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(D) The area of an isosceles triangle with base $b$ and equal sides $a$ is given by the formula: $	ext{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$.Given $

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None of these

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(D) The area of an isosceles triangle with base $b$ and equal sides $a$ is given by the formula: $	ext{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$.Given $b = 16 \, cm$ and $a = 10 \, cm$,we have:$	ext{Area} = \frac{16}{4} \sqrt{4(10)^2 - (16)^2} = 4 \sqrt{400 - 256} = 4 \sqrt{144} = 4 	imes 12 = 48 \, cm^2$.Let the side of the equilateral triangle be $s$. The area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$.Equating the areas: $\frac{\sqrt{3}}{4} s^2 = 48$.$s^2 = \frac{48 	imes 4}{\sqrt{3}} = \frac{192}{1.732} \approx 110.85$.$s = \sqrt{110.85} \approx 10.53 \, cm$.Since $10.53$ is not exactly $10.5$,the correct option is $D$ (None of these).

If the area of an equilateral triangle is equal to the area of an isosceles triangle whose base and equal sides are $16 \, cm$ and $10 \, cm$ respectively,then the side of the equilateral triangle is......$cm$

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