We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is $\frac{a}{\sqrt{3}}=R$ .

Upon substituting our given values, we will get:

$\frac{5x}{\sqrt{3}}=6r$

Let us solve for r.

$r=\frac{5x}{6\sqrt{3}}$

$\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}$

We know that area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}s^2$ , where s represents side length of triangle.

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2$

The area within circle and outside the triangle would be difference of area of circle and triangle as:

$A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}$

We can make a common denominator as:

$A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}$

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Therefore, our required expression would be $A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$ .

7 0

3 years ago

How many numbers are in the list 1,4,7 .......... 2005, 2008 URGENT! [If you answer correctly, I will mark you brainliest]

My name is Ann [436]

Answer:

669.333333333

Step-by-step explanation:

divide 2008 by 3

4 0

3 years ago