JeffTargetTestPrep wrote:
Bunuel wrote:
Attachment:
Untitled.png
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}-32\pi\), what is the radius of each circle?
(A) 4
(B) 8
(C) 16
(D) 24
(E) 32
We can let each radius = r, and so the side of each triangle = 2r.
Notice that the area of the equilateral triangle consists of the central shaded region and three identical circular sectors, each of which is a 60-degree sector from its circle. Using this information, we can create the following equation:
(Area of equilateral triangle) - (3 x area of 1/6 of each circle) = area of shaded region
(2r)^2√3/4 - 3(1/6 x π r^2) = 64√3 − 32π
[(4r^2)√3]/4 - (πr^2)/2 = 64√3 − 32π
(r^2)√3 - (πr^2)/2 = 64√3 − 32π
Multiplying both sides by 2, we have:
2(r^2)√3 - πr^2 = 128√3 − 64π
r^2(2√3 - π) = 128√3 − 64π
Dividing both sides by (2√3 - π), we have:
r^2 = 64
r = 8
Answer: B
hi
generis its been raining here... so after the rain i feel like solving algebra
regarding this " \(r^2(2√3 - π) = 128√3 − 64π\) divide both sides by (2√3 - π) "
\(r^2= \frac{128√3 − 64π}{(2√3 - π)}\) after this step i am getting confused (when i dvide numerator by denominator)
so this is what i get in numerator first i smplify numerator 128√3 − 64π ( i subtract 128 - 64 = 64) --->
\(r^2=\frac{64√3-π}{2√3 - π}\) now when i divide numerator by denominator i divide 64 by 2
\(r^2= 32\) and get this
why
have a great sunny weekend
La France va gagner la coupe du monde
3:1