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

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In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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New post 09 Jan 2019, 03:01
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A
B
C
D
E

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  5% (low)

Question Stats:

85% (01:26) correct 15% (02:07) wrong based on 107 sessions

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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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2019-01-09_1359.png [ 41.05 KiB | Viewed 1499 times ]

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Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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New post 09 Jan 2019, 04:17
Bunuel wrote:
Image
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


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]

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New post 09 Jan 2019, 07:54
Bunuel wrote:
Image
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


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]

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New post 09 Jan 2019, 11:00
Bunuel wrote:
Image
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


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]

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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]

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New post 09 Jan 2019, 12:33
Bunuel wrote:
Image
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


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]

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New post 22 Nov 2019, 01:08
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] 22 Nov 2019, 01:08
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