Gujarat Board GSEB Solutions Class 9 Maths Chapter 10 Circles Ex 10.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions 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.

Solution:

Let two circles with centres O and Oâ€™ intersect each other at P and Q.

Therefore, OP = OQ = 5cm, Oâ€™P = Oâ€™Q = 3cm and OOâ€™ = 4 cm

In Î”OOP, OO’^{2} + OP^{2} = 4^{2} + 3^{2}

= 16 + 9 = 25

= 52 = OP^{2}

âˆ´ AOOâ€™P = 90Â° (Using converse of Pythagoras Theorem)

But OO’ âŠ¥ PQ

As if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.

âˆ´ OOâ€™ coincides 0M i.e., PQ passes through the centre Oâ€™.

âˆ´ Length of the common chord

PQ = 2Oâ€™P = 2 x 3 cm = 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.

Solution:

Given: A circle with centre O. Its two equal

chords AB and CD intersect at E.

To Prove: AE = DE and CE = BE.

Construction: Draw OM âŠ¥Â AB and ON âŠ¥ CD. Join OE.

Proof: In Î”OME and Î”ONE,

OM = ON

[âˆ´ Equal chords of a circle are equidistant from the centre.]

OE = OE [Common]

âˆ´ Î”OME = Î”ONE [RHSI

âˆ´ ME = NE [CPCT]

â‡’ AM + ME = DN + NE

|âˆ´ AM = DN = \(\frac {1}{2}\) AB = \(\frac {1}{2}\) CD|

AE = DE

Now, AB – AE = CD – DE [Given AB = CD]

â‡’ BE = CE

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.

Solution:

Given: Two equal chords AB and CD of a circle with centre O intersect within the circle. Their point of intersection is E.

To Prove: âˆ OEA = âˆ OED.

Construction: Join OA and OD

Proof: In Î”OEA and Î”OED,

OE = OE [Common]

OA = OD [Radii of a circle]

AE = DE [Proved in 2 above]

âˆ´ Î”OEA. = Î”OED [SSS Rule]

âˆ´ âˆ OEA = âˆ OED [CPCT]

Note: Also it can be drawn OPâŠ¥ AB, OQ âŠ¥ CD and shown Î”OPE = Î”OQE to prove âˆ OEP = âˆ OEQ.

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 figure].

Solution:

Construction: Draw OM âŠ¥ BC.

Proof: The perpendicular drawn from the centre of a circle to a chord bisects the chord.

âˆ´ AM = DM ……….(1)

and BM = CM ………..(2)

Subtracting (2) from (1), we get

AM – BM = DM – CM

âˆ´ AB = CD.

Question 5.

Three girls Reshma, Salina 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 Salina and between Salina and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Solution:

Let KR = xm

ar(Î”ORS) = ar(Î”ORK) + ar(Î”SRK)

= \(\frac {(0K) (KR)}{2}\) + \(\frac {(KS) (OS)}{2}\)(KS) (KR)

= \(\frac {(x) (5)}{2}\) ………..(1)

Again, ar(Î”ORS)

From equations (1) and (2),

\(\frac {(x) (5)}{2}\) = 12

â‡’ x = \(\frac {12 x 2}{5}\) = \(\frac {24}{5}\) = 4.8 m

â‡’ KR = 4.8m

RM = 2 KR = 2 x (4.8) = 9.6km

âˆ´ Hence, the distance between Reshma and Mandip is 9.6m.

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 each other. Find the length of the string of each phone.

Solution:

Let BD = xm

Then in right triangle ODB,

OB^{2} = OD^{2} + BD^{2} [By Pythagoras Theorem]

Take AB = BC = AC

â‡’ (20)^{2} = OD^{2} + x^{2}

â‡’ OD^{2} = 400 – x^{2}

â‡’ OD = \(\sqrt{400-x^{2}}\)

Area of equilateral Î”ABC

ABC = \(\frac{\sqrt{3}}{4}\) (side)^{2} = \(\frac{\sqrt{3}}{4}\)BC^{2}

= \(\frac{\sqrt{3}}{4}\) (2BD)^{2} = \({\sqrt{3}\)BD^{2} = \({\sqrt{3}\)x^{2} ………..(1)

Again, area of equilateral triangle ABC

= Area of Î”OBC + Area of Î”OCA + Area of Î”OAB

= 3 Area of Î”OBC = 3 \(\frac{(BC) (OD)}{2}\)

= \(\frac{3(2BD) (OD)}{2}\) = 3 (BD) IAD)

= 3 x \(\sqrt{400-x^{2}}\) ……….(2)

From equations (1) and (2),

3 x \(\sqrt{400-x^{2}}\) = \(\sqrt{3}\) x^{2} ………….(2)

â‡’ \(\sqrt{3}\)\(\sqrt{400-x^{2}}\) = x

Squaring both sides,

3(400 – x^{2}) = x^{2}

â‡’ 1200 – 3x^{2} = x^{2}

â‡’ 4 x ^{2} = 1200 â‡’ x^{2} = 300

â‡’ x = 10\(\sqrt{3}\) â‡’ BD = 10\(\sqrt{3}\)

2BD = 20\(\sqrt{3}\) â‡’ BC = 20 \(\sqrt{3}\)

Hence, the length of string of each phone is 20\(\sqrt{3}\) m.

Alternative method: In an equilateral triangle length of altitude = \(\frac{\sqrt{3}}{4}\) (side)

AD = \(\frac{\sqrt{3}}{4}\) BC ………..(1)

Also, centroid of a triangle divides its median in the ratio 2: 1

âˆ´ AO : OD = 2 : 1 or AO : AD = 2 : 3

AD = \(\frac{3}{4}\) AO ………..(2)

From (1) and (2), \(\frac{\sqrt{3}}{2}\)BC = \(\frac{3}{4}\) AO

â‡’ BC = \(\sqrt{3}\) AO

= 20 \(\sqrt{3}\) m