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Math Expert V
Joined: 02 Sep 2009
Posts: 59725
In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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Difficulty:   5% (low)

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

### HideShow timer Statistics 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

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

Attachment: 2019-01-09_1402.png [ 23.49 KiB | Viewed 1649 times ]

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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5483
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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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

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
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9876
Location: Pune, India
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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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

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

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Karishma
Veritas Prep GMAT Instructor

Manager  G
Joined: 16 Oct 2011
Posts: 110
GMAT 1: 570 Q39 V41 GMAT 2: 640 Q38 V31 GMAT 3: 650 Q42 V38 GMAT 4: 650 Q44 V36 GMAT 5: 570 Q31 V38 GPA: 3.75
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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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

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)
Manager  S
Joined: 18 Oct 2018
Posts: 90
Location: India
GMAT 1: 710 Q50 V36 GPA: 4
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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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
Manager  S
Joined: 28 Mar 2017
Posts: 66
Location: Sweden
Concentration: Finance, Statistics
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

### Show Tags

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

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!
Manager  B
Joined: 10 Sep 2019
Posts: 53
Location: India
Concentration: Social Entrepreneurship, Healthcare
GPA: 2.6
WE: Project Management (Non-Profit and Government)
Re: In the figure above, three squares and a triangle have areas of A, B,  [#permalink]

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