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Bunuel

Rectangle RSTU has a perimeter of 42. The half circle with diameter RS has an area of \(8\pi\). What is the area of the unshaded part of the figure?

A. \(104- \pi\)
B. \(104- 2\pi\)
C. \(104- 4\pi\)
D. \(104- 6\pi\)
E. \(104- 8\pi\)

Attachment:
2020-10-27_12-40-56.png

\(\frac{πr^2}{2} = 8π\)

So, \(r = 4\)

Thus, RS = 8

Further \(2 ( 8 + b ) = 42\)

So, b = 13

Thus, Area of Unshaded region is \(13*8 - 8π\) , Thus Answer must be (E)
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Rectangle RSTU has a perimeter of 42. The half circle with diameter RS has an area of 8π. What is the area of the unshaded part of the figure?

Area of half circle = π(r^2)/2 = 8π
so, r = 4
2r = 8, which is equal to the two of the sides of the rectangle.

Perimeter of rectangle = 42
two of the sides are 8 units each. So, remaining two sides = (42 - 8*2)/2 = 13 each

Area of rectangle = 13*8 = 104

Unshaded area = 104 - 8π

Option E
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Quote:
Rectangle RSTU has a perimeter of 42. The half circle with diameter RS has an area of 8π. What is the area of the unshaded part of the figure?
Step 1: Understanding the question
Area of semicircle = \(\frac{1}{2}*(π*r^2) = 8π\)
\(r^2= 16\);
r =4 (ignoring -4)
RS = d = 8

Perimeter of the rectangle = 2(RS + ST) = 42;
ST = 13

Step 2: Calculation
Unshaded portion = Area of rectangle - Area of the semicircle
= (13*8) - 8π
= 104 - 8π

IMO E
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