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In the figure above, three squares and a triangle have areas of A, B,

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Bunuel
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In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 03:01
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In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36

PS76402.01

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KarishmaB
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 07:54
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Bunuel wrote:

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36


Show Spoiler
Attachment:
2019-01-09_1359.png

Attachment:
2019-01-09_1402.png


Since A, B and C are squares with area as 144, 81 and 225, the side in each case is 12, 9 and 15. Note that the sides are in the ratio 3:4:5.
So X must be a right triangle.
Area X = (1/2)*12*9 = 54

Answer (D)
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 04:17
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Bunuel wrote:

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36


Show Spoiler
Attachment:
2019-01-09_1359.png

Attachment:
2019-01-09_1402.png



A=12
B= 9
C= 15

triangle X : 3:4:5
9:12:15
so 0.5*9*12 = 54 IMO D
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 11:00
Bunuel wrote:

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36


Show Spoiler
Attachment:
2019-01-09_1359.png

Attachment:
2019-01-09_1402.png


Since each rectangle is stated to be a square, the sides of A B and C are 12,9 and 15 respectively. Since these values are in the ratio 3:4:5, the triangle must be a right triangle, therefore the lengths of A and B will be our length and our height. A=1/2 *B*h = 1/2 (12*9) = (1/2)(108) = 64 (D)
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 11:49
Since Area of square A=144, side of A=12
Since Area of square B=81,side of B= 9
Since Area of square C= 225,side of C=15

Now, in triangle X :
Sides are 9,12,15 which shows that the sides are in ratio of 3:4:5
Hence Triangle X is a right angled triangle.

so area of triangle X= 0.5*9*12 = 54
IMO D
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   09 Jan 2019, 12:33
Bunuel wrote:

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36


Show Spoiler
Attachment:
2019-01-09_1359.png

Attachment:
2019-01-09_1402.png


The size of the boxes is actually part of the proof for pythagoras' theorem, so if you recognized that this question was easy!
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   22 Nov 2019, 01:08
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There is a formula to calculate area of a triangle with 3 known sides. Though in this case it turns out to be a right triangle (which might be the case with most questions in GMAT), doesn't hurt to know it: Area = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s= a+b+c/2

Using this formula, we can directly get the area as 54.
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   13 Jan 2020, 15:36
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Bunuel wrote:

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36

PS76402.01

Show Spoiler
Attachment:
2019-01-09_1359.png

Attachment:
2019-01-09_1402.png


Based on the given areas, we see that the sides of triangle x are 12, 9, and 15, so we have a 9-12- 15 right triangle. Thus, the area of the triangle is 9 x 12 x 1/2 = 54.

Answer: D
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   12 Nov 2021, 05:29
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We know the area of a square is (side)* (side)

Since the given areas area 144,81,225 unit square we have the sides as 12,9 and 15 respectively.
(Sqrt of the area= side)

Hence the side ratio is 9:12:15 or 3:4:5

So X must be a right triangle,

Area X = 1/2 * b * h

=(1/2)*12*9 = 54

(option d)

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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   22 Nov 2021, 03:50
GMATBusters

1) when we are given sides of a triangle in a ratio of 3:4:5 (or any triplets) can we safely conclude that the given triangle is a right-angled triangle?

2) Similarly, if we are told that two sides of a right angled triangle are in the ratio of 3:4 we cannot conclude that the third side is a multiple of 5 until we are given that the third side is the longest. Right?
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Re: In the figure above, three squares and a triangle have areas of A, B, [#permalink] New post   21 Feb 2024, 05:05
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