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Geometry - Quadrilateral question

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by Xeb » Mon Oct 10, 2011 12:49 am
I came along this question and want to share it with you. I believe there are different ways to solve it.

A quadrilateral has a perimeter of 68. Inside the quadrilateral there is a square, its corners touch the middle point of each side length of the quadrilateral. What is the area of the square?

(A) 289
(B) 168
(C) 144.5
(D) 124.5
(E) 68

This is how I solved it.
68/4 = 17 = length of one side of the quadrilateral

17/2 = 8.5 = 1/2 length of one side of the quadrilateral.

Remember the corners of the square inside the quadrilateral touch the middle point of each side length of the quadrilateral. Imagine four triangles with base = 8.5, and height = 8.5. Now we can calculate the hypotenuse to find out one length of the side of the square, or calculate the area of the triangle, which is easier:

1/2 * 8.5 * 8.5 = 36.125

36.125 * 4 = 144.50 = the sum of the area of all four triangles.

Area of the quadrilateral = 17^2 = 189 - 144.5 = 144.5 [spoiler](C)[/spoiler]
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by shankar.ashwin » Mon Oct 10, 2011 1:45 am
Since connecting the mid-points of the quad form a square, The outside quad would be a square too (or possibly a rhombus?-not sure)

Hence side of square = 68/4 = 17

When the mid points are connected the inscribed square would have half the area of the outer square.

So inscribed square area = 17^2/2 = 144.5
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by rohit_gmat » Mon Oct 10, 2011 2:28 am
i like ur method (using the triangles) but i did it the hard way and found out the hyp of the isoscles triangle -> 8.5 * sqrt2 .... and then squared 8.5 and multiplied it with 2
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by GMATGuruNY » Mon Oct 10, 2011 3:39 am
Xeb wrote:I came along this question and want to share it with you. I believe there are different ways to solve it.

A quadrilateral has a perimeter of 68. Inside the quadrilateral there is a square, its corners touch the middle point of each side length of the quadrilateral. What is the area of the square?

(A) 289
(B) 168
(C) 144.5
(D) 124.5
(E) 68
Let the quadrilateral be a square with a perimeter of 68:

Image

Area of the outer square = 17² = 289.
Each of the 4 triangles = 1/8 of the area of the outer square.
Thus, the inner square = 1/2 of the area of the outer square:
(1/2)*289 = 144.5.

The correct answer is C.
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