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 O’ intersect each other at P and Q.

Therefore, OP = OQ = 5cm, O’P = O’Q = 3cm and OO’ = 4 cm

In ΔOOP, OO’^{2} + OP^{2} = 4^{2} + 3^{2}

= 16 + 9 = 25

= 52 = OP^{2}

∴ AOO’P = 90° (Using converse of Pythagoras Theorem)

But OO’ ⊥ 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.

∴ OO’ coincides 0M i.e., PQ passes through the centre O’.

∴ Length of the common chord

PQ = 2O’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 ⊥ AB and ON ⊥ CD. Join OE.

Proof: In ΔOME and ΔONE,

OM = ON

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

OE = OE [Common]

∴ ΔOME = ΔONE [RHSI

∴ ME = NE [CPCT]

⇒ AM + ME = DN + NE

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

AE = DE

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

⇒ 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: ∠OEA = ∠OED.

Construction: Join OA and OD

Proof: In ΔOEA and ΔOED,

OE = OE [Common]

OA = OD [Radii of a circle]

AE = DE [Proved in 2 above]

∴ ΔOEA. = ΔOED [SSS Rule]

∴ ∠OEA = ∠OED [CPCT]

Note: Also it can be drawn OP⊥ AB, OQ ⊥ CD and shown ΔOPE = ΔOQE to prove ∠OEP = ∠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 ⊥ BC.

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

∴ AM = DM ……….(1)

and BM = CM ………..(2)

Subtracting (2) from (1), we get

AM – BM = DM – CM

∴ 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(ΔORS) = ar(ΔORK) + ar(ΔSRK)

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

= \(\frac {(x) (5)}{2}\) ………..(1)

Again, ar(ΔORS)

From equations (1) and (2),

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

⇒ x = \(\frac {12 x 2}{5}\) = \(\frac {24}{5}\) = 4.8 m

⇒ KR = 4.8m

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

∴ 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

⇒ (20)^{2} = OD^{2} + x^{2}

⇒ OD^{2} = 400 – x^{2}

⇒ OD = \(\sqrt{400-x^{2}}\)

Area of equilateral ΔABC

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 ΔOBC + Area of ΔOCA + Area of ΔOAB

= 3 Area of ΔOBC = 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)

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

Squaring both sides,

3(400 – x^{2}) = x^{2}

⇒ 1200 – 3x^{2} = x^{2}

⇒ 4 x ^{2} = 1200 ⇒ x^{2} = 300

⇒ x = 10\(\sqrt{3}\) ⇒ BD = 10\(\sqrt{3}\)

2BD = 20\(\sqrt{3}\) ⇒ 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

∴ 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

⇒ BC = \(\sqrt{3}\) AO

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