Question

This is Q6 ex 10.4 class 9 ncert maths
Please answer​

Answer 1

Answer:

the answer is 41.7

I hope it will help you

Answer 2

Given:-

• Radius of circular park = 20m
• Let 3 boys be denoted by points A, S and D
• AS = SD = AD

ToFind:-

• Length of string of each phone

Solution:-

• Let AS = SD = AD = 2x
• In ASD all sides are equal
• So, ASD is equilateral triangle

We draw OP ⊥ SD

$\rm\implies \: SP = DP = \frac{1}{2}SD$

$\rm\implies \:SP = DP = \frac{2x}{2} = x$

• Join OS and AO

★ In ∆OPS

• (By phythagoras theorem)

$\rm\implies \:OS^2 = OP^2 + PS^2$

$\rm\implies \:20^2 = OP^2 + x^2$

$\rm\implies \:400 - x^2 = OP^2$

$\rm\implies \:OP^2 = 400 - x^2$

$\rm\implies \:OP = \sqrt{400 - x^2}$

★ In ∆APS

• (By phythagoras theorem)

$\rm\implies \:AS^2 = AP^2 + PS^2$

$\rm\implies \:2x^2 = AP^2 + x^2$

$\rm\implies \:4x^2 - x^2 = AP^2$

$\rm\implies \:AP = \sqrt{3}x$

★ Now,

$\rm\implies \:AP = AO + OP$

$\rm\implies \:\sqrt{3}x=20-\sqrt{400-x^2}$

$\rm\implies \:\sqrt{400-x^2} = \sqrt{3}x-20$

★ Squaring on both sides

$\rm\implies \:(\sqrt{400-x^2} )^2 = (\sqrt{3}x-20)^2$

$\rm\implies \:400-x^2 = (\sqrt{3}x)^2+(20)^2 - 2 \times (\sqrt{3}x)\times 20$

$\rm\implies \:4x^2 = 40\sqrt{3}x$

$\rm\implies \:x = 10\sqrt{3}m$

$\rm\therefore\:AS = SD = AD = 2x=2\times 10\sqrt{3}= {\rm{\pink{20\sqrt{3}m}}}$

★ Hence,

• $\rm{Length \:of \:string \:of \:each \:phone\: is \:20\sqrt{3}m}$