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Rectangle inscribed in a radius

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Rectangle inscribed in a radius

by Amrabdelnaby » Tue Dec 29, 2015 9:05 am
Dear Pros,

Could you please help me with this?

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?

A) 2r rt3
B) 2r (rt3 + 1)
C) 4r rt2
D) 4r rt3
E) 4r (rt3 + 1)

I solved it using the following method:

since the rectangle is inscribed in the circle, its diagonals are the diameter of the circle

hence the square of its length plus the square of its width is the square of the diagonal/diameter.

I plugged in the following:

L = 8
W = 6
D = 10

Hence R (radius) will be 5

according to the above the perimeter should be 28

Afterwards i plugged in the numbers in the answer choices giving me the following.

A) 10 x 1.7 --> much less than desired value
B) 10 x 2.7 --> 27 which is very close to desired value, hence keep it for now
C) 20 x 1.4 --> 28 which is exactly the same value

Here i stopped and automatically chose answer C since it yielded to the value I wanted specially that answers D & E will yield to greater values than 28.

However the OA is B!

Can you please explain? How come?!
Last edited by Amrabdelnaby on Tue Dec 29, 2015 9:16 am, edited 1 time in total.
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by GMATGuruNY » Tue Dec 29, 2015 9:09 am
A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be equal to the perimeter of the rectangle?

A. 2r(Sqrt 3)
B. 2r(Sqrt 3 + 1)
C. 4r(Sqrt 2)
D. 4r(Sqrt 3)
E. 4r(Sqrt3 + 1)
When shapes overlap, look for what they have IN COMMON.
When a rectangle is inscribed in a circle, the DIAGONAL of the rectangle is also the DIAMETER of the circle.

Almost every answer choice here includes √3.
√3 implies a 30-60-90 triangle.
The sides of a 30-60-90 triangle are in the following ratio:
1 - √3 - 2.

Let r=1, implying a diameter of 2.
Draw the following figure:

Image

The perimeter of the rectangle above = 2 + 2√3. This is our target.
Now we plug r=1 into the answers to see whether one of them yields our target of 2 + 2√3.

Answer choice B:
2r(√3 + 1) = 2(1)(√3 + 1) = 2√3 + 2.
Success!

The correct answer is B.
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by Amrabdelnaby » Tue Dec 29, 2015 11:50 am
Thanks alot Matt for the explanation.

Could you tell me what exactly i missed in doing the plug ins? or why they did not work?

Thanks
GMATGuruNY wrote:
A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be equal to the perimeter of the rectangle?

A. 2r(Sqrt 3)
B. 2r(Sqrt 3 + 1)
C. 4r(Sqrt 2)
D. 4r(Sqrt 3)
E. 4r(Sqrt3 + 1)
When shapes overlap, look for what they have IN COMMON.
When a rectangle is inscribed in a circle, the DIAGONAL of the rectangle is also the DIAMETER of the circle.

Almost every answer choice here includes √3.
√3 implies a 30-60-90 triangle.
The sides of a 30-60-90 triangle are in the following ratio:
1 - √3 - 2.

Let r=1, implying a diameter of 2.
Draw the following figure:

Image

The perimeter of the rectangle above = 2 + 2√3. This is our target.
Now we plug r=1 into the answers to see whether one of them yields our target of 2 + 2√3.

Answer choice B:
2r(√3 + 1) = 2(1)(√3 + 1) = 2√3 + 2.
Success!

The correct answer is B.
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by MartyMurray » Wed Dec 30, 2015 10:12 pm
Amrabdelnaby wrote:Could you tell me what exactly i missed in doing the plug ins? or why they did not work?
You assumed that the right triangles that make up the rectangle are 6 - 8 - 10 triangles. Then you sought to work from that assumption.

There is no reason to believe that the triangles are actually 6 - 8 - 10 triangles. They could be 1 - √3 - 2 triangles, 5 - 12 - 13 triangles or any other right triangles.

The point of the question is to figure out which choice can equal a perimeter via analyzing the relationships between the answer choices and 2r. The correct answer has a relationship to 2r that fits the constraints of some, unknown until you find the correct answer, right triangle.

Rather than figuring out what the relationship between 2r and the perimeter is by seeing which answer choice works, you sought to answer the question by first deciding what the relationship between 2r and the perimeter is.

Here's another way to answer the question. Let's look at the relationships between the answer choices and the length of the diagonal, 2r, to see which answer choice can work as sides of right triangles with hypotenuse 2r.

Since we are looking for the perimeter of the rectangle, which has four sides, the answer choices will be double the sides of the right triangle with hypotenuse 2r. So 1/2 the perimeter will somehow fit the right triangle equation A² + B² = (2r)² = 4r².

(A) 2r(√3) - Divide this by 2 to get half the perimeter, r(√3). The largest possible result for A² + B² would come from making the length of one side r(√3) and the length of the other 0.

A² + B² = 3r² + 0 = 3r² < 4r²

So we can't get a perimeter that works by using choice A.

(B) 2r(√3 + 1) - Divide by 2 to get 1/2 the perimeter, r(√3 + 1). With √3 and 1, r(√3 + 1) looks like the sides of a 30 - 60 - 90 triangle. Let's try sides √3r and 1r.

A² + B² = 3r² + 1r² = 4r²

That works.

NOW we know that the triangle is a 30 - 60 - 90, or 1 - √3 - 2, triangle.

The correct answer is B.
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