The radius of the circle is $4$ cm. Find length of the side of equilateral triangle(whose vertex lie on the circle) inside circle if the section inside circle and outside the triangle has area $29.05 c m^{2}$

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Solution

The correct option is C $7$
Area of remaining section=Area of circle-area of equilateral triangle
$\Rightarrow 29.05 = \frac{22}{7} \times 4 \times 4 - \frac{\sqrt{3}}{4} \times a^{2}$
$= \frac{\sqrt{3}}{4} \times a^{2} = \frac{352}{7} - 29.05$
$\Rightarrow a^{2} = [\frac{352 - (29.05 \times 7)}{7}] \times \frac{4}{\sqrt{3}} \approx 49 \Rightarrow a = \sqrt{49} = 7 c m$

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The radius of the circle is 4 cm. Find length of the side of equilateral triangle(whose vertex lie on the circle) inside circle if the section inside circle and outside the triangle has area 29.05 cm2

The correct option is C 7Area of remaining section=Area of circle-area of equilateral triangle⇒29.05=227×4×4−√34×a2=√34×a2=3527−29.05⇒a2=[352−(29.05×7)7]×4√3≈49⇒a=√49=7cm

The radius of the circle is $4$ cm. Find length of the side of equilateral triangle(whose vertex lie on the circle) inside circle if the section inside circle and outside the triangle has area $29.05 c m^{2}$