Q. 63.7( 408 Votes )

A circular park o

Answer :

The figure is shown as:



According to the given question,

AS = AD = SD

Let AS = AD = SD = 2x

In ASD, all sides are equal,

∴ ASD is an equilateral triangle.

Now draw OP⊥ SD

So, SP = DP = ½ SD

⇒ SP = DP    = x

Join OS and AO.

In ΔOPS,

OS2 = OP2 + PS2

(20)2 = OP2 + x2

400 = OP2 + x2

In ΔAPS,

AS2 = AP2 + PS2

(2x)2 = AP2 + x2

4x2 = AP2 + x2

Now,

AP = AO + OP

√3x = 20 + √400 – x2

√3x - 20 = √400 – x2

Squaring both sides, we get,

(√3x – 20)2 = (√400 – x2)2

(√3x)2 + (20)2 – 2×(√3x)×(20) = 400 – x2

3x2 + 400 – 40√3x = 400 – x2

40√3x = 4x2

x = 10 √3 m

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