A Rectangle is Inscribed in a Circle of Radius r. If the Rectangle is GMAT Problem Solving

Question: A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?

1. 2r√3
2. 2r(√3+1)
3. 4r√2
4. 4r√3
5. 4r(√3+1)

“A rectangle is inscribed in a circle of radius r. If the rectangle is”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “The Official Guide for GMAT Reviews”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:
Approach Solution 1:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle.

Now, since each option has √3 in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard triangle for the GMAT), because its sides are always in the ratio 1: √3: 2. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

So, in this case we would have:

The perimeter of the rectangle is

2r √3 + 2r =2 r(√3 +1).

Correct Answer: B

Approach Solution 2:
Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b,
(2r)^2 = a^2 + b^2 = 4^r^2

So when you square a and b and sum them, you should get 4r^2

The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b.
The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1.
a+b = r√3+r

Now check:
(√3r)^2 + r^2 = 4r^2

That is what we wanted.

Correct Answer: B

Approach Solution 3:

We need to find the perimeter of the rectangle i.e. 2(a + b)
We know that a^2 + b^2 = (2r)^2= 4r^2

So what can a and b be?

3r^2 + r^2 = 4r^2 (So a = √3r,b = r)
or
2r^2 + 2r^2 = 4r^2 (So a = √2r, b = √2r)
etc

Now, looking at the options, we see that
2(a + b) can be 2(√3r + r)

Correct Answer: B

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