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Question Number 194560 by Denis last updated on 09/Jul/23

Answered by MM42 last updated on 10/Jul/23

$$\mathrm{\Delta }XABisequilatraltriangle$$$$\mathrm{\angle }XAD=\mathrm{\angle }XBC={30}^0$$$$S_1=S_2\Rightarrow S_1+S_2=\frac{1}{6}\pi \times {21}^2=11\times 21$$$$S_3=\frac{\sqrt{3}}{4}\times {21}^2$$$$\Rightarrow ans={21}^2-(11\times 21+\frac{\sqrt{3}}{4}\times {21}^2)\approx 19$$

Commented by MM42 last updated on 10/Jul/23

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