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, OA^{2} = 5^{2} = 25 and

OP^{2} + AP^{2} = 4^{2} + 3^{2} = 16 + 9 = 25

Thus, in ∆ OAP, OA^{2} = OP^{2} + AP^{2}

∴ ∆ 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°

∴ RO^{2} = OL^{2} + RL^{2}

∴ 5^{2} = OL^{2} + 3^{2}

∴ 25 = OL^{2} + 9

∴ OL^{2} = 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 = √3x^{2} …………. (1)

In ∆ OMS, ∠M = 90°

∴ OM^{2} = OS^{2} – SM^{2} = (20)^{2} – (x)^{2} = 400 – x^{2}

∴ 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 ∙ \(\sqrt{400-x^{2}}\)

∴ x^{2} = 3(400 – x^{2}2)

∴ x^{2}= 1200 – 3x^{2}

∴ 4x^{2} = 1200

∴x^{2} = 300

∴x= 10 √3

SD = 2x = 2 × 10 √3 = 20 √3 m

Thus, the length of the string of each phone is 20 √3m.