Question

An equilateral triangle is inscribed in a circle of radius 2r. Express the area A within the circle but outside the triangle as a function of the length 7x of the side of the triangle. Pearson

Answer 1

Answer:

$A=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.$

Step-by-step explanation:

An equilateral triangle with the side 7x units is inscribed in a circle of radius 2r units.

1. The area of the circle is

$A_{circle}=2\pi (2r)^2=2\pi \cdot 4r^2=8\pi r^2\ un^2.$

2. The area of the equilateral triangle is

$A_{\triangle }=\dfrac{(7x)^2\sqrt{3}}{4}=\dfrac{49\sqrt{3}x^2}{4}\ un^2.$

3. The area A within the circle but outside the triangle is

$A=A_{circle}-A_{\triangle}=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.$