Question

class 9 chapter number 10 exercise 10.4 question number 5 NCERT solution ​

Answer 1

Answer:

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

AB = 6cm and BC = 6cm.

Radius OA = 5cm.  (given)

Draw BM ⊥ AC

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

 BM is perpendicular bisector of AC and thus it passes through the centre of the circle.

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

 

 In ΔOAM,

OA²=OM²+AM² ( by Pythagoras theorem)

 

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 = 25

y² = 25 – 49/25

y² = (625 – 49)/25

y²= 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 Mandip is 9.6 m.

Answer 2

Answer:

Let the positions of Reshma, Salma and Mandip be represented as A, B and C respectively.

From the question, we know that AB = BC = 6cm.

So, the radius of the circle i.e. OA = 5cm

Now, draw a perpendicular BM ⊥ AC.

Since AB = BC, ABC can be considered as an isosceles triangle. M is mid-point of AC. BM is the perpendicular bisector of AC and thus it passes through the centre of the circle.

Now,

let AM = y and

OM = x

So, BM will be = (5-x).

By applying Pythagorean theorem in ΔOAM we get,

OA2 = OM2 +AM2

⇒ 52 = x2 +y2 — (i)

Again, by applying Pythagorean theorem in ΔAMB,

AB2 = BM2 +AM2

⇒ 62 = (5-x)2+y2 — (ii)

Subtracting equation (i) from equation (ii), we get

36-25 = (5-x)2 +y2 -x2-y2

Now, solving this equation we get the value of x as

x = 7/5

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

y2 +(49/25) = 25

⇒ y2 = 25 – (49/25)

Solving it we get the value of y as

y = 24/5

Thus,

AC = 2×AM

= 2×y

= 2×(24/5) m

AC = 9.6 m

So, the distance between Reshma and Mandip is 9.6 m.