius of circle = 7 cm
Diameter of smaller circle = 7 cm
∴ Radius of smaller circle = \(\frac{7}{2}\) cm
Since AB and CD are to perpendicular the diameters of a circle,
∴ AO ⊥ CD
Area of bigger circle = πR² × 7 × 7 = 154 cm²
Area of bigger semicircle = \(\frac{154}{2}\) = 77 cm²
Area of smaller circle = πr²
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= 38.50 cm²

Area of ∆ABC = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 14 × 7 = 49 cm²
∴ Shaded Area = Area of bigger semi circle + Area of smaller circle – Area of triangle
= (77 – 49 + 38.5) cm² = 66.5 cm²

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see Fig.). Find the area of the shaded region.
(Use n = 3.14 and ,√3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm²

\(\frac{\sqrt{3}}{4}\) (side)² = 17320.5

(side)² = \(\frac{17320.5 \times 4}{1.73205}\)

(side)² = \(\frac{173205}{10} \times \frac{100000 \times 4}{173205}\)

side = \(\sqrt{4 \times 100 \times 100}\)
side = 2 × 100 = 200 cm
AB = BC = AC
Radius of circle (R) = \(\frac{A B}{2}=\frac{200}{2}\) = 100 cm
Sector angle, θ = 60°
Area of sector APN = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{3.14 \times 100 \times 100 \times 60}{360}\)

= \(\frac{15700}{3}\)

Area of three sector = 3 × \(\frac{15700}{3}\) cm²
∴ Required shaded Area = Area of triangle – Area of three sectors
= 17320.5 – 15700 = 1620.5 cm²
∴ Hence, Required shaded Area = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of circle (R) = 7 cm
Diameter of circle = 2 × R = 2 × 7 = 14 cm
Since there are three circles along a side of square
∴ side of squrae = 3 [14] = 42 cm
Total area of handkerchief = Area of square = (side)²
= (42)² = 1764 cm².
Area of 9 circular designs = 9πR²
= 9 × \(\frac{22}{7}\) × (7)2
= 9 × \(\frac{22}{7}\) × 7 × 7
= 9 × 154 = 1386 cm²
∴ Required area of remaining portion = Area of square – Area of 9 circular designs
= 1764 – 1386 = 378 cm²
∴ Required area of remaining portion = 378 cm².

Question 12.
In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
(ii) shaded region.

Solution:

Radius of quadrant (R) = 3.5 cm
Angle of sector (θ) = 90°
OD = 2 cm.

(i) Area of quadrant OACB = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90}{360}\) = 9.625 cm².

(ii) Area of ODB = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 3.5 × 2 = 3.5 cm²

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB
= 9.625 – 3.5 = 6.125 cm²
∴ Hence, Shaded Area = 6.125 cm².

Question 13.
In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution:
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA

OB² = OA² + AB²

OB = \(\sqrt{(20)^{2}+(20)^{2}}\)

= \(\sqrt{400+400}\)

= \(\sqrt{800}=\sqrt{400 \times 2}\)
OB = 20√2 cm
Area of square OABC = (side)² = (20)²
∴ Area of square = 400 cm²
Radius of quadrant (R) = 20√2 cm
Sector angle (θ) = 90°
∴ Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 20 \sqrt{2} \times 20 \sqrt{2} \times 90}{360}\)
= 2 × 314 cm² = 628 cm²
Required shaded Area = Area of sector – Area of square
= (628 – 400) cm² = 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ZAOB = 30°, find the area of the shaded region.

Solution:
Radius of sector OBA (R) =21 cm
Radius of sector ODC (r) 7 cm
Sector angle (θ) = 30°
Area of bigger sector (OAB) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{21 \times 21 \times 30}{360}\) = 115.5 cm²

Area of smaller sector (ODC) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 30}{360}\) = 12.83 cm²

Area of smaller sector (ODC) = 12.83 cm²
Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD
= 115.5 – 12.83 = 102.66
Hence, Shaded Area = 102.66 cm².

Question 15.
In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of quadrant ACPB (r) = 14 cm
Sector angle (θ) = 90°
AB = AC = 14 cm

Area of triangle = \(\frac{1}{2}\) AB × AC
= \(\frac{1}{2}\) × 14 × 14
= 98 cm²

Area of sector ACPB = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{14 \times 14 \times 30}{360}\) = 154 cm²

∴ Area of BOCPB = Area of sector ABPC – Area of \ABC
= 154 cm² – 98 cm² = 56 cm²
In ∆BAC, AB² + AC² = BC²
(14)² + (14)² = BC²
BC = \(\sqrt{196+196}=\sqrt{2(196)}\) = 14√2

∴ Radius of semi circle BOCR = \(\frac{14 \sqrt{2}}{2}\) = 7√2

Area of semi circle = \(\frac{\pi \mathrm{R}^{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \sqrt{2} \times 7 \sqrt{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 2}{2}\)
= 154 cm²

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]
= 154 – [154 – 98]
= (154 – 56) cm² = 98 cm²
Hence, Shaded Area = 98 cm².

Question 16.
Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Solution:
Side of square = 8 cm
Area of square = (8)² = 64 cm²
Line BD divides square ABCD into the equal parts
Area of ∆ABD = ar of ∆BDC
Sector angle θ = 90°
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{8 \times 8 \times 90}{360}\)

Area of sector = 50.28 cm²
Area of ∆ABD = \(\frac{1}{2}\) × AB × AD
= \(\frac{1}{2}\) × 8 × 8
= 32 cm²

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD
= 50.28 – 32 = 18.28 cm²
Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm²

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:

Radius of first circle (r₁) = 19 cm
Radius of second circle (r₂) = 9 cm
Let radius of third circle be R cm
According to condition
circumference of first circle + circumference of second circle = circumference of third circle
2πr₁ + 2πr₂ = 2πR
2π (r₁ + r₂] = 2πR
19 + 9 = R
∴ R = 28
∴ Radius of third circle (R) = 28 cm.

Question 2.
The Radii of two circles are 8 cm and 6 cm respectively. Find radius of circle which is having area equal to sum of the area of two circles.
Solution:
Radius of first circle (r₁) = 8 cm
Radius of second circle (r₂) = 6 cm
Let radius of third circle be R cm
According to question
Area of third circle = Area of first circle + Area of second circle
πR² = πr₁² + πr₂²
πR² = π[r₁² + r₂²]
R² = (8)² + (6)²
R = \(\sqrt{64+36}=\sqrt{100}\)
R = 10 cm
∴ Radius of required circle (R) = 10 cm.

Question 3.
Fig. depicts an archery target marLed with its five scoring areas from the centre ‘utwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score ¡s 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:
Diameter of Gold region = 21 cm
Radius of Gold region (R₁) = 10.5 cm
∴ Area of gold region = πR₁²
= \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}=\frac{690}{2}\) cm²
= 346.5 cm²
Width of each band = 10.5 cm
∴ Radius of Red and Gold region (R₂) = (10.5 + 10.5) = 21 cm
Combined radius of Blue, Red and Gold region (R₃) = R₂ + 10.5 cm
= 21 cm + 10.5 cm = 31.5 cm
Combined radius of Black, Blue, Red and Gold (R₄) = R₃ + 10.5
= 31.5 + 10.5 = 42cm
Area of circle having black radius = (Combined area of Gold, Red, Blue and Black radius) – (Combined Area of Gold, Red and Blue radius)
= πr₄² – πr₃²
= π [(42)² – (31.5)²]
= \(\frac{22}{7}\) [1764 – 992.25]
= \(\frac{22}{7}\) [771.75] = 2425.5 cm²

Combined Radius of white, black, blue, red, gold region (R₅) = R₄ + 10.5
R₅ = 42 + 10.5 = 52.5 cm
Combined radius of black, blue, red and gold = (R₄) = 42 cm.
Area of circle white scoring region = (Combined area of white, bLack, red, blue, gold region) – (Combined Area of Black, blue and gold region)
= πR₅² – πR₄²
= π[R₅² – πR₄²]
= \(\frac{22}{7}\) × [(52.5)² – (42)²]
= \(\frac{22}{7}\) [2756.25 – 1764]
= \(\frac{22 \times 992.25}{7}=\frac{21829.5}{7}\)
= 3118.5 cm²
∴ Area of white scoring region = 3118.5 cm²

∴ Area of red region = Area of red and gold region – Area of gold region
= πR₂² – πR₁²
= π [(21)² – (\(\frac{21}{2}\))²]
= \(\frac{22}{7}\) × 441 [1 – \(\frac{1}{4}\)]
= 22 × 63 \(\frac{3}{4}\)
= \(\frac{11 \times 189}{4}=\frac{2079}{4} \mathrm{~cm}^{2}\)
= 1039.5 cm²

∴ Area of Red region = 1039.5 cm²
Combined Radius of Gold, Red and Blue region R₃ (10.5 + 10.5 + 10.5) = 31.5 cm

Area of blue scoring region = (Combined area of red, blue and gold region) – (Combined area of Gold and red region)
= πR₃² – πR₂²
= π[R₃² – πR₂²]
= \(\frac{22}{7}\) × [(31.5)² – (21)²]
= \(\frac{22}{7}\) [992.25 – 441]
= \(\frac{22}{7}\) × 551.25 = \(\frac{121275}{7}\)
= 1732.5 cm²

Hence, area of gold ring; red ring; blue ring : black ring; white ring are 3465 cm² ; 1039.5 cm²; 1732.5 cm²; 24255 cm² ; 3118.5 cm² respectively.

Question 4.
The wheels of a ca are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a spel of 66 km per hour?
Solution:
Diameter of wheel = 80 cm
Radius of wheel (R) = 40 cm
Circumference of whed = 2πr
= 2 × \(\frac{22}{7}\) × 0.04
= \(\frac{22}{7}\) × 0.08 m
Let us suppose wheel of cr complete n revolutions of the wheel in 10 minutes = n[0.08 × \(\frac{22}{7}\)]
Speed of car = 66 km/hr. = 66 × 1000 m
Distance covered in 60 minutes = \(\frac{66 \times 1000}{60} \times 10\) = 11000 m
According to question.
∴ n[\(\frac{22}{7}\) × 0.08] = 11000
n = \(\frac{11000}{0.08} \times \frac{7}{22}\)
n = 4375
Hence, number of complete revolutions made by wheel in 10 minutes = 4375.

Question 5.
Tick the correct answer in the following and justify your choice : If the perimeter and area of a clrde are numerically equal, then the radius of the circle Is
(A) 2 units
(B) π units
(C) 4 units
(D) n units
Solution:
Perimeter of circle = Area of circle
2πR = πR²
2R = R²
⇒ R = 2
∴ Correct option A is (R) = 2 unit.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of the construction.

Question 1.
Draw a circle of radius 6 cm from a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:

Steps of construction:
1. Draw a circle (1) of radius 6 cm.
2. Take a point ‘P’ at a distance of 10 m. from the centre of the circle. Join OP.
3. Draw perpendicular bisector of OP. Let ‘M’ be the mid point OP.
4. With ‘M’ as centre and radius MO, draw a circle (II) which intersects the circle (I) at T and T’.
5. Then FT and PT’ are two required tangents.

Justification of construction:
We know that tangent at a point is always perpendicular to the radius at the point. Now
we have to prove that ∠PTO = ∠PT’O = 90°.
OT is joined.
Now, PMO is the diameter of circle (II) and ∠PTO is in the semicircle.
∴ ∠PTO = 90° [Angle in semicircle is a right angle].
Similarly, ∠PT’O = 90°
∴ PT and PT’ has to be the tangents to the circle at T and T’.
(On measuring, the lengths of tangents
i.e., PT = 8.1 cm
PT’ = 8.1 cm.
Co-centric circles. Two or more circles having same centre but different radii are called CO-CENTRIC circles.

Question 2.
Construct a tangent to a circle of radivs 4 cm from a point on the co-centric circle of radIus 6 cm and measure its length.
Also, erify the measurement by actual calculation.
Solution: Steps of construction:
STEPS OF CONSTRUCTION:
1. Draw a circle with cente O’ and radius 4 cm. Mark it as 1
2. Draw another circle with same centre ‘O’ and radius 6 cm and mark it as II.
3. Take any point ‘P’ on circle II. Join OP.
4. Draw pependicu1ar bisector of OP. Let it intersects ‘OP’ at M.
5. With M is centre and radius MO’ or ‘MP’, draw a circie III which intersects the circle ‘1’ at T and T’.
6. Join PT.
PT is the required tangent.

Justification of the construction :
Join OT.
Now OP is the diameter of the circle III.
∠OTP is in the semicircle.
∴ ∠OTP = 90° …………….(1)
[∴ Angle in a semicircle isa right angle]
Now OT ⊥ PT [using (I)]
∵ A line which makes an angle of 900 with radius at any point on the circle, the line is tangent to the circle.
∴ PT is tangent to the circle ‘I’
i.e. PT is tangent to the circle of radius 4.5 cm.

To calculate the length of tangent:
Consider ∆OTP,
∠OTP = 90° [using (i)]
∴ ∆OTP is a right angled triangle.
OT = 4 cm [Radius of I circle (given)]
OP = 6 cm [Radius of the II circle (given)]
PT = ? [to be calculated]
In rt. triangle ∆OTP,
By Pythagoras theorem
OP² = OT² + PT²
[(Hyp)² = (Base)² + (Perp.)²]
or PT² = OP² – OT²
= 6² – 4²
= 36 – 16 = 20
PT = \(\sqrt{20}\) cm
= 2√5 = 2 × 2.24 = 4.48 cm.
So, length of tangent by actual calculation = 4.48 cm = 4.5 cm.
Length of tangent by measurement = 4.5 cm
Hence, the length of tangent ‘PT’ is verified.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points ‘P’ and ‘Q’.
Solution:
Steps of construction:
1. Draw a circle of radius 3 cm and centre ‘O’.

2. Draw its diameter ‘AB’ and extend it in both directions as OX and OX’.
3. Take a point P’ on OX” direction and ‘Q’ on OX’ direction such that OP = OQ = 7 cm.
4. Draw perpendicular bisectors of OP and OQ which intersects OP and OQ at ‘M’ and ‘M” respectively.
5. With ‘M’ as centre and radius = ‘MO’ or MP, draw a circle ‘II’ which intersects the circle ‘I’ at T and T’.
6. Similarly with ‘M’’ as centre and radius = M’O or MQ, draw a circle (III) which intersects the circle ‘I’ at S’ and ‘S’’.
7. Join PT, PT’ and QS and QS’.

Justification of construction :
Join OT’ and ‘OT” and ‘OS’ and OS’.
To prove ‘PT & PT’ tangents to the circle
we will prove that ∠PTO = ∠PT’O = 90°.
Now ‘OP’ acts as the diameter of circle ‘II’ and ∠OTP is in the semicircle.
∴ ∠OTP = 90° …………….(1) [∵ Angle in semicircle is 90°]
But ‘OT’ is the radius of circle ‘I’ and line ‘PT’ touches the circle at T’.
∵ The line which touches the circle at a point and makes an angle of 90° with radius at that point, is tangent to the circle.
∴ PT is tangent to the circle I at point T through point ‘P.
Similarly PT’, QS and QS’ are tangents to the circle I.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°
Solution:
Steps of construction:
1. Draw the rough sketch of required figure.

∵ the tangents make an angle of 60° with each other.
∠OTP = ∠OQT = 90°
[Tangent is perpendicular to the radius of circle]
1. To find inclination of radii with each other
∠TOQ + ∠OTP + ∠OQT + ∠TPQ = 360° [Angle sum property of quad.]
or ∠TOQ + 90° + 90° + 60 = 360°
or ∠TOQ = 360 – 90° – 90° – 60° = 120°
2. Draw a circle of radius 5 cm.
3. Draw two radii of circle which make an angle of 120° with each other.
4. The radii intersect the circle at ‘A’ and
5. Make an angle of 90° at each point A and B, which intersect each other at ‘P’.

6. PA and PB are the required tangents.

Question 5.
Draw a line segment AB of length 8 cm. Taking ‘A’ as centre, draw a circle of radius 4 cm and taking ‘B’ as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:

Steps of construction :
1. Draw a line segment AB = 8 cm.
2. With ‘A’ as centre and radius 4 cm, draw a circle (I)
3. With ‘B’ as centre and radius 3 cm, draw a circle ‘I’.
4. Draw the perpendicular bisector of line segment AB which inersects ‘AB’ at ‘M’.
5. With ‘M as centre and radius MA or MB. draw a circle (III) which intersects the circle (I) at ‘S’ and ‘T’ and circle (II) at ‘P’ and ‘Q’.
6. Join ‘AP’ and AQ’. These are required tangents to the circle with radius 3 cm. from point ‘A’.
7. Join ‘BS’ and ‘BT’. These are required tangents to the circle with radius 4 cm from point ‘B’.

Justification of Construction:
In circle (III), AB acts as diameter then ∠ASB and ∠BPA are in semicircle.
∴ ∠ASB = 90° ………………(1) [Angle in semicircle]
and ∠BPA = 90° .
But ∠ASB is angle between radius of circle (I) and line segment BS’ and ∠BPA is angle between radius of circle (II) and line segment ‘AP’.
∵ Line segment which is perpendicular to the radius of circle, is tangent to the circle through that point.
∴ BS is tangent to circle (I) at point ‘S’ and AP is tangent to circle (II) at point ‘P’.
Similarly AQ and BT are tangents to the circle (II) and (I) respectively.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 5 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from ‘A’ to this circle.
Solution:

Steps of construction:
1. Construct rt. angled triangle. ABC according to given conditions and measurements.
2. Draw BD ⊥ AC.
3. Take mid point of side BC take it as
4. Take ‘M’ as centre and BC as diameter,
draw a circle through B. C, D using property, angle in semicircle is 90° (∠BDC 90°). Take this circle as I.
5. Now join ‘A’ and ‘M.
6. Draw perpendicular bisector of AM intersecting AM in point N. Now with ‘N’ as centre and ‘NA or ‘NM’ as radius, draw a circle (II) which intersects the circle (I) at ‘B’ and ‘P’.
7. Join AP.
8 AP and AB are the required tangents.

Justification of construction:
Line segment AM’ is diameter of circle (II)
∠APM is in semicircle
∴ ∠APM = 90° [Angle in semicircle]
i.e., MP ⊥ AP
But ‘MP’ is the radius of circle (I)
∴ AP is tangent to the circle (II)
[∵ Any line ⊥ to radius of circle at any point on the circle is tangent to the circle.]
Similarly AB is tangent to circle (I).

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this poiñt
to the circle.
Solution:
To draw circle with bangle means the centre of circle is unknown. First find the centre of circle.

Steps of construction:
1. Draw a circle. using a bangle (I).
2. Take any two chords AB and CD (non parallel) on circle.
3. Draw the perpendicular bisectors of chords AB and CD. The perpendicular bisectors intersect each other
[∵ any point lying on perpendicular bisector of line segment is equidistant from its end points
[∵ ‘O’ lies on ⊥ bisector of AH and CD]
∴ OA = OB and OC = OD
∴ OA = OB = OC = OD (Radii of circle)
∴ ‘O’ is the centre of circle.
4. Take any point ‘P’ out side the circle.
5. Join OP.
6. Draw the perpcndicular bisector of OP let ‘M’ the mid point of OP.
7. With ‘M’ as centre and radius ‘MP’ or ‘MO’, draw a circle II which intersects the circle (I) at T and T’.
8. Join PT and PT’, which is required pair of tangents.

Justification of construction:
Tangent at a point is always perpendicular to the radius at the point. Now, we have to prove
that ∠PTO = PT’O = 90°
Join OT.
Now ∠PTO is in the semicircle I.
∵ ∠PTO = 90° [Angle in semicircle is a right angle]
Similarly ∠PT’O = 90°
∴ PT and PT’ has to be the tangents to the circle at T and T’.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

In each of the questions, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide ¡tin the ratio 5 : 8. Measure the two parts.
Solution:
Given: A line segment of length of 7.6 cm.
Steps of construction:
1. Take a line segment AB = 7.6 cm.
2. Draw any ray AX, making an acute angle ∠BAX.
3. Locate 5 + 8 = 13 (given ratio 5: 8) points A₁, A₂, A₃, A₄, A₅, ………….., A₁₀, A₁₁, A₁₂, A₁₃ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = …………. = A₁₁A₁₂ = A₁₂ A₁₃.
4. Join BA₁₃.
5. Through point A5, draw a line A₅C || A₁₃B (by making an angle equal to ∠A₁₃B) at A₅ intersecting AB at ‘C’. Then AC : CB = 5 : 8;

Justification:
Let us see how this method gives us the required division.
In ∆AA₁₃B,
Since A₅C || A₁₃B
∴ By Basic Proportionality Theorem
\(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{\mathrm{AC}}{\mathrm{CB}}\)

By construction, \(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}=\frac{5}{8}\)

∴ \(\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{5}{8}\)
This shows that ‘C’ divides AB in the ratio 5 : 8.
On measuring the two parts, AC = 2.9 cm and CB = 4.7 cm.

Alternative Method:
Steps of construction:
1. Take a line segment AB = 7.6 cm
2. Draw any acute angle ∠BAX
3. Draw angle ∠ABY such that ∠ABY = ∠BAX.
4. Locate the points A₁, A₂, A₃, A₄, A₅ on ray AX such that A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
5. Locate the points B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈ on ray BY such that B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈
6. Join A₅B₈ let it intersects AB at point Then AC : CB = 5 : 8.

justification:
In ∆ACA₅ and ∆BCB₈,
∠ACA₅ = ∠BCB₈ [vertically opp. ∠s]
∠BAA₅ = ∠ABB₈ [construction]
∴ AACA₅ ~ ABCB₈ [AA-similarity cond.]
∴ Their corresponding sides must be in the same ratio. ,
\(\frac{A C}{B C}=\frac{C A_{5}}{C B_{8}}=\frac{A_{5} A}{B_{8} B}\)
(I)(II) (III)
From I and III, \(\frac{A C}{B C}=\frac{A_{5} A}{B_{8} B}\)

But, \(\frac{A_{5} A}{B_{8} B}=\frac{5}{8}\) [construction]

\(\frac{A C}{C B}=\frac{5}{8}\).

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle.
Solution:
Steps of construction:
1. Construct a triangle ABC with given measurements. AB = 5 cm, AC = 4 cm and BC = 6 cm.
2. Make any acute angle ∠CBX below the side BC.

3. Locate three points (greater of 2 and 3 in \(\frac{2}{3}\))B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C.
5. Through B₂ (smaller of 2 and 3 in \(\frac{2}{3}\) draw a line parallel to B₃C, which intersect BC in C’.
6. Through C’, draw a line parallel to CA meeting BA is A’.
Thus ∆A’BC’ is the required triangle whose sides are of corresponding sides of ∆ABC.

Justification of construction :
First we will show that first triangle and constructed triangle are similar.
i.e. ∆A’BC’ ~ ∆ABC.
Consider ∆A’BC’ and ∆ABC.
∠B = ∠B [Common]
∠A’C’B= ∠ACB [By construction]
∆A’C’B ~ ∆ACB [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) …………….(1)
Now, consider ∆B₂BC’ and ∆B₃BC,
∠B = ∠B [common]
∠B₂C’B = ∠B₂CB [construction]
∴ ∆B₂BC’ ~ ∆B₃BC [AA -similarity]
∴ Their corresponding sides must be in the same ratio.

\(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{2}}{\mathrm{CB}_{3}}\)

I II III

Taking (I) and (II).
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{\mathrm{B}_{2} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{2}{3}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{2}{3}\) ……………(2)

From (1) & (2),
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{2}{3}\)

⇒ A’B = \(\frac{2}{3}\) AB and BC’ = \(\frac{2}{3}\) BC; C’A’ = \(\frac{2}{3}\) CA.
Hence, the construction is Justified.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
Solution:

Steps of construction :
1. Construct a triangle ABC in which AB = 7 cm, BC 6 cm and AC =5 cm.
2. Make any acute angle ∠BAX below the base AB.
3. Locate seven points A₁, A₂, A₃, A₄, A₅, A₆, A₇ on the ray AX such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
4. Join BA₅.
5. Through A₇, draw a line parallel A₅B. Let it meets AB at B’ on being produced such that AB’= \(\frac{7}{5}\) AB.
6. Through B’, draw a line parallel to BC which meets AC at C’ on being produced.
∆AB’C’ is the required triangle.

Justification of the construction.
In ∆ABC and ∆AB’C’,
∠A = ∠A [common]
∠ABC = ∠AB’C’ [corresponding ∠s]
∴ ∠ABC – ∠AB’C’ [AA-similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}\) ……………..(1)

Again, in ∆AA₅B and AA₇B’
∠A = ∠A [common]
∠AA5B = ∠AA7 B’ [corresponding ∠s]
∴ ∆AA5B ~ ∆AA7B’ [AA – similarity]
∴ Their corresponding sides must be in the same ratio.
\(\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}=\frac{\mathrm{A}_{5} \mathrm{~B}}{\mathrm{~A}_{7} \mathrm{~B}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}\) [construction]

But, \(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{5}{7}\) …………….(2)

From (1) and (2),

\(\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{CA}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{5}{7}\)

or \(\frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{C^{\prime} A}{C A}=\frac{7}{5}\)

⇒ AB’ = \(\frac{7}{5}\) AB; B’C’ = \(\frac{7}{5}\) BC and C’A’ = \(\frac{7}{5}\) CA

Hence, the sides of ∆AB’C’ are \(\frac{4}{4}\) of ∆ABC.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 12 times
the corresponding sides of the isosceles triangle.
Solution:
Given: Base of isosceles triangle is 8 cm and Altitude = 4 cm
To construct: A triangle whose sides are times the sides of isosceles triangle.
Steps of construction:
1. Take base AB = 8 cm.
2. Draw perpendicular bisector of AB. Let it intersect AB at ‘M’.
3. With M as centre and radius = 4 cm, draw an arc which intersects the perpendicular bisector at ‘C’
4. Join CA and CB.
5. ∆ABC is an isosceles with CA = CB.
6. Make any acute angle ∠BAX below the side BC.
7. Locate three (greater of ‘2’ & ‘3’ in 1\(\frac{1}{2}\) or \(\frac{3}{2}\))
A₁, A₂, A₃ on ‘AX’ such that A A₁ = A₁ A₂ = A₂ A₃.

8. Join A₂ (2nd point smaller of ‘2 and ‘3’ in ) and B.
9. Through A₃, draw a line parallel to A₂B meet AB is B’ cm being produced.
10. Through B’, draw a line parallel to BC which meets AC in C’ on being produced. ∆AB’C’ is the required triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of ∆ABC.

Justification of construction :
First we will prove ∆AB’C’ are ∆ABC and similar.
Consider ∆ AB’C’ and ∆ ABC
∠A = ∠A [Common]
∠AB’C’ = ∠ABC [By construction]
∠AB’C’ ~ ∠ABC [By AA – similarityj
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}\) ……………(1)

Now consider ∆ A₃AB’ and ∆ A,AB
∠A = ∠A [common]
∠B’A₃A = ∠BA₂A [By construction]
∴∆ A₃A B’ – ∆A₂AB [AA – similarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{A_{3} A}{A_{2} A}=\frac{A B^{\prime}}{A B}=\frac{B^{\prime} A_{3}}{B A_{2}}\)
I II III
Taking (I) & (II),
\(\frac{A B^{\prime}}{A B}=\frac{A_{3} A}{A_{2} A}\)

But, \(\frac{A_{3} A}{A_{2} A}=\frac{3}{2}\) [construction]
⇒ \(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{3}{2}\) ……………..(2)
From (1) & (2)m
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}}{\mathrm{CA}}=\frac{3}{2}\left(1 \frac{1}{2}\right)\)

⇒ AB’ = 1\(\frac{1}{2}\) (AB); B’C’ = 1\(\frac{1}{2}\) BC and C’A’ = 1\(\frac{1}{2}\) (CA)
Hence, given result is justified.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB =5 cm and ¿ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
1. Take a line segment BC = 6 cm
2. Construct an angle of measure 60° at point B. i.e., ∠CBX = 60°.
3. With B as centre and radius 5 cm draw an arc intersecting BX at ‘A’

4. Join A and C.
5. At B, make any acute angle ∠CBY below the side BC.
6. Locate four points (greater of 3 and 4 in \(\frac{3}{4}\)) B₁, B₂, B₃, B₄ on BY such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄. .
7. Join B₄ and C.
8. Draw a line through B₃ (smaller of 3 and 4 in ) parallel to B₄C making corresponding angles. Let the line through B₃ intersects BC in C’.
9. Through C’, draw a line parallel to CA which intersects BA at A’.
The ∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) of sides of ∆ABC.

Justification of the construction:
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [commoni
∠A’C’B = ∠ACB [corresponding ∠s]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio.

∴ \(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Now consider ∆B₃B C’ and ∆B₄BC.
∠B = ∠B [common]
∠ C’ B₃B = ∠CB₄B [corresponding ∠s]
∆B₃BC’ ~ ∆B₄BC [AA – similarity con.]
Their corresponding sides must be in the same ratio.
\(\frac{B_{3} B}{B_{4} B}=\frac{B C^{\prime}}{B C}=\frac{C^{\prime} B_{3}}{C B_{4}}\)
(I) (II) (III)

From (I) and (II),
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}\)

But, = \(\frac{\mathrm{B}_{3} \mathrm{~B}}{\mathrm{~B}_{4} \mathrm{~B}}=\frac{3}{4}\) [construction]

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}\) ………..(3)

From (1) and (3)
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{3}{4}\)

and C’A’= \(\frac{3}{4}\) CA.
∆A’BC’ is the required triangle whose sides are \(\frac{3}{4}\) sides of ∆ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°. ∠A = 105°. Then construct a triangle whose sides are j- times
the corresponding sides of ∆ABC.
Solution:
Steps of construction:
1. Construct the triangle ABC with the given measurements.
BC = 7 cm; ∠B = 45, ∠A = 105°
By angle sum property of triangle
∠A + ∠B + ∠C= 180°
105° + 45° + ∠C = 180°
∠C = 180 – 150° = 30°
2. Make any acute angle ∠CBX at point B, below the sides BC.

3. Locate four points (greater of 3 and 4 in \(\frac{4}{3}\)) B₁, B₂, B₃, B₄ on ‘BX’ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
4. Join B₃C (smaller of 3 and 4 in \(\frac{4}{3}\)).
5. Through B₄, draw a line parallel to B₃C meeting BC in C’ on being produced.
6. Through C’, draw another line parallel to CA meeting BA in A’ on being produced.
7. ∆A’BC’ is the required triangle whose sides are times the triangle ABC.

Justification of construction:
Consider the ∆ A’BC’ and ∆ ABC,
∠B = ∠B [common]
∠A’C’B = ∠ACB [construction]
∴ ∆A’BC’ ~ ∆ABC [AA – similarity]
∴ Their corresponding sides must be in the same ratio

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ………….(1)

Again, consider ∆B₄B C’ and ∆B₃BC,
∠B = ∠B [common]
∠C’B4B = ∠CB3B [By consiruction]
∴ BB C’ AB3BC [AA-si niilarity]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{B}_{4} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}_{4}}{\mathrm{CB}_{3}}\)
I II III

Taking I and II members.

\(\frac{B C^{\prime}}{B C}=\frac{B_{4} B}{B_{3} B}\)

But, \(\frac{B_{4} B}{B_{3} B}=\frac{4}{3}\) (construction)

or \(\frac{B C^{\prime}}{B C}=\frac{4}{3}\) ………….(2)

From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{4}{3}\)

⇒ A’B = \(\frac{4}{3}\) AB; BC’ = \(\frac{4}{3}\) BC and C’A’ = \(\frac{4}{3}\) CA
Hence the construction is justified.

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of construction:
1. Draw a right triangle using given conditions. Consider the triangle as ABC in which BC = 4 cm; AB = 3 cm and
∠B = 90°.
2. Make any acute angle ∠CBX below the line BC.
3. Locate five points (greater of 5 and 3 in \(\frac{5}{3}\)) B₁, B₂, B₃, B₄. B₅ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
4. Join B₃ (smaller of ‘5’ and ‘3’ in \(\frac{5}{3}\)) and ‘C’.

5. Through B₅. draw a line parallel to BC meeting BC is C’ on being produced.
6. Again draw a line through C’ parallel to CA meeting BA in A’ on being produced.
∆A’BC’ is the required triangle whose sides are \(\frac{5}{3}\) times the sides of ∆ABC.

Justification of construction :
Consider ∆A’BC’ and ∆ABC
∠B = ∠B [common]
∠A’C’B = ∠ACB [By construction]
∴ ∆A’BC’ ~ ∆ABC [AA-similarity condition]
∴ Their corresponding sides must be in the same ratio
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}\) ……………..(1)

Again, in ∆B₅C’B and ∆XB₃CB,
∠B = ∠B [common]
∠C’B₅B = ∠CB₃B [By construction]
∴ ∆B₅C’B ~ ∆B₃CB [AA-similarityj
∴ Their corresponding sidcs must be in the same ratio.

\(\frac{\mathrm{B}_{5} \mathrm{C}^{\prime}}{\mathrm{B}_{3} \mathrm{C}}=\frac{\mathrm{C}^{\prime} \mathrm{B}}{\mathrm{CB}}=\frac{\mathrm{BB}_{5}}{\mathrm{BB}_{3}}\)

I II III

Taking II and III members.
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{B}_{5} \mathrm{~B}}{\mathrm{~B}_{3} \mathrm{~B}}\)

But, \(\frac{B_{5} B}{B_{3} B}=\frac{5}{3}\) [construction]

\(\frac{B C^{\prime}}{B C}=\frac{5}{3}\) ……………(2)
From (1) and (2),

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{C}^{\prime} \mathrm{A}^{\prime}}{\mathrm{CA}}=\frac{5}{3}\)

⇒ A’B = \(\frac{5}{3}\) AB; BC’ = \(\frac{5}{3}\) BC and C’A’ = \(\frac{5}{3}\) CA
Hence the construction is justified.