. Ifa line intersects two concentric circles (circleswith the same cen

1.	. Ifa line intersects two concentric circles (circleswith the same centre) with centre O atA, B, C andD, prove that AB- CD (see Fig. 10.25).Three girls Reshma, Salma and Mandip areplaying a game by standing on a circle of radius5m drawn in a park. Reshma throws a ball toSalma, Salma to Mandip, Mandip to Reshma. Ifthe distance between Reshma and Salma andetween Salma and Mandip is 6m each, what ise distance between Reshma and Mandip?AFio. 1025
Answer» Let O bethe centre of the circle. A, B and C represent the positions of Reshma, Salmaand Mandip. AB = 6cm and BC = 6cm. Radius OA = 5cm.(given) Draw BM ⊥ AC ABC is an isosceles triangle as AB = BC, M ismid-point of AC. BM isperpendicular bisector of AC and thus it passes through the centre of thecircle. Let AM = y and OM = x then BM = (5-x). In ΔOAM, OA²=OM²+AM² ( by Pythagorastheorem) 5²=x²+y²—(i) In ΔAMB, AB²=BM²+AM²(by Pythagoras theorem) 6²= (5-x)²+y²— (ii) Subtracting (i) from (ii), 36 – 25 = (5-x)² -x² 11 = (25+x²– 2×5×x) - x² 11= 25+x²-10x - x² 11= 25-10x 10x = 14 x= 7/5 Substituting the value of x in (i), we get y²+ 49/25 = 25y²=25 – 49/25y²=(625 – 49)/25y²=576/25 y = 24/5 Thus, AC = 2×AM AC = 2×y AC= 2×(24/5) m = 48/5 m = 9.6 m Hence, the Distance between Reshma and Mandipis 9.6 m.

. Ifa line intersects two concentric circles (circleswith the same centre) with centre O atA, B, C andD, prove that AB- CD (see Fig. 10.25).Three girls Reshma, Salma and Mandip areplaying a game by standing on a circle of radius5m drawn in a park. Reshma throws a ball toSalma, Salma to Mandip, Mandip to Reshma. Ifthe distance between Reshma and Salma andetween Salma and Mandip is 6m each, what ise distance between Reshma and Mandip?AFio. 1025

Answer»

Let O bethe centre of the circle. A, B and C represent the positions of Reshma, Salmaand Mandip.

AB = 6cm and BC = 6cm.

Radius OA = 5cm.(given)

Draw BM ⊥ AC

ABC is an isosceles triangle as AB = BC, M ismid-point of AC.

BM isperpendicular bisector of AC and thus it passes through the centre of thecircle.

Let AM = y and OM = x then BM = (5-x).

In ΔOAM,

OA²=OM²+AM² ( by Pythagorastheorem)

5²=x²+y²—(i)

In ΔAMB,

AB²=BM²+AM²(by Pythagoras theorem)

6²= (5-x)²+y²— (ii)

Subtracting (i) from (ii),

36 – 25 = (5-x)² -x²

11 = (25+x²– 2×5×x) - x²

11= 25+x²-10x - x²

11= 25-10x

10x = 14

x= 7/5

Substituting the value of x in (i), we get

y²+ 49/25 = 25y²=25 – 49/25y²=(625 – 49)/25y²=576/25

y = 24/5

Thus,

AC = 2×AM

AC = 2×y

AC= 2×(24/5) m

= 48/5 m = 9.6 m

Hence, the Distance between Reshma and Mandipis 9.6 m.