Answer:
[tex]A=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.[/tex]
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
[tex]A_{circle}=2\pi (2r)^2=2\pi \cdot 4r^2=8\pi r^2\ un^2.[/tex]
2. The area of the equilateral triangle is
[tex]A_{\triangle }=\dfrac{(7x)^2\sqrt{3}}{4}=\dfrac{49\sqrt{3}x^2}{4}\ un^2.[/tex]
3. The area A within the circle but outside the triangle is
[tex]A=A_{circle}-A_{\triangle}=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.[/tex]
