Question

class 9 ncert chapter 10 circles ex 10.4 questions 6​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Length of each string is 20√3 metres

Step-by-step explanation:

Let ankur,syed and David are seating on points A,B and C;as shown in the figure attatched.

Radius of the park is 20m.

Let AB=BC=AC=a (given in question,they are seating at equal distance)

Thus triangle ABC is an equilateral triangle.

Area of ∆ABC

$= \frac{ \sqrt{3} }{4} {a}^{2}$

Area of ∆ABC= Area of ∆AOB+Area of ∆AOC+Area of ∆BOC

(equal chords are equidistant from center of circle)

Draw perpendicular on AB,BC and AC from centre,let the length is h metre

$\frac{ \sqrt{3} }{4} {a}^{2} = \frac{1}{2} ah + \frac{1}{2} ah + \frac{1}{2} ah \\ \\ \frac{ \sqrt{3} }{4} {a}^{2} = \frac{3}{2} ah \\ \\ h = \frac{ \sqrt{3} }{4} \times \frac{2}{3}a \\ \\ \boxed{ h = \frac{1}{2 \sqrt{3} } a}$

We know that chord is bisected by perpendicular drawn from centre of circle.

In Right triangle BDO

${BO}^{2} = {BD}^{2} + {DO}^{2} \\ \\ ( {20)}^{2} = { (\frac{a}{2} )}^{2} + ( { \frac{a}{2 \sqrt{3} } })^{2} \\ \\ 400 = \frac{ {a}^{2} }{4} + \frac{ {a}^{2} }{12} \\ \\ 400 = \frac{3 {a}^{2} + {a}^{2} }{12} \\ \\ 400 = \frac{4 {a}^{2} }{12} \\ \\ {a}^{2} = 1200 \\ \\ a = \sqrt{1200} \\ \\ a = \sqrt{12 \times 100} \\ \\ a = 10 \sqrt{12} \\ \\ = 10 \sqrt{2 \times 2 \times 3} \\ \\ a= 20 \sqrt{3}$

Length of each string is 20√3 metres.

Hope it helps you.