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by theCodeToGMAT » Tue Oct 08, 2013 8:19 pm
If ABCD is a square with area 625, and CEFD is a rhombus with area 500, then the area of the shaded region is

Note: Figure not drawn to scale

(A)125
(B)175
(C)200
(D)250
(E)275

Answer [spoiler]{E}[/spoiler]
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by vipulgoyal » Tue Oct 08, 2013 8:38 pm
Area of square = 625 so side = 25
Area of rhombus = 500. So altitude = 500/25 = 20
base of triangle = root (25^2 - 20^2) = 15
Area of triangle = (1/2) * 15 * 20 = 150

Area of shaded region = 625 - (500 - 150) = 275
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by theCodeToGMAT » Tue Oct 08, 2013 8:42 pm
Vipul, that is pretty much understandable from the Figure that Square and Rhombus share the same base
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by mevicks » Tue Oct 08, 2013 8:48 pm
Area of a Square = S^2
Area of a Rhombus = b*h

Image

Given:

For the Square:-
S^2 = 625
S = 25
All the sides of the rhombus are 25 each (Since CD is a shared side)

For the Rhombus:-
b * h = 500
25 * h = 500
h = 20(EP in the figure)

Triangle ECP is a 3-4-5 Triangle
Thus CP = 15 and PD = 10

Shaded region area = 625 - (1/2 * 15 * 20 + 20 * 10) = 275

[spoiler]Answer : E[/spoiler]

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Vivek
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by theCodeToGMAT » Tue Oct 08, 2013 8:54 pm
Another Approach to this question:

Triangle REC & Triangle KFD are equal

So, Area of ABKR = 625-500 = 125

125 = 25 * BK => BK = 5
So, RC = 20

RE = sqrt(25^2 - 20^2) = 15

Area of RCE = 1/2 * 15 * 20 = 150

So, 150 + 125 = 275
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by vipulgoyal » Tue Oct 08, 2013 9:38 pm
i didnt see the figure, well I have edited my post mean while you promptly posted
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by theCodeToGMAT » Tue Oct 08, 2013 9:45 pm
vipulgoyal wrote:i didnt see the figure, well I have edited my post mean while you promptly posted
ohk :)
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by mevicks » Tue Oct 08, 2013 11:36 pm
theCodeToGMAT wrote:
Triangle REC & Triangle KFD are equal

So, Area of ABKR = 625-500 = 125

...
Very nicely done!

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Vivek
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by GMATGuruNY » Wed Oct 09, 2013 3:05 am
theCodeToGMAT wrote:If ABCD is a square with area 625, and CEFD is a rhombus with area 500, then the area of the shaded region is

Note: Figure not drawn to scale

(A)125
(B)175
(C)200
(D)250
(E)275

Answer [spoiler]{E}[/spoiler]
Image

Square ABCD:
Since area = s² = 625, AD=25.

Rhombus AHGD:
Since area = bh = (AD)(EH) = 500, we get:
25(EH) = 500
EH = 500/25 = 20.

∆AHE:
Since AD=25 in rhombus AHGD, AH=25.
Since AH=25 and EH=20, ∆AHE is a 3:4:5 triangle in which each side is multiplied by 5:
3:4:5 = 15:20:25.
Thus, AE = 15.
Area of ∆AHE = (1/2)bh = (1/2)(AE)(EH) = (1/2)(15)(20) = 150.

Rectangle EHFD:
Since ED = 25-15 = 10, area = bh = (ED)(EH) = 10*20 = 200.

Shaded region = Square ABCD - ∆AHE - Rectangle EHFD = 625-150-200 = 275.

The correct answer is E.
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