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,

OS^{2} = OP^{2} + PS^{2}

(20)^{2} = OP^{2} + x^{2}

400 = OP^{2} + x^{2}

In ΔAPS,

AS^{2} = AP^{2} + PS^{2}

(2x)^{2} = AP^{2} + x^{2}

4x^{2} = AP^{2} + x^{2}

Now,

AP = AO + OP

√3x = 20 + √400 – x^{2}

√3x - 20 = √400 – x^{2}

Squaring both sides, we get,

(√3x – 20)^{2} = (√400 – x^{2})^{2}

(√3x)^{2} + (20)^{2} – 2×(√3x)×(20) = 400 – x^{2}

3x^{2} + 400 – 40√3x = 400 – x^{2}

40√3x = 4x^{2}

x = 10 √3 m

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