In the figure above, ABCD is a rectangle inscribed in a circle. If the

Question

https://gmatclub.com/forum/download/file.php?id=27336 In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.) A. 1:2 B. 3:2π C. 2:5 D. 4:3π E. 6:5π Kudos for a correct solution . rectangle-circle.gif

BrentGMATPrepNow · Accepted Answer

Let's say AD has length 1 and AB has length 3 (this also means that side CD also has length 3)
 https://i.imgur.com/YXFwtHG.png

Since the blue triangle is a right triangle, we can use the Pythagorean Theorem to find the length of the 3rd side (the hypotenuse)
 https://i.imgur.com/e6wmHMi.png 
If we let x = the length of the hypotenuse, we can write: 1² + 3² = x²
Simplify: 1 + 9 = x²

So, x² = 10, which means x = √10
 https://i.imgur.com/d1m8ZRg.png

At this point, we can see that the diameter of the circle is √10, which means the radius is  (√10)/2

What is the ratio of the area of the rectangle to the area of the circle?  
Area of rectangle = (length)(width) 
= ( 1 )( 3 )
= 3

Area of circle = π(radius)²
= π( √10)/2 )²
= π(10/4)
= π(5/2)
= 5π/2

So, the RATIO = 3/(5π/2)
= (3)(2/5π)
= 6/5π
= 6 : 5π

Answer: E

Cheers,
Brent

ENGRTOMBA2018 · Answer

Let AD = x ---> AB = 3x

Let radius = r---> AC = diameter of the circle = 2r

Now in right angled triangle ADC , right angled at D,

AD^2+CD^2 = AC^2  ---->  x^2+(3x)^2 = (2r)^2  --->  r^2 = \frac{5x^2}{2}

Thus the ratio of rect area / circle area = Ratio = \frac{x*3x}{\frac{5\pi*x^2}{2}} ----> Ratio =  \frac{6}{5\pi} . E is the correct answer.

GMATinsight · Answer

Let, AB = 3
and AD = 1
i.e. BD =  \sqrt{3^2 + 1^2}  =  \sqrt{10}  = Diameter of Circle

Area of Rectangle = AB x BD = 3 x 1 = 3

Area of Circle = (π/4)*Diameter^2 = (π/4)*10 = 5π/2

Area of Rectangle / Area of Circle = 3 / (5π/2) = 6/5π

Answer: Option E

ashokk138 · Answer

Given: AB = 3 * AD

Let AD = x => AB = 3x. Let the diameter be d

d^2 = x^2 + 9x^2

d = ( x *  \sqrt{10}  )/2

radius = d/2 = (x *  \sqrt{10}  )/4

Area of rectangle : area of circle

x * 3x : π * (x/4 \sqrt{10} ) ^ 2

3x^2 : π * x^2 * (10/4)

12: 10 π => 6 : 5π

Option E

aimtoteach · Answer

Lets assume AD = 2 so AB = 6 (3 times AD)
 
By applying Pyth. Theory we know that AC ( which is the diameter of the circle) is -

6^2 +2^2 = AC^2

AC = 2√10

Ratio = 2*6 / pi (2√10)^2

=   6/5pi is the answer --> option E

GMATinsight · Answer

The highlighted part is a mistake which has led to several mistakes in the three steps in between.

Bunuel · Answer

800score Official Solution:

Let AD = x. AB = 3x. Area of the rectangle is (AD) × (AB) = x × (3x) = 3x².

Using the Pythagorean theorem, take AB and AD to get the diameter AC of the circle.
x² + (3x)² = AC²
x² + 9x² = AC²
10x² = AC²
AC = x√10 and the radius is   / 2.

The area of a circle is πr², so the area is π  ² = (10πx²)/4 = (5πx²)/2.

The ratio is:
3x² : (5πx²)/2 = 3 : 5π/2 = 6 : 5π

The correct answer is E.

gota900 · Answer

For my understanding:

Is the statement "figure not drawn to scale" the obvious indicator that we can not assume the properties of a 90-60-30 triangle? That was what I tried to do.

So the trick here is to recognize that? I thought that:

Since the diagonal of the rectangle passes the midpoint, I could conclude that it functions as hypothenuse which yields us the information that we will definitely have 90 degrees in the rectangle. Furthermore, I based my following assumptions (90-60-30) triangle on the fact that the line that passes the midpoint would work as some sort of bisector to the 90 degree angles of the rectangle. (Is this where it all went wrong?)

Is the only thing I can trust in this prompt the fact that we have a rectangle inscribed in a circle? All the previously mentioned conclusions can't be drawn with certainty, is that correct? I'm a little confused here...

GMATinsight · Answer

Adding video solution to the list

https://www.youtube.com/watch?v=ExTtxjB20vo

bumpbot · Answer

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In the figure above, ABCD is a rectangle inscribed in a circle. If the

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