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Rectangle ABCD is comprised of 4 right triangles and rectangle FGHI.

Expert replies
by Brent@GMATPrepNow » Mon Apr 20, 2020 5:31 am
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Rectangle ABCD is comprised of 4 right triangles and rectangle FGHI. Triangles ADI and CBG are identical, and each has an area of 96. Triangles AFB and CHD are identical, and each has an area of 150. If the sides of each triangle have integer lengths, what is the area of rectangle FGHI?

A) 8
B) 10
C) 12
D) 14
E) 16

Answer: A
Source: www.gmatprepnow.com
Brent Hanneson - Creator of GMATPrepNow.com
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Source: — Problem Solving |

Brent@GMATPrepNow wrote: ↑
Mon Apr 20, 2020 5:31 am
Image
Rectangle ABCD is comprised of 4 right triangles and rectangle FGHI. Triangles ADI and CBG are identical, and each has an area of 96. Triangles AFB and CHD are identical, and each has an area of 150. If the sides of each triangle have integer lengths, what is the area of rectangle FGHI?

A) 8
B) 10
C) 12
D) 14
E) 16

Answer: A
Source: www.gmatprepnow.com
Important: Since each right triangle has integer lengths, we're looking for Pythagorean triples that satisfy the given information.
Aside: Pythagorean triples are sets of three integers that could be the measurements of a right triangle.

Some Pythagorean triples include:
3-4-5
5-12-13
7-24-25
8-15-17
etc

Also note that we can take the above Pythagorean triples and create additional Pythagorean triples by multiplying all values by the same integer.
For example, we can take the Pythagorean triple 3-4-5, and multiply of the three values by various integers to create additional triples such as:
6-8-10
9-12-15
12-16-20
15-20-25
etc

Notice that the largest number in a triple will be the length of the hypotenuse. So the first two values must be the lengths of the two legs of the right triangle.

Okay, let's first find a Pythagorean triple for the blue right triangle.
Image
Area of triangle = (base)(height)/2
So, if we let x and y be the lengths of the two legs, we can write: xy/2 = 96

Now let's find a Pythagorean triple that meets the condition that xy/2 = 96
Well, if we have a 12-16-20 right triangle, then the area = (12)(16)/2 = 96. PERFECT!
So the blue right triangle looks like this
Image


Now let's first find a Pythagorean triple for the red right triangle.
If we let j and k be the lengths of the two legs, we can write: jk/2 = 150
Among the various Pythagorean triples, we can see that, if we have a 15-20-25 right triangle, then the area = (15)(20)/2 = 150. PERFECT!
So the red right triangle looks like this
Image


When we add the relevant information to our diagram we get:
Image


We know that: (area of small rectangle FGHI) = (area of big rectangle ABCD) - (area of the 4 triangles)
Substitute values to get: area of small rectangle FGHI = (20)(25) - (150 + 150 + 96 + 96)
= 500 - 492
= 8

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Brent@GMATPrepNow wrote: ↑
Mon Apr 20, 2020 5:31 am
Image
Rectangle ABCD is comprised of 4 right triangles and rectangle FGHI. Triangles ADI and CBG are identical, and each has an area of 96. Triangles AFB and CHD are identical, and each has an area of 150. If the sides of each triangle have integer lengths, what is the area of rectangle FGHI?

A) 8
B) 10
C) 12
D) 14
E) 16

Answer: A
Source: www.gmatprepnow.com
As all the sides of the triangles are integers, we have to look for Pythagorean triplets

\(96= 1\cdot 96 = 2\cdot 48=3 \cdot 32=4 \cdot 24=6 \cdot 16= 8 \cdot 12\)

Length of sides of triangle \(AID\) are \(12, 16\) and \(20\) cm, respectively.

\(150=1\cdot 150 = 2 \cdot 75=3 \cdot 50= 5\cdot 30=6 \cdot 25= 10 \cdot 15\)

Length of sides of triangle \(DHC\) are \(15, 20\) and \(25\) cm, respectively.

Area of \(ABCD = 20\cdot 25=500\)
Area of 4 triangles \(= 150+150+96+96= 492\)

Area of \(FGHI = 500-492 = 8\)

Regards!
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