Bhagya

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 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.1

1. Write the fraction representing the shaded portion:

Solution:

2. Colour the part according to the given fraction:

Solution:

3. Write the fraction for each of the following:

Question (i)
(i) Three-Fourth
(ii) Seven-Tenth
(iii) A Quarter
(iv) Five-Eighth
(v) Three-Twelvth.
Solution:
(i) \(\frac {3}{4}\)
(ii) \(\frac {7}{10}\)
(iii) \(\frac {1}{4}\)
(iv) \(\frac {5}{8}\)
(v) \(\frac {3}{12}\)

4. Write the fraction for the followings:

Question (i)
numerator = 5
denominator = 9
Answer:
\(\frac {5}{9}\)

Question (ii)
numerator = 2
denominator = 11
Answer:
\(\frac {2}{11}\)

Question (iii)
numerator = 6
denominator = 7
Answer:
\(\frac {6}{7}\)

5. Write the numerator and the denominator for the followings:

Question (i)
\(\frac {2}{3}\)
Solution:
Given fraction is \(\frac {2}{3}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 2
and Denominator = 3

Question (ii)
\(\frac {1}{4}\)
Solution:
Given fraction is \(\frac {1}{4}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 1
and Denominator = 4

Question (iii)
\(\frac {5}{11}\)
Solution:
Given fraction is \(\frac {5}{11}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 5
and Denominator = 11

Question (iv)
\(\frac {9}{13}\)
Solution:
Given fraction is \(\frac {9}{13}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 9
and Denominator = 13

Question (iv)
\(\frac {17}{16}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 17 and
Denominator = 16

6. Express:

Question (i)
1 day as a fraction of 1 week.
Solution:
We know 1 week = 7 days
∴ Required fraction= \(\frac {1}{7}\)

Question (ii)
40 seconds as a fraction of 1 minute.
Solution:
We know 1 minute = 60 seconds
∴ Required fraction = \(\frac {40}{60}\) or \(\frac {2}{3}\)
(Dividing both terms by 20)

Question (iii)
15 hours as fraction of 1 day.
Solution:
We know 1 day = 24 hours
∴ Required fraction = \(\frac {15}{24}\) or \(\frac {5}{8}\)
(Dividing both terms by 3)

Question (iv)
2 months as a fraction of 1 year.
Solution:
We know 1 year = 12 months
∴ Required fraction = \(\frac {2}{12}\) or \(\frac {1}{6}\)
(Dividing both terms by 2)

Question (v)
45 cm as a fraction of 1 metre.
Solution:
We know 1 metre = 100 cm
∴ Required fraction = \(\frac {45}{100}\) or \(\frac {9}{20}\)
(Dividing both terms by 5)

7. Write the numbers from 1 to 25.

Question (i)
What fraction of them are even numbers?
Solution:
Numbers from 1 to 25 are:
1,2, 3, 4,5,6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 i.e. 25 in number.
Even numbers out of these numbers are:
2, 4, 6, 8, 10, 12, 14, 16, 18; 20, 22, 24 i.e. 12 in number
∴ Required fraction = \(\frac {12}{25}\)

Question (ii)
What fraction of them are prime numbers?
Solution:
Prime number out of these numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. 9 in number
∴ Required fraction = \(\frac {9}{25}\)

Question (iii)
What fraction of them are multiples of 3?
Solution:
Multiples of 3 out of these numbers are:
3, 6, 9, 12, 15, 18, 21, 24 i.e. 8 in number
∴ Required fraction = \(\frac {8}{25}\)

8. In class 6th, there are 24 boys and 18 girls. What fraction of total students represent boys and girls?
Solution:
Boys = 24
Girls = 18
Total students = 24 + 18 = 42
Fraction which represents boys
= \(\frac {24}{42}\) or \(\frac {4}{7}\)
(Dividing both terms by 6)
Fraction which represents girls
= \(\frac {18}{42}\) or \(\frac {3}{7}\)
(Dividing both terms by 6)

9. A bag contains 6 red balls and 7 blue balls. What fraction of balls represent red and blue colour?
Solution:
Red balls = 6
Blue balls = 7
Total number of red and blue balls = 6 + 7 = 13
Fraction which represents red balls = \(\frac {6}{13}\)
Fraction which represents blue balls = \(\frac {7}{13}\)

10. Sidharth has a cake. He cuts it into 10 equal parts. He gave 2 parts to Naman, 3 parts to Nidhi, 1 parts to Seema and the remaining four parts he kept for himself. Find:

Question (i)
What fraction of cake, he gave to Naman?
Solution:
Total parts = 10
Fraction of cake, he gave to Naman = \(\frac {2}{10}\) or \(\frac {1}{5}\)

Question (ii)
What fraction of cake, he gave to Nidhi?
Solution:
Fraction of cake, he gave to Nidhi = \(\frac {3}{10}\)

Question (iii)
What fraction of cake, he kept for himself?
Solution:
Fraction of cake, he kept for himself = \(\frac {4}{10}\) or \(\frac {2}{5}\)

Question (iv)
Who has more cake than others?
Solution:
Sidharth has more cake than others

11. In a box, there are 12 apples, 7 oranges and 5 guavas. What fraction of fruits in box represents each?
Solution:
Apples = 12, Oranges = 7
and Gauvas = 5
Total fruits = 12 + 7 + 5 = 24.
Fraction which represents apples
= \(\frac {12}{24}\) or \(\frac {1}{2}\)
Fraction which represents oranges
= \(\frac {7}{24}\)
Fraction which represents Gauvas
= \(\frac {5}{24}\)

12. Dishmeet has 20 pens. He gives one-fourth to Balkirat. How many pens Dishmeet and Balkirat have?
Solution:
Total pens = 20
Pens Balkirat has = One-fourth of 20
= \(\frac {22}{7}\) × 20 = 5
Pens Dishmeet has = 20 – 5 = 15

13. Represent the following fraction on the number line?

Question (i)
\(\frac {2}{5}\)
Solution:
In order to represent \(\frac {2}{5}\) on number line, we divide the gap between 0 and 1 into 5 equal parts, which are, (as shown)

Question 2.
\(\frac {5}{7}\)
Solution:
In order to represent \(\frac {5}{7}\) on number line, we divide the gap between 0 and 1 into 7 equal parts, which are, (as shown)

Question 3.
\(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\)
Solution:
In order to represent \(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\) on number line, we divide the gap between 0 and 1 into 10 equal parts, which are (as shown)

Question 4.
\(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\)
Solution:
In order to represents \(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\) on number line, we divide the gap between 0 and 1 into 8 equal parts, which are (as shown)

14. Find:

(i) \(\frac {3}{5}\) of 20 books
(ii) \(\frac {5}{8}\) of 32 pens
(iii) \(\frac {1}{6}\) of 36 copies 6
(iv) \(\frac {4}{7}\) of 21 apples
(v) \(\frac {3}{4}\) of 28 pencils.
Solution:

15. Balkirat had a box of 36 erasers. He gave \(\frac {1}{2}\) of them to Rani, \(\frac {2}{9}\) of them to Yuvraj and keeps the rest.

Question (i)
(i) How many erasers does Rani get?
(ii) How many erasers does Yuvraj get?
(iii) How many erasers does Harnik keep?
Solution:
Total erasers = 36

(iii) Erasers Harnik gets = 36 – (18 + 8)
= 36 – 26
= 10

16. State True/False

Question (i)

Solution:
(i) False
(ii) True
(iii) True
(iv) True

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Answer:

In ∆ ABC, draw l, the perpendicular bisector of side AB and m, the perpendicular bisector of side BC. Name the point of intersection of l and m as P.
P is a point on the perpendicular bisector of AB. Hence, P is equidistant from A and B.
∴ PA = PB
P is a point on the perpendicular bisector of BC. Hence, P is equidistant from B and C.
∴ PB = PC
Thus, PA = PB = PC
Hence, P is the required point which is equidistant from all the vertices of ∆ ABC.
Note: Since ∆ ABC given here is an acute angled triangle, point P lies in the interior of ∆ ABC. If ∆ ABC is a right angled Mangle, point P lies on the hypotenuse. Actually, in that case, point P will be the midpoint of the hypotenuse. Lastly, if ∆ ABC is an obtuse angled triangle, point P lies in the exterior of ∆ ABC. This point P is called the circumcentre of ∆ ABC.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:
In ∆ ABC, draw the bisectors of ∠B and ∠C to intersect each other at point I.
I is a point on the bisector of ∠ B. Hence, I is equidistant from sides BA and BC. Similarly, I is a point on the bisector of ZC. Hence, I is equidistant from sides BC and CA.
Thus, point I is the required point which is equidistant from all the three sides AB, BC and CA of ∆ ABC.
This point I is called the incentre of ∆ ABC. It always lies in the interior of ∆ ABC irrespective of the type of ∆ ABC.

Question 3.
In a huge park, people are concentrated at three points (see the given figure):

A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it ?
(Hint: The parlour should be equidistant from A, B and C.)
Answer:
First of all, construct ∆ ABC with the given points A, B and C as vertices. Then, as shown in example 1, draw the perpendicular s bisectors of any two sides of ∆ ABC and name their point of intersection as P.

Now, the ice cream parlour should be set up at the location given by point P as it is equidistant from all the three places (points) A, B and C.

Question 4.
Complete the hexagonal and star shaped < Rangolies [see figure (1) and (2)] by filling them with as many equilateral triangles of > side 1 cm as you can. Count the number < of triangles in each case. Which has more triangles?

Answer:
In figure (1), if we join the opposite vertices, we get three longest diagonals of hexagon ABCDEE By the intersection of these diagonals we get point O and six equilateral triangles – ∆ OAB, ∆ OBC, ∆ OCD, ∆ ODE, ∆ OEF and ∆ OFA. Each side of all these equilateral triangles will measure 5 cm. In each of these six triangles, we can fill 25 (1 + 3 + 5 + 7 + 9) equilateral triangles with side 1 cm each. Hence in the hexagonal Rangoli ABCDEF, we can fill 25 × 6 = 150 triangles.

Similarly, in figure (2), we can fill 150 triangles in the inner hexagon and 150 triangles in the six triangles on the boundary of the hexagon. Thus, in figure (2), 150 + 150 = 300 triangles can be filled.
Hence, more triangles can be filled in figure (2).

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Answer:

In ∆ ABC, ∠ B is a right angle.
∴ ∠ B = 90° and AC is the hypotenuse.
In ∆ ABC,
∠A + ∠B + ∠C = 180°
∴ ∠ A + 90° + ∠ C = 180°
∴ ∠ A + ∠ C = 90°
Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.
∴ ∠ A < 90° and ∠ C < 90°
∴ ∠ A < Z B and ∠ C < ∠ B
∴ BC < AC and AB < AC (Theorem 7.7)
Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.

Question 2.
In the given figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠ QCB. Show that AC > AB.

Answer:
∠ PBC < ∠ QCB (Given)
∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)
∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)
Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.
∴ ∠ ABC = 180° – ∠ PBC and
∠ ACB = 180° – ∠ QCB
Substituting these values in (1), we get
∠ ABC > ∠ ACB
Now, in ∆ ABC, ∠ ABC > ∠ ACB A
∴ AC > AB (Theorem 7.7)

Question 3.
In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Answer:
In ∆ OAB, ∠ B < ∠ A
∴ OA < OB (Theorem 7.7) …………….. (1)
In ∆ OCD, ∠ C < ∠ D
∴ OD < OC (Theorem 7.7) ……………… (2)
Adding (1) and (2),
OA + OD < OB + OC
∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD ,(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

Answer:

Construction: In quadrilateral ABCD, draw diagonal AC.
AB is the smallest side of ABCD and CD is the longest side of ABCD.
∴ AB < BC and AD < CD.
In ∆ ABC, AB < BC
∴ ∠ ACB < ∠ BAC …… (1)
In ∆ CDA, AD < CD
∴ ∠ DCA < ∠ DAC ……… (2)
Adding (1) and (2),
∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC
∴ ∠ BCD < ∠ BAD
∴ ∠ BAD > ∠ BCD
Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.

Question 5.
In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Answer:
PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS = \(\frac{1}{2}\) ∠QPR ……………. (1)
∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.
∴ PSR = ∠ Q + ∠QPS and
∠PSQ = ∠R + ∠RPS …………….. (2)
Now, in A PQR, PR > PQ
∴ ∠ Q > ∠ R
∴ ∠Q + \(\frac{1}{2}\) ∠ QPR > ∠R + \(\frac{1}{2}\) ∠ QPR
∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]
∴ ∠PSR > ∠PSQ

Question 6.
Show that of all line segments drawn from a given point not on a given line, the perpendicular line segment is the shortest.
Answer:

AB is a line and P is a point not on AB.
PM is the perpendicular line segment drawn from P to line AB.
N is any point on AB, other than M.
In ∆ PMN, ∠ M = 90°
∴ ∠ N < 90°
Thus, in ∆ PMN, ZN < ZM.
∴ PM < PN
This is true for any location of point N.
Hence, of all the line segments drawn from a point not on a given line, the, perpendicular line segment is the shortest.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 7 Triangles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
In ∆ ABC, ∠A = ∠C, AC = 5 and BC = 4. Then, the perimeter of ∆ ABC is ……………. .
A. 9
B. 14
C. 13
D. 15
Answer:
C. 13

Question 2.
In ∆ PQR, PQ = PR, QR is extended to S and ∠PRS = 110°. Then, ∠PQR = ……………… .
A. 30°
B. 50°
C. 80°
D. 70°
Answer:
D. 70°

Question 3.
In ∆ ABC and ∆ DEF, AB = DE, BC = EF and ∠B = ∠E. If the perimeter of ∆ ABC is 20, then the perimeter of ∆ DEF is ……………. .
A. 10
B. 20
C. 15
D. 40
Answer:
B. 20

Question 4.
In ∆ ABC and ∆ PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. If ∠A + ∠C = 130°, then ∠Q = …………….. .
A. 65°
B. 130°
C. 50°
D. 100°
Answer:
C. 50°

Question 5.
In ∆ PQR, ∠P = ∠Q = ∠R. If PQ = 6, then the perimeter of ∆ PQR is ………………. .
A. 12
B. 9
C. 18
D. 24
Answer:
C. 18

Question 6.
In ∆ ABC, AB < AC. Then …………… holds good.
A. ∠A < ∠B
B. ∠B < ∠C
C. ∠C < ∠A
D. ∠C < ∠B
Answer:
D. ∠C < ∠B

Question 7.
In ∆ PQR, ∠R > ∠Q. Then, ………………. holds good.
A. PQ > PR
B. QR > PQ
C. PR > PQ
D. PQ > QR
Answer:
A. PQ > PR

Question 8.
In ∆ ABC, AB > BC and BC > AC. Then, the smallest angle of ∆ ABC is …………………. .
A. ∠A
B. ∠C
C. ∠B
D. ∠A or ∠C
Answer:
C. ∠B

Question 9
………………….. cannot be the measures of the sides of a triangle.
A. 10, 12, 14
B. 2, 3, 4
C. 8, 9, 10
D. 2, 4, 10
Answer:
D. 2, 4, 10

Question 10.
In ∆ PQR, PQ = 4, QR = 6 and PR = 5. Then, …………….. is the angle with greatest measure in ∆ PQR.
A. ∠P
B. ∠Q
C. ∠R
D. ∠QRP
Answer:
A. ∠P

Question 11.
In ∆ XYZ, ∠X = 45° and ∠Z = 60°. Then, …………….. is the longest side of ∆ XYZ.
A. XY
B. YZ
C. XZ
D. XY or YZ
Answer:
C. XZ

Question 12.
In ∆ ABC, ∠B = 30° and BC is extended to D. If ∠ACD = 110°, then the longest side of ∆ ABC is ………………. .
A. AB
B. BC
C. CA
D. AB or AC
Answer:
B. BC

Question 13.
In ∆ ABC, AB = 4 and BC = 7, Then, ……………… holds good.
A. AC < 7
B. AC > 4
C. 4 < AC < 7
D. 3 < AC < 11
Answer:
D. 3 < AC < 11

Question 14.
In ∆ PQR, PQ = 3 and QR = 7. Then, …………….. holds good.
A. PR = 4
B. PR = 10
C. 10 > PR > 4
D. 7 > PR > 3
Answer:
C. 10 > PR > 4

Question 15.
In ∆ ABC, the bisectors of ∠B and ∠C intersect at I. If ∠A = 70°, then ∠BIC = ………………… .
A. 35°
B. 75°
C. 100°
D. 125°
Answer:
D. 125°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :
(i) Length = 28 cm, Breadth = 15 cm
(ii) Length = 9.4 cm Breadth = 2.5 cm
Solution:
(i) Given length of rectangle = 28 cm
Breadth of rectangle = 15 cm
Perimeter of rectangle = 2 [length + Breadth]
= 2 [28 + 15]
= 2 × 43
= 86 cm

Area of rectangle = length × Breadth
= 28 × 15
= 420 cm²

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]
= 2 × 11.9
= 23.8 cm
Area of rectangle = 9.4 × 2.5
= 23.5 cm²

2. Find the perimeter and the area of a square whose side measures
(i) 29 cm
(ii) 8.3 cm
Solution:
(i) Given side of square = 29 cm
Perimeter of square = 4 × side
= 4 × 29
= 116 cm
Area of square = (side)²
= (29)²
= 841 cm²

(ii) Perimeter of square = 4 × 8.3
= 33.2 cm
Area of square = 8.3 × 8.3
= 68.89 cm²

3. The perimeter of a square park is 148 m. Find its area.
Solution:
Given the perimeter of square park = 148 m
Side of the square park = \(\frac{perimeter}{4}\)
= \(\frac {148}{4}\)
Area of the square park = (side)²
= (37)²
= 1369 m²

4. The area of a rectangle is 580 cm². Its length is 29 cm. Find its breadth and also, the perimeter.
Solution:
Given area of rectangle = 580 cm²
Length of the rectangle = 29 cm
Let breadth of the rectangle = b cm
Area of the rectangle = length × breadth
580 = 29 × b
\(\frac {580}{29}\) = b
b = 20 cm
Perimeter of rectangle = 2 [length + breadth]
= 2 [29 + 20]
= 2 × 49
= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?
Solution:
Given length of the rectangle = 48 cm
Breadth of the rectangle = 32 cm
Perimeter of the rectangle = 2 [length + breadth]
= 2 [48 + 32]
= 2 × 80
= 160 cm
Let side of square = a cm
Perimeter of the square = 4 × a
Since wire is rebent into the shape of a square
Perimeter of square = Perimeter of rectangle
4 a = 160
Therefore, a = \(\frac {160}{4}\)
= 40 cm
Area of square = (side)²
= 40 × 40
= 1600 cm²
Area of rectangle = length × breadth
= 48 × 32
= 1536 cm²
∴ Square encloses more area by 64 cm²

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.
Solution:
Given side of square park = 75 m
Area of square park = (75)²
= 75 × 75
= 5625 m²
Length of rectangular park = 125 m
Let breadth of rectangular park = b m
Area of rectangular park = length × breadth
= 125 × b m²
Given that
Area of rectangular park = Area of square park
125 × b = 5625
= 45 m
Perimeter of rectangular park = 2 [length + breadth]
= 2 [125 + 45]
= 2 × 170
= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m².

Solution:
Length of door = 2.5 m
Breadth of door = 1.5 m
Area of door = length × breadth
= 2.5 × 1.5
= 3.75 m²
Area of wall = 9 × 6
= 54 m²
Area of wall painting = Area of wall including door – Area of door
= 54 – 3.75
= 50.25 m²
Cost of painting 1 m² of wall = ₹ 30
Cost of painting 50.25 m² of wall = ₹ 50.25 × 30
= ₹ 1507.50

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m².
Solution:
Area of door = 3 × 2 = 6 m²
Area of window = 2.5 m × 1.5 m
= 3.75 m²
Area of wall = 7.8 m × 3.9 m
= 30.42 m²
Area of painting the wall = Area of wall – Area of door – Area of window
= 30.42 – 6 – 3.75
= 20.67 m²
Cost of painting the wall = ₹ 25 × 20.67
= ₹ 516.75

9. Find the area and the perimeter of the following figures.

Solution:
(i) Perimeter of the given figure
= AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9
= 38 cm²
Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH
= 2 × 3.5 + 5 × 2 + 10 × 3.5
= 7 + 10 + 35
= 52 cm²

(ii) Perimeter of the given figure
= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm
= 29 cm
Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III
= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm
= 12 cm² + 5.25 cm² + 2.25 cm
= 19.5 cm²

10. Multiple Choice Questions :

Question (i).
What is the area of a rectangle of dimensions 12 cm × 10 cm ?
(a) 44 cm²
(b) 120 cm²
(c) 1200 cm²
(d) 1440 cm²
Answer:
(b) 120 cm²

Question (ii).
Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.
(a) 6 cm
(b) 3 cm
(c) 9 cm
(d) 12 cm
Answer:
(a) 6 cm

Question (iii).
If each side of a square is 1 m then its area is ?
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²
Answer:
(d) 10000 cm²

Question (iv).
Find the area of a square whose perimeter is 96 cm.
(a) 576 cm²
(b) 626 cm²
(c) 726 cm²
(d) 748 cm².
Answer:
(a) 576 cm²

Question (v).
The area of a rectangular sheet is 500 cm². If the length of the sheet is 25 cm, what is its breadth ?
(a) 30 cm
(b) 40 cm
(c) 20 cm
(d) 25 cm.
Answer:
(c) 20 cm

Question (vi).
What happens to the area of a square, if its side is doubled ?
(a) The area becomes 4 times, the area of original square.
(b) The area becomes \(\frac {1}{4}\) times, the area of original square.
(c) The area becomes 16 times, the area of original square.
(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.
Answer:
(a) The area becomes 4 times, the area of original square.

Punjab State Board PSEB 7th Class English Book Solutions English Reading Comprehension Conversation / Dialogue Based Exercise Questions and Answers, Notes.

PSEB 7th Class English Reading Comprehension Conversation / Dialogue Based

Read the following conversation carefully and answer the questions that follow:

1. Akbar : Birbal make me a painting. Use your imagination in it.
Birbal : I am not an artist, how can I paint ?
Akbar : If I don’t get a good painting, you shall be punished.
Birbal : (Next day) I am here with the painting.
Akbar : I am happy to see that you obeyed me. Please show me the painting.
Birbal : Huzoor, have a look. I am opening the covered frame.
Akbar : This painting is nothing but only ground and sky. There is a little grass on the ground. What is this?
Birbal : A cow eating grass, Huzoor!
Akbar : Where is the cow and grass ?
Birbal : I used my imagination. The cow ate the grass and returned to its shed.

Question 1.
The given conversation is between ……………
(a) Birbal and Begum
(b) Birbal and his Son
(c) Akbar and his Son
(d) Akbar and Birbal.
Answer:
(d) Akbar and Birbal.

Question 2.
What did Akbar want Birbal to make for him?
(a) a cake
(b) a painting
(c) a dish
(d) all of the above.
Answer:
(b) a painting.

Question 3.
Who ate the grass ?
(a) cow
(b) cat
(c) camel
(d) goat.
Answer:
(a) cow.

Question 4.
What was in the painting ?
(a) ground and sky
(b) boy and sky
(c) girl and kite
(d) cat and dog.
Answer:
(a) ground and sky.

Question 5.
Birbal carried a ……………. with him.
(a) covered glass
(b) covered frame
(c) covered bowl
(d) covered box.
Answer:
(b) covered frame.

2. Sadab : Good afternoon.
Ritu : Good afternoon.
Sadab : So, where are you going now?
Ritu : I am going to meet my friend Garima in Kolkata Market.
Sadab : Oh, so nice, I have to buy a new school uniform.
Ritu : Will you accompany me ?
Sadab : Yes, it will be great fun for me.
Ritu : So, can we move now so that we can reach on time.
Sadab : Sure.

Question 1.
This conversation is between ………..
(a) Sadab and Garima
(b) Ritu and Garima
(c) Ritu and Sadab
(d) Garima and Kolkata Market.
Answer:
(c) Ritu and Sadab.

Question 2.
Where is Ritu going ?
(a) to meet her sister Garima
(b) to meet her friend Garima
(c) to meet her teacher
(d) to buy a school uniform.
Answer:
(b) to meet her friend Garima.

Question 3.
Garima will be found ……………
(a) in Kolkata market
(b) in Bombay market
(c) in Kolkata street
(d) in Sadab market.
Answer:
(a) in Kolkata market.

Question 4.
What does Sadab want to buy ?
(a) some new books
(b) a new mobile phone
(c) a new school uniform
(d) new shoes.
Answer:
(c) a new school uniform.

Question 5.
What was the time of conversation ?
(a) early morning
(b) afternoon
(c) late evening
(d) night.
Answer:
(b) afternoon.

3. Headman Ant asked, “What happened ? Why are you running ?” Rabbit said, “We’re running for our lives. I heard the news of Tsunami, very high sea waves.”

Headman Ant asked everyone to stay calm and explained that they were just testing the warning system. It didn’t happen for real. Let us all be prepared to deal with disasters.

Next day at the village meeting, everyone gave their ideas.

Elephant : “We should always follow news carefully.”
Butterfly : “I have already put my valuable things away in a safe place.”
Rats : “I have prepared survival bag which will always be close to me. I have kept a bottle of water, medicines, dried food, some beans, the radio for news and a family photo.”
Mother Frog : “My kids will not go out to play by themselves in hazardous places.”
Mother Goat : “We must run to higher grounds or to the safe areas. I shall make a map of the village.
Headman Ant : “Why didn’t you run like asked Grandpa others ?
OX : ……..
Grandpa OX : “We are too old. We don’t want to be a burden to anyone.”
Headman Ant : (said politely) “No one is a burden. What would your grandchildren do without you ??

Question 1.
Rabbit was running for life because he had heard the news of ……………
(a) Earthquake
(b) Tsunami
(c) Flash flood
(d) Land slide.
Answer:
(b) Tsunami.

Question 2.
What did the Headman Ant tell the animals about the news ?
(a) It was real
(b) It was to create panic only
(c) It was not real
(d) It was to entertain the old people.
Answer:
(c) It was not real.

Question 3.
Who is most careful about news ?
(a) Grandpa Ox
(6) Rabbit
(c) Elephant
(d) Frog.
Answer:
(c) Elephant.

Question 4.
Which one of the following is not true according to the conversation ?
(a) The elephant was not present at the meeting
(6) Grandpa Ox does not want to be a burden to anyone
(c) Mother Goat wants to run to the higher places
(d) Rats have prepared survival bags.
Answer:
(a) The elephant was not present at the meeting.

Question 5.
Tsunami is a ……………..
(a) safe place from high sea wave
(b) very high sea wave
(c) cool breeze coming from sea
(d) dry wind blowing towards sea.
Answer:
(b) very high sea wave.

4. Ali : Hello ! I’m Ali. What’s your name?
Ravi : Hello ! My name’s Ravi. I’m ten years old.
Ali . : I’m ten, too. Do you play cricket ?
Ravi : Not very well. I’m fond of football.
Ali : Oh, good ! Will you help me with my homework?
Ravi : Of course, I will. Come to me during the break.
Ali : Thanks. I’ll help you play cricket.
Ravi : That will be great. Thank you !
Ali : : Where do you live ?
Ravi : I live near the school and Ali, where do you live?
Ali : I live in the school hostel.

Question 1.
What is common between Ali and Ravi ?
(a) both are ten years old
(b) both play cricket very well
(c) both live in the school hostel
(d) both are fond of playing football.
Answer:
(a) both are ten years old.

Question 2.
Where does Ali live ?
(a) near the school
(b) in the school hostel
(c) near the football club
(d) near the cricket club.
Answer:
(b) in the school hostel.

Question 3.
Ali will help Ravi ……………
(a) play football
(b) do his homework
(c) play cricket
(d) all the above.
Answer:
(c) play cricket.

Question 4.
Ravi will help Ali ………
(a) play cricket
(b) play football
(c) do his homework
(d) none of these.
Answer:
(c) do his homework.

Question 5.
When will Ravi help Ali with his homework?
(a) after playing cricket
(b) after playing football
(c) when the school starts
(d) during the break.
Answer:
(d) during the break.

5. Ravi : Do you know what time it is now, please ?
Aman : No, I’m really very sorry, I don’t have a watch.
Rama : Could you tell me the time, please ?
Radha : It’s about eleven-thirty.
Rama : Thanks a lot.
Radha : At what time do you come to school ?
Rama : At 8:30 in the morning.
Radha : At what time does the school get over ?
Rama : At 3 o’clock in the afternoon.
Ajay : What time is the next bus, please ?
Anil : It leaves at 5:30 p.m.
Ajay : Is there any bus after 10 p.m. ?
Anil : The last bus leaves at 10:30 p.m.
Ajay : Thanks a lot for the information.

Question 1.
The main point in this conversation is …………… .
(a) bus
(b) time
(c) watch
(d) school.
Answer:
(b) time.

Question 2.
Aman is unable to tell Ravi time because …………….
(a) he is in a hurry
(b) he is very sorry
(c) he has no watch
(d) he has no clock.
Answer:
(c) he has no watch.

Question 3.
When does Rama go to school ?
(a) at 8:30
(b) at 3 o’clock
(c) at 11:30
(d) at 10 o’clock.
Answer:
(a) at 8:30.

Question 4.
What does Ajay want to know ?
(a) times about school
(b) times about buses
(c) time by Aman’s watch
(d) none of these.
Answer:
(b) times about buses.

Question 5.
Who tells Ajay time for last bus ?
(a) Rama
(b) Anil
(c) Radha
(d) Aman.
Ans.
(b) Anil.

6. Raj : Could you help me, please ?
Shopkeeper : Certainly.
Raj : Thank you.
Shopkeeper : What can I get you ?
Raj : A black shoe polish.
Rani : Excuse me please, where are the pens ?
Shopkeeper : They are in the second row. Let me help you.
Rani : How much do the two pens cost ?
Shopkeeper : The price is written on them. They cost Rs. 10 each.
Rani : Thank you.

Question 1.
Whose help does Raj ask for ?
(a) Rani’s
(b) shopkeeper’s
(c) nobody’s
(d) his own.
Answer:
(b) shopkeeper’s.

Question 2.
Raj wants to buy a ……………..
(a) brown shoe polish
(b) black shoe polish
(c) some pens and a polish
(d) none of these.
Answer:
(b) black shoe polish.

Question 3.
The pens are lying
(a) in a box
(b) in the first row
(c) in the second row
(d) on the upmost shelf.
Answer:
(c) in the second row.

Question 4.
How many pens does Rani want to buy ?
(a) two
(b) three
(c) one.
(d) four.
Answer:
(a) two.

Question 5.
How much do the two pens cost ?
(a) Rs. 20
(6) Rs. 10
(c) Rs. 30
(d) Rs. 15.
Answer:
(a) Rs. 20.

7. Aryan : Why are you wearing this funny dress, Praveen ?
Praveen : Aryan, what is funny about it?
Aryan : Look at your clothes ! They look so tight, and your shoes have wheels. And what’s that you have on your head ?
Praveen : (laughs) Oh, are you talking about this dress ? I’m going for my skating class and this is my helmet to save me from injuries.
Aryan : Skating ?
Praveen : Yes, skating. That’s why I’m wearing these roller skating shoes.
Aryan : Where are you going for your skating class ?
Praveen : I’m going to the ‘Roller Mall’. It has a skating rink.
Aryan : Could I come with you to see what you do there?
Praveen : Sure, come along. (They reach the skating rink.) Aryan (amazed) Oh, so many people moving on the wheels !
Praveen : They’re all skaters moving on roller skates.
Aryan : Look ! How they all are gliding on the floor. I wish I could do the same. Will you teach me how to skate ?
Praveen : Oh yes, whenever you are ready.
Aryan : Thanks.

Question 1.
Praveen’s dress looks funny to Aryan. Why?
(a) It is very tight.
(b) His shoes have wheels.
(c) There is a, helmet on his head.
(d) all these.
Answer:
(d) all these.

Question 2.
Praveen is going to …………
(a) his school to take part in a drama
(b) a park for a picnic
(c) perform a magic show
(d) his skating class.
Answer:
(d) his skating class.

Question 3.
A helmet saves us from …………
(a) foot injuries
(b) severe cold on ice
(c) head injuries
(d) all the above.
Answer:
(c) head injuries.

Question 4.
What does Aryan want to learn ?
(a) skating
(b) use of wheeled shoes
(c) using a helmet
(d) wearing a tight dress.
Answer:
(a) skating.

Question 5.
Who will teach Aryan how to skate ?
(a) skating teacher
(b) his elder brother
(c) Praveen
(d) class teacher.
Answer:
(c) Praveen.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

1. Construct ΔABC in which AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
Solution:
Given : Three sides of a triangle as AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
To construct : A triangle with these three sides.
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment BC = 5 cm.

Step 3 : From B; point A is at a distance 3.5 cm. So, with B as centre, draw an arc of radius
3.5 cm (Now point A will be some where on this arc. Our job is to find where exactly A is.)

Step 4. From C, point A is at a distance of 7 cm. So, with C as centre; draw an arc of radius 7 cm. (A will be some where on this arc. We have to fix it).

Step 5. A has to be on both the arcs drawn. So it is the point of intersection of arcs.
Mark the point of intersection of arcs as A. Join AB and AC.
Thus we obtain ΔABC.

2. Construct a triangle ABC in which AB = BC = 6.5 cm and CA = 4 cm. Also name the kind of triangle drawn.
Solution:
Given : Three sides of triangle as AB = BC = 6.5 cm. and CA = 4 cm.
To construct : A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment AC = 4 cm.

Step 3. From A; point B is at a distance of 6.5 cm. So, with A as centre, draw an arc of radius 6.5 cm. (Now point B will be somewhere on this arc. Our job is to find where exactly A is.)

Step 4. From C; point B is at a distance of 6.5 cm. So; with C as centre; draw an arc of radius 6.5 cm. (B will be some where on this arc. We have to fix it.)

Step 5. B has to be on both the arcs drawn. So it is the point of intersection of arcs. Mark the point of intersection of arcs as B. Join AB and BC.

We observe that AB = BC = 6.5 cm.
Since two sides are of equal length. Thus we obtain an isosceles ΔABC.

3. Construct a triangle XYZ such that length of each side is 5 cm. Also name the kind of triangle drawn.
Solution:
Given : A triangle XYZ in which XY = YZ = ZX = 5 cm..
To Construct. A triangle XYZ with each side 5 cm.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle XYZ and indicate the lengths of the three sides.

Step 2. Draw a line segment YZ = 5 cm.

Step 3. From Y; point X is at a distance of 5 cm. So, with Y as centre, draw an arc of radius 5 cm. (Now point X will be somewhere on this arc. Our job is to find where exactly X is.)

Step 4. From Z, point X is at a distance of 5 cm. So, with Z as centre, draw an arc of radius 5 cm. (X will be somewhere oh this arc. We have to fix it.)

Step 5. Point X has to be on both the arcs drawn. So, it is the point of intersection of arcs.
Mark the point of intersection of arcs as X. Join XY and XZ.

Thus we obtain an equilateral ΔXYZ each of whose side is 5 cm.

4. Construct a triangle PQR such that PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm. Measure ∠PQR and also name the kind of triangle drawn.
Solution:
Given. Three sides of triangle as PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm.
To construct. A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the lengths of the three sides.

Step 2. Draw a line segment QR of length 6 cm.

Step 3. From Q; point P is at a distance of 2.5 cm. So, with Q as centre, draw an arc of radius 2.5 cm. (Now point P will be some where on this arc. Our job is to find where exactly P is.)

Step 4. From R; point P is at a distance of 6.5 cm. So; with R as centre; draw an arc of radius 6.5 cm. (P will be somewhere on this arc. We have to fix it.)

Step 5. Point P has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark a point of intersection of arcs as P. Join PQ and PR.

Thus we obtain ΔPQR; on measuring ∠PQR, we observe that ∠PQR = 90° so it is a right angled triangle.

5. Construct a triangle ABC, in which AB = 6 cm, BC = 2 cm, CA = 3 cm. (If possible). If not possible give the reason.
Solution:
Since AB = 6 cm, BC = 2 cm, CA = 3 cm
Here BC + CA = 2 cm + 3 cm
= 5 cm < 6 cm < AB
Which is not possible because the sum of two sides of a triangle is always greater than third side of the triangle.

6. Question (i).
Which of the following can be used to construct a triangle ?
(a) The lengths of the three sides
(b) The perimeter of the triangle
(c) The measures of three angles
(d) The name of three vertices
Answer:
(a) The lengths of the three sides

Question (ii).
A triangle can be constructed by taking its sides as :
(a) 1.8 cm, 2.6 cm, 4.4 cm
(b) 3 cm, 4 cm, 8 cm
(c) 4 cm, 7 cm, 2 cm
(d) 5 cm, 4 cm, 4 cm.
Answer:
(d) 5 cm, 4 cm, 4 cm.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 3 Playing with Numbers MCQ Questions

Multiple Choice Questions

Question 1.
Which number is a factor of every, number?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 2.
How many even numbers are prime?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 3.
The smallest composite number is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 4.
Which of the following number is a perfect number?
(a) 8
(b) 6
(c) 12
(d) 18.
Answer:
(b) 6

Question 5.
Which of the following is not a multiple of 7?
(a) 35
(b) 48
(c) 56
(d) 91.
Answer:
(b) 48

Question 6.
Which of the following is not a factor of 36?
(a) 12
(b) 6
(c) 9
(d) 8.
Answer:
(d) 8.

Question 7.
The number of prime numbers upto 25 are:
(a) 9
(b) 10
(c) 8
(d) 12.
Answer:
(a) 9

Question 8.
Which mathematician gave the method to find prime and composite numbers?
(a) Aryabhatta
(b) Ramayan
(c) Eratosthenes
(d) Goldbach.
Answer:
(c) Eratosthenes

Question 9.
The statement “Every even number greater than 4 can be expressed as the sum of two odd prime numbers” is given by:
(a) Goldbach
(b) Eratosthenes
(c) Aryabhatta
(d) Ramanujan.
Answer:
(a) Goldbach

Question 10.
Which of the following is a prime number?
(a) 221
(b) 195
(c) 97
(d) 111.
Answer:
(c) 97

Question 11.
Which of the following number is divisible by 4?
(a) 52369
(b) 25746
(c) 21564
(d) 83426.
Answer:
(c) 21564

Question 12.
Which of the following is not true?
(a) If a number is factor of two numbers then it is also factor of their sum.
(b) If a number is factor of two numbers then it is also factor of their difference.
(c) 15 and 24 are co-prime to each other.
(d) 1 is neither prime nor composite.
Answer:
(c) 15 and 24 are co-prime to each other.

Question 13.
Which of the following pair is co-prime?
(a) (12, 25)
(b) (18, 27)
(c) (25, 35)
(d) (21, 56).
Answer:
(a) (12, 25)

Question 14.
Which of the following number is divisible by 8?
(a) 123568
(b) 412580
(c) 258124
(d) 453230.
Answer:
(a) 123568

Question 15.
Prime factorisation of 84:
(a) 2 × 2 × 3 × 2 × 7
(b) 7 × 2 × 3 × 3
(c) 2 × 3 × 7 × 2
(d) 3 × 2 × 3 × 2 × 7.
Answer:
(c) 2 × 3 × 7 × 2

Question 16.
H.C.F. of 25 and 45 is:
(a) 15
(b) 5
(c) 225
(d) 135.
Answer:
(b) 5

Question 17.
If L.C.M. of two numbers is 36 then which of the following can not be their H.C.F.?
(a) 9
(b) 12
(c) 8
(d) 18.
Answer:
(c) 8

Question 18.
The L.C.M. of two co-prime numbers is 143. If one number is 11 then find other number.
(a) 132
(b) 154
(c) 18
(d) 13.
Answer:
(d) 13.

Question 19.
Find the greatest number which divides 145 and 235 leaving the remainder 1 in each case.
(a) 24
(b) 18
(c) 19
(d) 17.
Answer:
(b) 18

Question 20.
The greatest 4 digit number which is divisible by 12,15 and 20.
(a) 9990
(b) 9000
(c) 9960
(d) 9999.
Answer:
(c) 9960

Question 21.
Which of the following is a prime number?
(a) 23
(b) 51
(c) 39
(d) 26.
Answer:
(a) 23

Question 22.
Which of die following is a prime number?
(a) 32
(b) 30
(c) 31.
(d) 33.
Answer:
(c) 31.

Question 23.
Which of the following is a composite number?
(a) 12
(b) 19
(c) 29
(d) 31.
Answer:
(a) 12

Question 24.
Which of the following is an even number?
(a) 13
(b) 15
(c) 16
(d) 19.
Answer:
(c) 16

Question 25.
Which of the following is an odd number?
(a) 12
(b) 13
(c) 14
(d) 20.
Answer:
(b) 13

Fill in the blanks:

Question 1.
…………… is an even prime number?
Answer:
2

Question 2.
…………… is the greatest prime number between 1 and 10.
Answer:
7

Question 3.
……………. is neither prime nor composite number.
Answer:
1

Question 4.
A number which has only two factors is called a …………….. number.
Answer:
prime number

Question 5.
A number which has more than two factors is called a ……………… number.
Answer:
composite number

Write True/False:

Question 1.
The sum of three odd number is even. (True/False)
Answer:
False

Question 2.
All prime numbers are odd. (True/False)
Answer:
False

Question 3.
All even numbers are composite numbers. (True/False)
Answer:
False

Question 4.
1 neither prime nor composite. (True/False)
Answer:
True

Question 5.
If a number is factor of two numbers then it is also factor of their sum. (True/False)
Answer:
True

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 6 Lines and Angles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The measure of the complementary angle of an angle with measure 40° is ………………….. .
A. 40°
B. 20°
C. 140°
D. 50°
Answer:
D. 50°

Question 2.
The measure of the supplementary angle of an angle with measure 70° is ………………… .
A. 20°
B. 35°
C. 70°
D. 110°
Answer:
D. 110°

Question 3.
∠ ABC and ∠ ABD form a linear pair. If ∠ ABC = 30°, then ∠ ABD = ………………. .
A. 30°
B. 60°
C. 150°
D. 15°
Answer:
C. 150°

Question 4.
∠P and ∠Q are supplementary angles such that ∠P = 2x – 5 and ∠Q = 3x + 10. Then ∠Q = …………….. .
A. 35°
B. 65°
C. 105°
D. 115°
Answer:
D. 115°

Question 5.
The measure of an angle is four times the measure of its complementary angle. Then, the measure of that angle is ………………… .
A. 18°
B. 72°
C. 40°
D. 10°
Answer:
B. 72°

Question 6.
The measures of two supplementary angles differ by 20°. Then, the measure of the acute angle among them is …………….. .
A. 50°
B. 80°
C. 100°
D. 20°
Answer:
B. 80°

Question 7.
The measure of an angle is twice the measure of its supplementary angle. Then, the measure of that angle is ………………. .
A. 60°
B. 120°
C. 50°
D. 100°
Answer:
B. 120°

Question 8.
∠ ACD is an exterior angle of ∆ ABC. If ∠ ACD = 110° and ∠ A = 60°, then ∠ B = ………………….. .
A. 50°
B. 60°
C. 70°
D. 55°
Answer:
A. 50°

Question 9.
In ∆ ABC, ∠ A = 70° and ∠ B = 60°. Then, the measure of an exterior angle of ∆ ABC can be ……………….. .
A. 50°
B. 110°
C. 100°
D. 70°
Answer:
B. 110°

Question 10.
In ∆ ABC, ∠ B = 55° and ∠ C = 65°. Then the measure of an exterior angle of ∆ ABC cannot be ………………. .
A. 125°
B. 120°
C. 115°
D. 110°
Answer:
D. 110°