(1st) notice that the the bottom side CD is shared by both the rhombus and the square
(2nd) we can find the side of the “square” that would be hidden behind the rhombus by taking the square root of the area = 625
Side CD = sqrt(625) = 25
(2nd) we are given that the area of the rhombus = 500
Area of rhombus = (Base side) * (Perpendicular Height)
We can draw 2 perpendicular height lines:
(1) drawn from vertex E perpendicular to side CD ——> call the intersection point X
(2) drawn from vertex D to side EF of the rhombus ———> call the intersection point Y
The area of the shaded portion = area of square - (area of right triangle CED) - (area of rectangle EXDY)
(3rd) find the length of the perpendicular heights = EX and DY ——> call H
Area of rhombus = 500 = (side CD) * (H)
500 = 25H
H = 20 = EX = DY
(4th)we can find the length of CX and XD, which together sum to equal side CD of 25, using Pythagorean Theorem
Let the leg CX = l
Using right triangle CEX:
(25)^2 = (20)^2 + (l)^2
(l))^2 = 225
l = 15 ——> which means
CX = 15
And
DX = CD - CX
DX = 25 - 15
DX = 10
(4th) we can find the area of the right triangle CEX and the rectangle EXYD now:
Area right triangle CEX = (1/2) (15) (20) = 150
Area of rectangle EXYD = (10) (20) = 200
150 + 200 = 350
350 is the area covering what would be the square ABCD
Area of shaded region = 625 - 350 = 275
(E)275
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