Class 9

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction :

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E
   Thus, ∠CAB is the required angle of 90°.

Justification:
Draw PX and PY.
In ∆ PAX and ∆ PAY,
AX = AY (Radii of same arc)
PX = PY (Radii of congruent arcs)
PA = PA (Common)
∴ By SSS rule, ∆ PAX ≅ ∆ PAY
∴ ∠PAX = ∠PAY (CPCT)
But, ∠PAX + ∠PAY = 180° (Linear pair)
∴ ∠PAY = \(\frac{180^{\circ}}{2}\) = 90°
∴ ∠CAB = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:

Steps of construction:

1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect at P on one side of line MAB.
4. Draw ray AC passing through E Thus, ∠CAB of 90° is received.
5. Name the point of intersection of the arc with centre A and ray AC as Z.
6. Taking Y and Z as centres and radius more than \(\frac{1}{2}\)YZ, draw arcs to intersect each other at Q.
7. Draw ray AQ.
   Thus, ∠QAB is the required angle of 45°.

Justification:
In example 1, we have already justified that ∠CAB = 90°. So, we do not repeat that part’ here.
Draw QZ and QY.
In ∆ AYQ and ∆ AZQ,
AY = AZ (Radii of same arc)
YQ = ZQ (Radii of congruent arcs)
AQ = AQ (Common)
∴ By SSS rule, ∆ AYQ ≅ ∆ AZQ
∴ ∠QAY = ∠QAZ (CPCT)
But, ∠QAY + ∠QAZ = ∠ZAY = ∠CAB = 90°
∴ ∠QAY = \(\frac{90^{\circ}}{2}\) = 45°
∴ ∠QAB = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
Answer:

Steps of construction:

1. Draw any ray AB. With centre A and any radius, draw an arc to intersect AB at X.
2. With centre X and the same radius [as in step (1)], draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AT, the bisector of ∠YAB.
   Thus, ∠TAB is the required angle of 30°.

(ii) 22\(\frac{1}{2}\)°
Answer:

Steps of construction:

1. Draw any ray AB. Produce AB on the side of A to get line CAB.
2. Taking A as centre and any radius, draw an arc of a circle to intersect line CAB at X and Y.
3. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 45°.
5. Draw ray AN, the bisector of ∠MAB. Then, ∠NAB = 22\(\frac{1}{2}\)°.
   Thus, ∠NAB is the required angle of 22\(\frac{1}{2}\)°.

(iii) 15°
Answer:

Steps of construction:

1. Draw any ray AB. Taking A as centre and any radius, draw an arc of a circle to intersect AB at X.
2. Taking X as centre and the same radius as before, draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
3. Draw ray AL, the bisector of ∠YAB. Then, ∠LAB = 30°.
4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 15°.
   Thus, ∠MAB is the required angle of 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75° and (ii) 105°
Answer:

Steps of construction:

1. Draw any ray AB and produce it on the side of A to get line CAB. Taking A as centre and any radius draw an arc of a circle to intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more than \(\frac{1}{2}\)XY, draw arcs to intersect each other at point L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
3. Taking X as centre and radius AX, draw an arc of a circle to Intersect the first arc (arc XY) with centre A at Z.
4. Draw ray AZ. Then, ∠ZAB = 60°.
5. Now, draw ray AM, the bisector of ∠LAZ. Then, ∠MAB = 75° and ∠MAC = 105°.
   Thus, ∠MAB and ∠MAC are the required angles of measure 75° and 105° respectively.

(iii) 135°
Answer:

Steps of construction:

1. Draw line CAB. Taking A as centre and any radius, draw an arc of a circle to Intersect line CAB at X and Y.
2. Taking X and Y as centres and radius more them \(\frac{1}{2}\)XY, draw arcs to intersect each other at P on one side of line CAB.
3. Draw ray AP Then, ∠PAB = ∠PAC = 90°.
4. Draw ray AQ, the bisector of ∠PAC.
5. Then, ∠QAB = 135°.
   Thus, ∠QAB is the required angle of 135°.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Answer:
Line segment XY is given. We have to construct an equilateral triangle with each side being equal to XY.

Steps of construction:

1. Draw any ray BM.
2. With centre B and radius XY, draw an arc of a circle to intersect BM at C.
3. Taking B and C as centres and rhdius XY, draw arcs to intersect each other at A on one side of line AC.
4. Draw AB and AC.
   Thus, ∆ ABC is the required equilateral triangle with each side being equal to XY.

Justification:
The arc drawn with centre B and radius XY intersects ray BM at C. ∴ BC = XY. The arcs drawn with centres B and C and radius XY intersect at A.
∴ AB = XY and AC = XY.
Thus, in ∆ ABC, AB = BC = AC = XY.
Hence, ∆ ABC is an equilateral triangle in which all sides are equal to XY.
Note: If the measure of sides are given numerically, e.g., 4 cm, 5 cm, etc., then we have to use graduated scale instead of straight edge.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 10 Circles 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 a circle with centre P, AB and CD are congruent chords. If ∠PAB = 40°, then ∠CPD = ………………..
A. 40°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 2.
In a circle with radius 5 cm, the length of a chord lying at distance 4 cm from the centre is …………………. cm.
A. 3
B. 6
C. 12
D. 15
Answer:
B. 6

Question 3.
In a circle with radius 13 cm, the length of a chord is 24 cm. Then, the distance of the chord from the centre is ……………….. cm.
A. 10
B. 5
C. 12
D. 6.5
Answer:
B. 5

Question 4.
In a circle with radius 7 cm, the length of a minor arc is always less than ………………… cm.
A. 11
B. 22
C. 15
D. π
Answer:
B. 22

Question 5.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠APB = 150°, then ∠ARB = …………… .
A. 150°
B. 75°
C. 50°
D. 100°
Answer:
B. 75°

Question 6.
In a circle with centre P, AB is a minor arc. Point R is a point other than A and B on major arc AB. If ∠ARB = 80°, then ∠APB = ……………. .
A. 40°
B. 80°
C. 160°
D. 60°
Answer:
C. 160°

Question 7.
In cyclic quadrilateral ABCD, ∠A – ∠C = 20°.
Then, ∠A = ………………. .
A. 20°
B. 80°
C. 100°
D. 50°
Answer:
C. 100°

Question 8.
In cyclic quadrilateral PQRS, 7∠P = 2∠R.
Then, ∠P = ………………….. .
A. 20°
B. 40°
C. 140°
D. 100°
Answer:
B. 40°

Question 9.
The measures of two angles of a cyclic quadrilateral are 40° and HOP. Then, the measures of other two angles of the quadrilateral are ……………….. .
A. 40° and 110°
B. 50° and 100°
C. 140° and 70°
D. 20° and 120°
Answer:
C. 140° and 70°

Question 10.
In cyclic quadrilateral PQRS, ∠SQR = 60° and ∠QPR = 20°. Then, ∠QRS = ……………… .
A. 40°
B. 60°
C. 80°
D. 100°
Answer:
D. 100°

Question 11.
In cyclic quadrilateral ABCD, ∠CAB = 30° and ∠ABC = 100°. Then, ∠ADB =
A. 50°
B. 100°
C. 75°
D. 60°
Answer:
A. 50°

Question 12.
Equilateral ∆ ABC is inscribed in a circle with centre P. Then, ∠BPC = ……………. .
A. 60°
B. 90°
C. 120°
D. 75°
Answer:
C. 120°

Question 13.
∆ ABC is inscribed in a circle with centre O and radius 5 cm and AC is a diameter of the circle. If AB = 8 cm, then BC = ………………… cm.
A. 10
B. 8
C. 6
D. 15
Answer:
C. 6

Question 14.
In cyclic quadrilateral ABCD, ∠A = 70° and ∠B + ∠C = 160°. Then, ∠B = ………………. .
A. 35°
B. 25°
C. 50°
D. 130°
Answer:
C. 50°

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Question 1.
Prove that the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:

Circles with centres O and P intersect each other at points A and B.
In ∆ OAP and ∆ OBR
OA = OB (Radii of circle with centre O)
PA = PB (Radii of circle with centre P)
OP = OP (Common)
∴ By SSS rule, ∆ OAP = ∆ OBP
∴ ∠OAP = ∠OBP (CPCT)
Thus, OP subtends equal angles at A and B. Hence, the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer:

Draw the perpendicular bisector of AB to intersect AB at M and draw the perpendicular bisector of CD to intersect CD at N.
Both these perpendicular bisectors pass through centre O and since AB || CD; M, O and N are collinear points.
Now, MB = \(\frac{1}{2}\)AB = \(\frac{5}{2}\) = 2.5 cm,
CN = \(\frac{1}{2}\)CD = \(\frac{11}{2}\) = 5.5 cm and MN = 6 cm.
Let ON = x cm s
∴ OM = MN – ON = (6 – x) cm
Suppose the radius of the circle is r cm.
∴ OB = OC = r cm
In ∆ OMB, ∠M = 90°
∴OB² = OM² + MB²
∴ r² = (6 – x)² + (2.5)²
∴ r² = 36 – 12x + x² + 6.25 ………….. (1)
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + CN²
∴ r² = (x)² + (5.5)²
∴ r² = x² + 30.25 ………………. (2)
From (1) and (2),
36 – 12x + x² + 6.25 = x² + 30.25
∴ – 12x = 30.25 – 6.25 – 36
∴- 12x = – 12
∴x = 1
Now, r² = x² + 30.25
∴ r² = (1)2 + 30.25
∴ r² = 31.25
∴ r = √31.25 (Approximately 5.6)
Thus, the radius of the circle is √31.25 (approximately 5.6) cm.
Note: If the calculations are carried out in simple fractions, then MB = \(\frac{5}{2}\) cm, CN = \(\frac{11}{2}\) cm and radius is \(\frac{5 \sqrt{5}}{2}\) (approximately 5.6) cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chords is at distance 4 cm from the cehtre, what is the distance of the other chord from the centre?
Answer:

In a circle with centre O, chord AB is parallel to chord CD, AB = 8 cm and CD = 6 cm.
Draw OM ⊥ AB, ON ⊥ CD, radius OB and radius OC.
Then, MB = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 8 = 4 cm,
NC = \(\frac{1}{2}\)CD = \(\frac{1}{2}\) × 6 = 3cm and ON = 4cm.
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + NC² = 4² + 3² = 16 + 9 = 25
∴ OC = 5 cm
∴ OB = 5 cm (OB = OC = Radius)
In ∆ OMB, ∠M = 90°
∴ OB² = OM² + MB²
∴ 5² = OM² + 4²
∴ 25 = OM² + 16
∴ OM² = 9
∴ OM = 3 cm
Thus, the distance of the other chord from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:

In ∆ ABE, ∠AEC is an exterior angle.
∴ ∠AEC = ∠ABE + ∠BAE
∴ ∠ABE = ∠AEC – ∠BAE
∴ ∠ABC = ∠AEC – ∠DAE ……………. (1)
Now, ∠AEC = \(\frac{1}{2}\) ∠AOC (Theorem 10.8)
and ∠ DAE = \(\frac{1}{2}\) ∠DOE (Theorem 10.8)
Substituting above values in (1),
∠ABC = \(\frac{1}{2}\) ∠AOC – \(\frac{1}{2}\)∠DOE
∴ ∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)
Here, ∠AOC is the angle subtended by chord AC at the centre and ∠DOE is the angle subtended by chord DE at the centre.
Thus, ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Note: There is no need for chords AD and CE to be equal.

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:

ABCD is a rhombus and its diagonals intersect at M.
∴ ∠BMC is a right angle.
A circle is drawn with diameter BC.
There are three possibilities for point M:
(1) M lies in the interior of the circle,
(2) M lies in the exterior of the circle.
(3) M lies on the circle.
According to (1), if M lies in the interior of the circle, then BM produced will intersect the circle at E. Then, ∠BEC is an angle in a semicircle and hence a right angle, i.e.,
∠MEC = 90°.
In ∆ MEC, ∠ BMC is an exterior angle.
∴ ∠ BMC > ∠ MEC, i.e., ∠ BMC > 90°. In this situation, ∠ BMC is an obtuse angle which contradicts that ∠ BMC = 90°.
Similarly, according to (2), if M lies in the exterior of the circle, then ∠BMC is an acute angle which contradicts that ∠BMC 90°. Thus, possibilities (1) and (2) cannot be true.
Hence, only possibility (3) is true, i.e., M lies on the circle.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary at) E. Prove that AE = AD.
Answer:

Here, the circle through A, B and C intersects CD at E.
∴ Quadrilateral ABCE is cyclic.
ABCD is a parallelogram.
∴ ∠ABC = ∠ADC
∴ ∠ABC = ∠ADE
In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180°
∴ ∠ADE + ∠AEC = 180° ……………… (1)
Moreover, ∠AEC and ∠AED form a linear pair.
∴ ∠AED + ∠AEC = 180° ………………. (2)
From (1) and (2),
∠ADE + ∠AEC = ∠AED + ∠AEC
∴ ∠ ADE = ∠ AED
Thus, in ∆ AED, ∠ADE = ∠AED.
∴ AE = AD (Sides opposite to equal angles)
Note: If the circle intersect CD produced, l then also the result can be proved in similar way.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer:

Chords AC and BD of a circle bisect each other at point O.
Hence, the diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD Is a parallelogram.
∴ ∠BAC = ∠ACD (Alternate angles formed by transversal AC of AB || CD)
Moreover, ∠ACD = ∠ABD (Angles in same segment)
∴ ∠BAC = ∠ABD
∴ ∠BAO = ∠ABO
∴ In A OAB, OA = OB.
But, OA = OC and OB = OD
∴ OA = OB = OC = OD
∴ OA + OC = OB + OD
∴ AC = BD
Thus, the diagonals of parallelogram ABCD are equal.
∴ ABCD is a rectangle.
∴ ∠ABC = 90°
Hence, ∠ABC is an angle in a semicircle and AC is a diameter.
Similarly, ∠BAD = 90°.
Hence, ∠BAD is an angle in a semicircle and BD is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\)A, 90° – \(\frac{1}{2}\)B and 90° – \(\frac{1}{2}\)C.
Answer:

The bisectors of ∠A, ∠B and ∠ C of ∆ ABC intersect the circumcircle of ∆ ABC at D, E and F respectively. .
∠FDE = ∠FDA + ∠EDA (Adjacent angles)
= ∠ FCA + ∠ EBA (Angles in same segment)
= \(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠B (Bisector of angles in ∆ ABC)
= \(\frac{1}{2}\)(∠ B + ∠ C)
= \(\frac{1}{2}\)(180° – ∠A) [∠A + ∠B + ∠C = 180°)
= 90° – \(\frac{1}{2}\) ∠A
Thus, ∠FDE = 90° – \(\frac{1}{2}\) ∠A.
Similarly, ∠ DEF = 90° – \(\frac{1}{2}\) ∠B and
∠ EFD = 90° – \(\frac{1}{2}\) ∠C.

Question 9.
Two congruent circles intersect each Other at points A and B. Through A any line segment PAQ is drawn so that P 9 lie-on. , the two circles. Prove that BP = BQ.
Answer:

Two congruent circles with centres X and Y intersect at A and B.
Hence, AB is their common chord.
In congruent circles, equal chords subtend equal angles at the centres.
∴ ∠AXB = ∠AYB
In the circle with centre X, ∠AXB = 2∠APB and in the circle with centre Y, ∠AYB = 2∠AQB.
∴ 2∠ APB = 2∠ AQB
∴ ∠APB = ∠AQB
∴ ∠QPB = ∠PQB
Thus, in ∆ BPQ, ∠QPB = ∠PQB
∴ QB = PB (Sides opposite to equal angles)
Hence, BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:

In ∆ ABC, the bisector of ∠A intersects the circumcircle of ∆ ABC at D.
∴∠BAD = ∠CAD
Aso, ∠BAD = ∠BCD and ∠CAD = ∠CBD (Angles in same segment)
∴ ∠BCD = ∠CBD
Thus, in ∆ BCD, ∠BCD = ∠CBD
∴BD = CD (Sides opposite to equal angles)
Thus, point D is equidistant from B and C.
Hence, D is a point on the perpendicular bisector of BC.
Thus, the bisector of ∠ A and the perpendicular bisector of side BC intersect at D and D is a point on the circumcircle of ∆ ABC.
Thus, in ∆ ABC, if the angle bisector of ∠A and the perpendicular bisector of side BC intersect, they intersect on the circumcircle of ∆ ABC.
Note: In ∆ ABC, if AB = AC, then the bisector of ∠A and the perpendicular bisector of side BC will coincide , and would not intersect in a single point.

Punjab State Board PSEB 9th Class English Book Solutions English E-mail Message Writing Exercise Questions and Answers, Notes.

PSEB 9th Class English E-mail Message Writing

Inviting friend to Watch a Play

Suppose you are Surjit Singh. Write an e-mail to your friend, Vipin Goyal, inviting him to watch a play.

Hi Vipin.
I am going to Government College for Women, Amritsar, to watch a play on 6 July, 20 – -.
Would you like to come? Let me know by Tuesday so that I can buy your ticket too.
Love
Surjit Singh.

House for Rent

Suppose you are Ramneek. Write an e-mail to your cousin, Darshan Pal, to put up a notice on his college notice-board to rent out your house.

Dear Pal
My father wants to rent our the Second floor of our house. There are two rooms, a kitchen and two attached bathrooms. He should like to have ₹ 2000 as rent. He will rake two months’ rent in advance. He warns to rent out the house to students. Please put up a notice on your college notice-board.
Regards
Ramneek.

Congratulation on Engagement

Suppose you are Shvinder Gill. Write an e-mail to your friend, Alok Wasn, congratulating him on his engagement.

Hi Alok
I have learnt that you are engaged. Congratulating! who is the lucky girl? who does she live and what does she do? Let me know when are you getting married? Is the dare fixed.
Love
Shivinder Gill.

Trip to the South

Suppose you are Varsha Gill. Write an e-mail to your friend, Asha Lakhpal, describing your visit to the south.

Hello Asha
Sorry, I couldn’t write to you earlier. I visited the south with my friend last month. We spent eight days there. We liked the Meenakshi Temple ar Madurai very much. The sunset at Kanyakumari was fascinating. We also went to Aurbindo Ashratn at Puducherry. It was very powerful there.
Love
Varsha.

E-mail (electronic mail) is the medium of communication that sends and receives messages through a specially designed computer network. With the revolution in information technology along with the rapid growth of the Internet, e-mail has become the most popular medium of communication. More and more people are using e-mail to send their messages. Due to its high speed, efficiency and low cost, e-mail has become one of the most important channels of communication. As e-mails are faster than letters, they are used for a quick transmission of all sorts of information.

Specimen of an Informal E-mail

Hi Paul
Sorry to say I’ll he a bit late for tonight’s rehearsal as something’s come up at home and I won’t be able to get away on time. I hope to make it by 7.15.
D. Paul.

Specimen of an Formal E-mail

Dear Ms. Maya
The books you ordered last week are now in stock and awaiting collection. I attach a list of the course books currently in stock at the bookshop.
Julie
Assistant Manage!

HFI Bookshop
Tel : 01123318301
Fax : 01123317931

Punjab State Board PSEB 9th Class English Book Solutions English Reading Comprehension  Picture/Poster Based Exercise Questions and Answers, Notes.

PSEB 9th Class English Reading Comprehension Picture/Poster Based

Answers have been given at the end of this set.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
Who did the fox invite to dinner ?
(a) The duck.
(b) The crane.
(c) The vixen.
(d) The deer.
Answer:
(b) The crane.

Question 2.
What did he serve his guest in the dinner ?
(a) Fruits.
(b) Meat.
(c) Soup.
(d) Eggs.
Answer:
(c) Soup.

Question 3.
The fox was very cunning. He placed a …….. before his guest.
(a) deep bowl
(b) flat dish
(c) narrow jar
(d) sound pitcher.
Answer:
(b) flat dish

Question 4.
What did the crane serve the fox when he invited the fox to dinner ?
(a) Cake.
(b) Milk.
(c) Rice.
(d) Chicken Curry
Answer:
(c) Rice.

Question 5.
The fox had to go hungry. Why ?
(a) Because the crane served the rice in a narrrow jar.
(b) Because the fox could not put his mouth in the narrow jar.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question.

Question 1.
An elephant and a ………. were very good friends.
(a) barber
(b) carpenter
(c) tailor
(d) cobbler
Answer:
(c) tailor

Question 2.
While going to the pond for water, the elephant would daily ………….
(a) stop at the tailor’s shop
(b) have a banana from the tailor
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(c) both (a) and (b)

Question 3.
One day when the elephant put his trunk into the shop,
(a) the tailor gave him a banana
(b) the tailor’s son gave him a banana
(c) the tailor pricked a needle into it
(d) the tailor’s son pricked a needle into it.
Answer:
(d) the tailor’s son pricked a needle into it.

Question 4.
The elephant had his revenge by ……….
(a) filling his trunk with muddy water
(b) throwing muddy water in the tailor’s shop
(c) both (a) and (b)
(d) neither (a) nor (b).
Answer:
(b) throwing muddy water in the tailor’s shop

Question 5.
The moral conveyed through these picture is ……….
(a) Might is right.
(b) Tit for tat.
(c) Do good have good.
(d) No pains no gains.
Answer:
(b) Tit for tat.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
The man is this picture is the famous cricketer
(a) Virat Kohli
(b) Irfan Pathan
(c) Sachin Tendulkar
(d) Mahendra Singh Dhoni.
Answer:
(c) Sachin Tendulkar

Question 2.
He is popularly known as ……. of cricket.
(a) Master Bowler
(b) Master Blaster
(c) Master Batsman
(d) Master Crickter.
Answer:
(b) Master Blaster

Question 3.
At the age of sixteen, he made his international debut against
(a) England
(b) Australia
(c) Sri Lanka
(d) Pakistan.
Answer:
(d) Pakistan.

Question 4.
Sachin took retirement from cricket in
(a) 2005
(b) 2018
(c) 2020
(d) 2013.
Answer:
(d) 2013.

Question 5.
Sachin Tendulkar was honoured with many prestigious awards like
(a) Arjuna Award and Rajiv Khel Ratna
(b) Padma Shri and Bharat Ratna
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(c) Both (a) and (b)

Look at this poster and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
For what purpose is this poster designed ?
(a) To promote education for boys.
(b) To promote education for girls.
(c) To provide employment for boys.
(d) To provide employment for girls.
Answer:
(b) To promote education for girls.

Question 2.
Girls and boys have ……….. to education.
(a) no right
(b) equal right
(c) no interest
(d) equal interest.
Answer:
(b) equal right

Question 3.
How can the nation’s progress be accelerated ?
(a) By educating the boys.
(b) By educating the girls
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
We should not …….. the girls their rights.
(a) excuse
(b) give
(c) accept
(d) deny.
Answer:
(d) deny.

Question 5.
This poster teaches us to stop ……….
(a) the evil of dowry
(b) the evil of female foeticide
(c) the evil of bride burning
(d) the evil of gender discrimation.
Answer:
(b) the evil of female foeticide

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Why did the fox jump into the well to drink water ?
(a) Because the water was very low.
(b) Because he wanted to eat the goat.
(c) Because he wanted to have a bath.
(d) None of these three.
Answer:
(a) Because the water was very low.

Question 2.
The fox drank water and ……….
(a) drenched his thirst
(b) quenched his thirst
(c) satisfied his thirst
(d) toasted his thirst.
Answer:
(b) quenched his thirst

Question 3.
How did the fox succeed in be fooling the goat who was passing that way ?
(a) He told the goat that it was very hot outside.
(b) He told the goat that it was very cold inside the well.
(c) He told that the water was very sweet.
(d) All of these three.
Answer:
(d) All of these three.

Question 4.
What happened when the foolished goat jumped into the well ?
(a) The fox at once climbed over her back.
(b) The fox jumped out of the well.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 5.
The moral of this story is
(a) As you sow so shall you reap
(b) Think before you speak.
(c) Look before you leap.
(d) All that glitters is not gold.
Answer:
(c) Look before you leap.

Look at these pictures and answer the questions given below :

Choose the correct option to answer each question :

Question 1.
Once upon a time a shepherd-boy ……… the sheep of the villagers.
(a) looked into
(b) looked after
(c) looked at
(d) looked for.
Answer:
(b) looked after

Question 2.
What mischief did he make one day ?
(a) He climbed up a tree.
(b) He started crying, “Wolf! Wolf!’’
(c) He shouted for help.
(d) All of these three.
Answer
(d) All of these three.

Question 3.
Why did the villagers who came to help the boy become cross with him ?
(a) Because they found no wolf there.
(b) Because the boy told them that he had shouted in fun only.
(c) Both (a) and (b).
(d) Neithor (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What happened when a wolf really came there.
(a) The boy shouted for help but nobody came.
(b) The villagers didn’t believe the boy’s cries as he had be fooled them once.
(c) The wolf sprang upon the boy and. tore him to pieces.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is —
(a) No pain no gain.
(b) Never give up hope in the hour of difficulty.
(c) Once a liar, always a liar
(d) Think before you speak.
Answer:
(c) Once a liar, always a liar

Look at these pictures and answer the questions given below :

Choose die correct option to answer each question:

Question 1.
An old farmer had three sons who were ………..
(a) very active
(b) very idle
(c) very naughty
(d) very hardworking.
Answer:
(b) very idle

Question 2.
What did the farmer said to sons before his death?
(a) He asked them to work hard in their uk.
(b) He asked them not to fight with each other after his death.
(c) He told them that there was a big treasure in his field.
(d) He asked them to live together happily.
Answer:
(c) He told them that there was a big treasure in his field.

Question 3.
What happened when the sons dig their field ?
(a) They found there a big treasure.
(b) They found there no treasure.
(c) They found there tools of farming.
(d) None of these three.
Answer:
(b) They found there no treasure.

Question 4.
What did the old man ask them to do?
(a) He asked them to sell that field to him.
(b) He asked them to sow seeds in their field.
(c) He asked them to dig their field more deeply.
(d) Any of these three.
Answer:
(b) He asked them to sow seeds in their field.

Question 5.
What happened when the Sons sow seeds in their field?
(a) There was a good crop that year.
(b) They got a lot of money for it.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 6.
The moral of this story is ……
(a) Do good find good.
(b) As you sow, so shall you reap.
(c) Hard work is the key to success.
(d) Hard work is man’s greatest treasure.
Answer:
(d) Hard work is man’s greatest treasure.

Look at these pictures and answer the questions given below:

Choose the correct option to answer each question :

Question 1.
What was the capseller doing in a forest ?
(a) He was passing through the forest to reach a village.
(b) He lay down under a tree to take some rest.
(c) He went there to sell his caps.
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question 2.
What happened when the capseller fell asleep ?
(a) Some monkeys came there.
(b) The monkeys untied the bundle of caps.
(c) The monkeys took away all the caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 3.
What did the capseller see when he woke up ?
(a) He found all his caps missing.
(b) He found the monkeys wearing his caps.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Answer:
(c) Both (a) and (b).

Question 4.
What did he do to recover his caps ?
(a) He took off his caps and threw it down.
(b) The monkeys imitated him.
(c) The monkey threw down their caps.
(d) All of these three.
Answer:
(d) All of these three.

Question 5.
The moral of the story is ………
(a) Tit for tat.
(b) A stitch in time saves nine.
(c) God helps those who help themselves.
(d) Never give up hope in the hour of difficulty.
Answer:
(d) Never give up hope in the hour of difficulty.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

Question 1.
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AGB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Answer:
∠AOC = ∠AOB + ∠BOC (Adjacent angles)
∴ ∠AQC = 60° + 30°
∴ ∠AOC = 90°
Now, 2 ∠ADC = ∠AOC (Theorem 10.8)
∴ ∠ADC = \(\frac{1}{2}\) ∠AOC
∴ ∠ADC = \(\frac{1}{2}\) × 90°
∴ ∠ADC = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Answer:

In the circle with centre O, chord AB is equal to radius PA.
∴ In ∆ PAB, PA = PB = AB
∆ PAB is an equilateral triangle.
∴ ∠ APB = 60°
Now, 2∠AYB = ∠APB (Theorem 10.8)
∴ ∠AYB = \(\frac{1}{2}\) ∠APB
= \(\frac{1}{2}\) × 60° = 30°
Quadrilateral AXBY is a cyclic quadrilateral.
∴ ∠X + ∠Y = 180° (Theorem 10.11)
∴ ∠X + 30°= 180°
∴ ∠X = 150°
Thus, the angle subtended by the chord at point X on the minor arc is 150° and the angle subtended by the chord at point Y on the major arc is 30°.

Question 3.
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer:
Here, reflex angle ∠POR = 2 × ∠PQR (Theorem 10.8)
∴ Reflex angle ∠POR = 2 × 100° = 200°
Now, ∠POR + Reflex angle ∠POR = 360°
∴ ∠POR + 200° = 360°
∴∠POR = 160°
In ∆ OPR. OP = OR (Radii)
∴ ∠OPR = ∠ORP
In ∆ OPR, ∠OPR + ∠ORP + ∠POR = 180°
∴ ∠OPR + ∠OPR + 160° = 180°
∴ 2∠OPR = 20°
∴ ∠OPR = 10°

Question 4.
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Answer:
In ∆ ABC, ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° + 31° + ∠BAC = 180°
∴ 100° + ∠BAC = 180°
∴ ∠BAC = 80°
Now, ∠BDC = ∠BAC (Theorem 10.9)
∴ ∠BDC = 80°

Question 5.
In the given figure, A, B, C and D are four s points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Answer:
In ∆ CDE, ∠BEC is an exterior angle.
∴ ∠BEC = ∠ECD + ∠EDC
∴ 130° = 20° + ∠BDC
∴ ∠BDC = 110°
Now, ∠BAC = ∠BDC (Theorem 10.9)
∴ ∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral Whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Answer:

∠DAC = ∠DBC (Theorem 10.9)
∴ ∠DAC = 70°
∠BAD = ∠BAC + ∠DAC (Adjacent angles)
∴ ∠BAD = 30° + 70°
∴ ∠BAD = 100°
In cyclic quadrilateral ABCD,
∠ BAD + ∠BCD = 180° (Theorem 10.11)
∴ 100° + ∠ BCD = 180°
∴ ∠BCD = 80°
In ∆ ABC, if AB = BC, then ∠ BAC = ∠ BCA
∴ 30° = ∠BCA
∴ ∠BCA = 30°
∠BCD = ∠BCA + ∠ACD (Adjacent angles)
∴ 80° = 30° + ∠ACD
∴ ∠ACD = 50°
∴ ∠ECD = 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Answer:

The vertices of cyclic quadrilateral ABCD lie on a circle with centre O and AC and BD are diameters of the circle.
As AC is a diameter, ∠ABC = ∠ADC = 90° (Angle in a semicircle)
As BD is a diameter, ∠BCD = ∠BAD = 90° (Angle in a semicircle)
Thus, all the four angles, ∠BAD, ∠ABC, ∠BCD and ∠ADC of quadrilateral ABCD are right angles.
Hence, quadrilateral ABCD is a rectangle.

Question 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer:

In trapezium ABCD, AB || CD and AD = BC.
Draw AM ⊥ CD and BN ⊥ CD, where M and N are points on CD.
In ∆ AMD and ∆ BNC,
∠AMD = ∠BNC (Right angles)
Hypotenuse AD = Hypotenuse BC (Given)
AM = BN (Distance between parallel lines)
∴ By RHS rule, ∆ AMD ≅ ∆ BNC
∴ ∠ADM = ∠BCN
∴ ∠ADC = ∠BCD
Now, AB || CD and AD is their transversal.
∴ ∠BAD + ∠ADC = 180° (Interior angles on the same side of transversal)
∴ ∠ BAD + ∠BCD = 180°
Thus, in quadrilateral ABCD, ∠A + ∠C = 180°.
Hence, ABCD is a cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Answer:

∠ACP and ∠ABP are angles in the same segment.
∴ ∠ACP = ∠ABP (Theorem 10.9) …………… (1)
∠QCD and ∠QBD are angles in the same segment.
∴ ∠QCD = ∠QBD (Theorem 10.9) …………….. (2)
Now, ∠ABP and ∠QBD are vertically opposite angles.
∴ ∠ABP = ∠QBD ………………… (3)
From (1), (2) and (3),
∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer:
Circles are drawn taking sides AB and AC of ∆ ABC as diameters. These circles intersect each other at points A and P.
Draw common chord AP.
Since AB is a diameter, ∠APB is an angle in a semicircle.
∴ ∠APB = 90°
Since, AC is a diameter, ∠APC is an angle in a semicircle.
∴ ∠APC = 90°
Then, ∠APB + ∠APC = 90° + 90° = 180°
∠APB and ∠APC are adjacent angles with common arm AP and their sum is 180°.
∴ ∠APB and ∠APC form a linear pair.
Hence, the point of intersection of the circles with two sides of a triangle as diameters lies on the third side of the triangle.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC., Prove that ∠CAD = ∠CBD.
Answer:

In figure (1), line segment AC subtends equal angles at two points B and D lying on the same side of AC. Hence, by theorem 10.10, all the four points lie on the same circle.
Now, ∠CAD and ∠CBD are angles in the same segment.
∴ ∠CAD = ∠CBD (Theorem 10.9)
In figure (2), in quadrilateral ABCD,
∠B = ∠D = 90°.
∴ ∠B + ∠D = 180°
Hence, ABCD is a cyclic quadrilateral.
Again, ∠CAD and ∠CBD are angles in the same segment.
∴ ∠CAD = ∠CBD (Theorem 10.9)

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Answer:
Suppose ABCD is a cyclic parallelogram.
ABCD is a cyclic quadrilateral! .
∴ ∠A + ∠C = 180°
and ∠ B + ∠ D = 180° …….. (1)
ABCD is a parallelogram.
∴ ∠A = ∠C and ∠B = ∠D ……….. (2)
From (1) and (2),
∠A = ∠B = ∠C = ∠D = 90°
Thus, all the angles of quadrilateral ABCD are right angles.
Hence, ABCD is a rectangle.
Thus, a cyclic parallelogram is a rectangle.

Punjab State Board PSEB 9th Class English Book Solutions English Note-Making & Messages Exercise Questions and Answers, Notes.

PSEB 9th Class English Note-Making & Messages

Note-making

का अर्थ है किसी पैरे की मुख्य बातों को संक्षिप्त और साफ-सुथरे ढंग से प्रस्तुत करना। अच्छे Notes में निम्नलिखित विशेषताएं होती है –
1. वे संक्षिप्त होते हैं।

2. केवल प्रासंगिक बातें ही उनमें दी जाती हैं।

3. केवल शब्दों या वाक्यांशों का प्रयोग ही किया जाता है। पूरे वाक्यों की आमतौर पर आवश्यकता नहीं होती। अन्य शब्दों में हम कह सकते हैं कि Notes बनाते समय प्रयुक्त भाषा व्याकरण की दृष्टि से पूरी तरह सही नहीं भी हो सकती।

4. सूचना को सूचीबद्ध ढंग से प्रस्तुत किया जाता है। इसे विभाजित व उपविभाजित किया जाता है। विभाजन निम्न प्रकार से हो सकता है –
मुख्य खण्ड : 1, 2, 3, इत्यादि।
उपखण्ड : a, b, c, इत्यादि।

Passage 1:

If the young students in schools and colleges do not learn discipline, they will never be able to extract’ obedience from others in society. In fact, society will never accept them as persons fit for any responsible position in life. A school or college without discipline can never impart? suitable education to students.

Such a school or college is no better than a factory that turns out imperfect’ men and women. Sense of discipline plays a very important part in the playground and the battlefield. A disciplined team is likely to win the match in spite of its weakness but a very good team may not fare well for want of discipline. The rule of discipline equally applies to soldiers in the battlefield.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Need for discipline in schools and colleges for good education.
2. Indisciplined students fail’ to win any respect or position later on in their life.
3. Importance of discipline, for players on the playground
4. Importance of discipline, for soldiers in the battlefield.

Passage 2

Early rising leads to health and happiness. The man who rises late, can have little rest in the course of the day. Anyone who lies in bed late is compelled to work till a late hour in the evening. He has to go without the morning exercise which is so necessary for his health. In spite of all efforts’, his work will not produce as good results as that of the early riser. The reason for this is that he cannot take advantage of the refreshing hours in the morning.

Some people say that the quiet hour of midnight is the best time for working. Several great thinkers say that they can write best only when they burn the midnight oil. Yet it is true to say that few men have a clear brain at midnight when the body needs rest and sleep. Those who work at that time soon ruin their health. Bad health must, in the long run, have a bad effect on the quality of their work.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Advantages of early rising :
(i) health
(ii) Disadvantages

2. Disadvantages of late rising
(i) work till late in the evening
(ii) go without morning exercise
(iii) work not done properly.

(3) (i) Burning midnight oil bad for health.
(ii) Bad health, poor quality of our work.

Passage 3

Games, though essential, should not become the be-all and end-all of student life. Generally, the sportsmen waste too much time on them, and fail in their examinations. One must never devote more than an hour to sports and after that, one should not even think about them. Again, if a player plays a game rashly’, there is every danger of breaking bones.

If it is played without the spirit of sportsmanship, it can lead to bad blood and quarrels. In some of the colleges, there is a tradition that if the visiting team is winning a match, the home team plays foul, picks a quarrel and breaks the bones of the visitors. But in spite of these minor defects, sports are very useful in keeping the students busy and in developing their personalities.

India expects its citizens to have the qualities of true sportsmen. If we all acquire these qualities, there will be no narrow-mindedness, no corruption, and no injustice. There will be independence in the real sense of the word.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.

1. Sports essential for students, but not the be-all and end-all.
2. Wasting too much time → failure in examinations.
3. Playing rashly → breaking of bones.
4. Lack of sportsmanship → quarrels between teams.
5. True sportsmanship can end narrow – mindedness, corruption and injustice.

Passage 4

Of all amusements which can possibly be imagined for a hard-working man after his daily toil, there is nothing like reading an entertaining book. It calls for no bodily exertion of which he has had enough. It relieves his home of its dullness. It transports him to a livelier and more interesting scene, and while he enjoys himself there, he may forget the evils of the present moment.

It accompanies him to his next day’s work and if the book he has been reading be anything above the very idlest and the dullest, it gives him something to think about besides the drudgery of his everyday occupation. If I were to pray for a taste which should stand me in good stead under every variety of circumstances and be. a source of happiness and cheerfulness through life, it would be a taste for reading.

Give a man this taste,.and the means of gratifying it, and you can hardly fail to make him happy unless indeed you put into his hand a most perverse selection of books.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Reading of an interesting book a good diversion after the day’s hard work.
2. (i) Removes dullness of home,
(ii) Transports one into a livelier world.
3. (i) A good book food for thought.
(ii) Stands in good srcad under every circumstance.
4. Taste for reading, a source of great happiness.

Passage -5

English is important not because a number of people know it in India, although it is a factor to be remembered. It is not important because it is the language of Milton and Shakespeare, although that has to be considered. English is important because it is the major window for us on the modern world. And we dare not close that window. If we close it, we imperil our future. We think of Industrialisation, scientific development, research and technology.

But every door of modern knowledge will be closed if we do not have one or more foreign languages. We need not have English : we can have Russian, French or German, if you like, but obviously it is infinitely simpler for us to deal with a language which we know than to shift over to Russian, French or German which will be a tremendous job.
Certainly, we want to learn foreign languages, because we deal with the people of those languages in business, trade and science. So in the present stage of our development, we cannot go ahead without English and other foreign languages.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
1. Mistakes do little harm

• when admitted.
• set tight before they a do any carnage

2. Delay in admitting mistakes

• harmful for the task in hand.
• harmful for the reputation.

3. Person who admits his mistakes

• is liked by everybody
• wins the confidence and respect of others.

4. Person who hides his mistakes

• is considered a fool.
• nobody likes him

Passage – 7

Teachers have a great responsibility at this time when our society is undergoing transformation. The future of the teaching profession in India will depend on the decision which the teachers take on vital questions relating to social change. In normal times, when society is comparatively more stable, the teachers’ primary task is transmitting culture. But in a period of transition, like the one through which we are passing, they have sometimes to set aside the culture in which they live, make a proper appraisal9 of it, pick out its salient features and reinterpret them for the new generation. The oncoming generations can rise to a high level of wisdom and cultivation only when teachers guide them carefully during this period of change.

Read the given passage carefully and make notes on it in points only, using headings and sub-headings.
Answer:
A. Teachers’ role in normal times :

• transmitting culture.

B. Modern rimes :

• not normal
• a period of transition.

C. teacher roles:

• proper appraisal of old culture
• pick out its salient features
•  re-interpreting them for future generations.

Passage – 8
Each one of us must realize that the only future for India and her people is one of tolerance and co-operation, which have been the basis of our culture for ages past. We have laid down in our constitution that Indians a Secular State. This does not mean irreligion. It means equal respect for all faiths and equal opportunities for those who profess different faiths. We have, therefore, always to keep in mind this vital aspect of our culture which is also of the highest importance in India today. Those who put up barriers between one Indian and another and who promote disruptive tendencies do not serve the cause of India and her culture. They weaken us at home and discredit us abroad.
Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
(A) Tolerance and co-operation

• basis of our past culture
• pillars of our future

(B) Secularism

• equal respect for all faiths
• equal opportunities
• of highest importance in present-day India.

(C) Disruptive tendencies

• serve no cause
• discredit the country

Passage – 9

As a result of a long series of discoveries, mans life has been altered more radically and more rapidly during the last one hundred and fifty years than during the whole of the preceding two thousand years. In what ways does this alteration chiefly show itself ? In the first place, most of the external enemies to which our species in the past had been exposed are either overcome or are in a fair way to being overcome.

Look back over mans life in the past and you cannot but realize what sordid, meagre, frightened affair it must have been. His crops and, therefore, his livelihood have been at the mercy of forces which he could neither understand nor control; forces of fire and flood, of earthquake and drought; his communities were swept by pestilence and famine; and with the sweat of his brow, he wrung meager sustenance from nature. Today, thanks to science, all these enemies to man’s well-being have either disappeared or have been reduced to comparative impotence.

Read the above passage carefully and make notes on it in points only, using headings and sub-headings.
Answer
1. Discoveries of science during the
2. Man’s life completely changed.
3. External enemies overpowered :

• floods and fires
• earthquakes
• famines.
• pestilence

4. Now getting one’s livelihood not so difficult as it used to bo’

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Answer:

The circle with centre O and radius 5 cm intersects the circle with centre P and radius 3 cm at points A and B.
Hence, AB is their common chord.
Then, OP = 4 cm (Given),
OA = 5 cm and PA = 3 cm.
In ∆ OAP, OA² = 5² = 25 and
OP² + AP² = 4² + 3² = 16 + 9 = 25
Thus, in ∆ OAP, OA² = OP² + AP²
∴ ∆ OAP is a right triangle in which ∠OPA is a right angle and OA is the hypotenuse.
Thus, in the circle with centre O, OP is perpendicular from centre O to chord AB.
∴ OP bisects AB.
AB = 2PA = 2 × 3 = 6 cm
Thus, the length of the common chord is 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer:

In the circle with centre O, equal chords AB and CD intersect at E.
Draw OM ⊥ AB and ON ⊥ CD.
∴ AM = BM = \(\frac{1}{2}\)AB and CN = DN = \(\frac{1}{2}\)CD.
But, AB = CD
∴AM = BM = CN = DN …………….. (1)
Chords AB and CD, being equal, are equidistant from the centre.
∴ OM = ON
In ∆ OME and ∆ ONE,
∠OME = ∠ONE (Right angles)
OE = OE (Common)
OM = ON
By RHS rule, ∆ OME ≅ ∆ ONE
∴ME = EN (CPCT) ……………… (2)
From (1) and (2),
AM + ME = CN + NE
∴ AE = CE
Similarly, BM – ME = DN – NE
∴ BE = DE
Thus, if two equal chords of a circle intersect within the circle, the segments of one chord are equal to corresponding segments of the other chord.

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer:
As the data of example 2 and example 3 are same, we use the proof of example 2 up to the required stage and do not repeat it here.
In example 2, we proved that,
∆ OME ≅ ∆ ONE ,
∴ ∠ OEM = ∠ OEN
∴ ∠ OEA = ∠ OEC
Thus, the line joining the point of intersection of two equal chords of a circle to the centre makes equal angles with the chords.

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see the given figure).

Answer:

From centre O, draw perpendicular OM to line AD.
In the outer circle, OM is the perpendicular drawn from centre O to chord AD.
Hence, M is the midpoint of AD.
∴ MA = MD …………… (1)
In the inner circle, OM is the perpendicular drawn from centre O to chord BC.
Hence, M is the midpoint of BC.
∴ MB = MC ………….. (2)
Subtracting (2) from (1),
MA – MB = MD – MC
∴ AB = CD

Question 5.
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Answer:

Here, OR = OM = OS = 5 m (Radius of the circle) and RS = SM = 6 m.
In quadrilateral ORSM, OR = OM = 5 m and RS = SM = 6 m.
∴ Quadrilateral ORSM is a kite.
∴ It diagonal OS bisects the diagonal RM at right angles.
∴ ∠RKO = 90° ………………. (1)
OK is perpendicular from centre O to chord RM.
Hence, K is the midpoint of RM.
∴ RM = 2RK ………………… (2)
From centre O, draw perpendicular OL to chord RS.
∴ RL = \(\frac{1}{2}\)RS = \(\frac{1}{2}\) × 6 = 3 m
In ∆ RLO, ∠ L = 90°
∴ RO² = OL² + RL²
∴ 5² = OL² + 3²
∴ 25 = OL² + 9
∴ OL² = 16
∴ OL = 4 m
Now, area of ∆ ROS = \(\frac{1}{2}\) × RS × OL
= \(\frac{1}{2}\) × OS × RK [by (1)]
∴RS × OL = OS × RK
∴ 6 × 4 = 5 × RK
∴ 24 = 5 × RK
∴ RK = \(\frac{24}{5}\) = 4.8 m
Then, RM = 2RK [by (2)]
∴ RM = 2 × 4.8
∴ RM = 9.6 m
Thus, the distance between Reshma and Mandip is 9.6 m.

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk with each other. Find the length of the string of each phone.
Answer:

Here, the circle with centre O represents the park and the points A, S and D represent the positions of Ankur, Syed and David respectively. Since Ankur, Syed and David are sitting at equal distances from the others, ∆ ASD is an equilateral triangle.

Then, drawing the perpendicular bisector of SD from its midpoint M, it will pass through O as well as A.
Suppose, SM = x m
∴ SD = 2SM = 2xm
Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) (side)²
∴ Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) × (2x)²
∴ Area of equilateral ∆ ASD = √3x² …………. (1)
In ∆ OMS, ∠M = 90°
∴ OM² = OS² – SM² = (20)² – (x)² = 400 – x²
∴ OM = \(\sqrt{400-x^{2}}\)
Now, area of ∆ OSD = \(\frac{1}{2}\) × SD × OM
∴ Area of ∆ OSD = \(\frac{1}{2}\) × 2x × \(\sqrt{400-x^{2}}\)
∴ Area of ∆ OSD = x\(\sqrt{400-x^{2}}\) …………….. (2)
Here, ∆ OAS, ∆ OSD and ∆ ODA are congruent triangles.
Area of ∆ ASD = Area of ∆ OAS + Area of ∆ OSD + Area of ∆ ODA
∴ Area of ∆ ASD = 3 × Area of ∆ OSD
∴ √3 ∙ x<sup2 = 3 × x\(\sqrt{400-x^{2}}\)
∴x = √3 ∙ \(\sqrt{400-x^{2}}\)
∴ x² = 3(400 – x²2)
∴ x²= 1200 – 3x²
∴ 4x² = 1200
∴x² = 300
∴x= 10 √3
SD = 2x = 2 × 10 √3 = 20 √3 m
Thus, the length of the string of each phone is 20 √3m.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Answer:

Thus, given a pair of circles, the maximum number of common points they have is 2.

Question 2.
Suppose you are given a circle. Give a construction to find its centre.
Answer:

• In the given circle, draw chords AB and BC with one endpoint B in common.
• Draw l-the perpendicular bisector of AB and m-the perpendicular bisector of BC.
• Let l and m intersect at O.
• Then, O is the centre of the given circle.

Note: Here, any two chords without an end-point in common can be drawn.

Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Answer:

Here, two circles with centre O and P intersect each other at points A and B.
AB and OP intersect at M.
In ∆ OAP and ∆ OBR
OA = OB (Radii of the circle with centre O)
PA = PB (Radii of the circle with centre P).
OP = OP (Common)
∴ ∆ OAP ≅ ∆ OBP (SSS rule)
∴ ∠ AOP = ∠ BOP (CPCT)
∴ ∠ AOM = ∠BOM
Now, in ∆ AOM and ∆ BOM,
AO = BO (Radii of the circle)
∠ AOM = ∠ BOM
OM = OM (Common)
∴ ∆ AOM = ∆ BOM (SAS rule)
∴ AM = BM and ∠ AMO = ∠ BMO (CPCT)
But, ∠AMO + ∠BMO = 180° (Linear pair)
∴ ∠ AMO = ∠ BMO = \(\frac{180^{\circ}}{2}\) = 90°
Thus, line OM is the perpendicular bisector of AB.
Hence, line OP is the perpendicular bisector of AB.
Thus, the centres O and P of the circle intersecting in points A and B lie on the perpendicular bisector of common chord AB.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.2

Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Answer:

Two circles with centres O and P are congruent. Moreover, chord AB of the circle with centre
O and chord CD of the circle with centre P are congruent.
In ∆ OAB and ∆ PCD,
OA = PC and OB = PD (Radii of congruent circles)
And, AB = CD (Given)
∴ ∆ OAB ≅ ∆ PCD (SSS rule)
∴ ∠AOB = ∠ CPD
Thus, equal chords of congruent circles subtend equal angles at their centres.

Question 2.
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Answer:

Two circles with centres O and P are congruent. Moreover, ∠ AOB subtended by chord AB of the circle with centre O and ∠CPD subtended by chord CD of the circle with centre P at their respective centres are equal.
In ∆ OAB and ∆ PCD,
OA = PC and OB = PD (Radii of congruent circles)
And, ∠AOB = ∠CPD (Given)
∴ ∆ OAB ≅ ∆ PCD (SAS rule)
∴ AB = CD
Thus, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.