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]

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 [/img]

Vipul, that is pretty much understandable from the Figure that Square and Rhombus share the same base

Area of a Square = S^2 Area of a Rhombus = b*h 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] Regards, Vivek

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

i didnt see the figure, well I have edited my post mean while you promptly posted

vipulgoyal wrote:i didnt see the figure, well I have edited my post mean while you promptly posted ohk

theCodeToGMAT wrote: Triangle REC & Triangle KFD are equal So, Area of ABKR = 625-500 = 125 ... Very nicely done! Regards, Vivek

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] 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.