SOLUTION: A rectangle whose length is twice as long as its width is inscribed in a circle of area π. What is the area of the rectangle?
(A) 2/5
(B) 4/5
(C) 8/5
(D) 5/4
(E) 5/8
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-> SOLUTION: A rectangle whose length is twice as long as its width is inscribed in a circle of area π. What is the area of the rectangle?
(A) 2/5
(B) 4/5
(C) 8/5
(D) 5/4
(E) 5/8
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Question 1201954: A rectangle whose length is twice as long as its width is inscribed in a circle of area π. What is the area of the rectangle?
(A) 2/5
(B) 4/5
(C) 8/5
(D) 5/4
(E) 5/8 Found 2 solutions by htmentor, mananth:Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! A circle with an area of pi has a radius of 1. Now consider the inscribed rectangle. The diagonal of this rectangle is the hypotenuse of a right triangle with the length and width as the other two sides. The hypotenuse of the rectangle is equal to 2, the diameter of the circle. Since the length is twice the width, by the Pythagorean theorem, we have 2^2 = W^2 + (2W)^2 = 5W^2.
Thus W = sqrt(4/5) = 2/sqrt(5)
Therefore, the area is W*L = 4/sqrt(5)*2/sqrt(5) = 8/5
You can put this solution on YOUR website! let x be width then 2x =length
Are of circle = pi*r^2
pi=pir^2
r^2=1
r=1
d=2
Apply pythagoras theorem in triangle
x^2+(2x)^2 = 2^2
5x^2=4
x^2= 4/5
x=(2)/sqrt(5)
width = (2)/sqrt(5)
length = 4/sqrt(5)
Area of rectangle = 2 * area of triangle
Area of rectangle = 2* 1/2 * 2/sqrt(5) * 4/sqrt(5)
=8/5
