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# What is the area of quadrilateral ABCD shown above?

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What is the area of quadrilateral ABCD shown above?  [#permalink]

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16 Jun 2020, 09:55
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What is the area of quadrilateral ABCD shown above?

A. 14
B. 16
C. 18
D. 20
E. 21

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What is the area of quadrilateral ABCD shown above?  [#permalink]

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16 Jun 2020, 11:42
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Invert the triangle BCD about BD as shown in the above figure.

Area of quadrilateral ABCD is equivalent to Area of isosceles right angle triangle ABC(in the right)

Area of isosceles right angle triangle ABC $$= \frac{1}{2}* 6*6 =18$$
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What is the area of quadrilateral ABCD shown above?  [#permalink]

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20 Jun 2020, 13:16
Bunuel wrote:

What is the area of quadrilateral ABCD shown above?

A. 14
B. 16
C. 18
D. 20
E. 21

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The attachment 2020-06-16_2151.png is no longer available

Attachment:

triangle.png [ 20.37 KiB | Viewed 280 times ]

We are going to draw triangle CFD (see diagram above) such that angle F is a right angle and angle FCD = angle BCD = 45 degrees. We see that triangle CFD is a 45-45-90 right triangle. We will argue that the area of quadrilateral ABCD is equal to the area of triangle CFD. The argument is as follows:

Both quadrilateral ABCD and triangle CFD have the quadrilateral region CBED in common. Outside of this common region, quadrilateral ABCD has a triangular region AED and triangle CFD has a triangular region BEF. However, both triangles AED and BEF are congruent since angle F and angle A are congruent (they are right angles), angle AED and angle BEF are congruent (they are vertex angles) and side BE and side DE are congruent (they are the congruent sides of isosceles triangle BED). Therefore, triangles AED and BEF have the same area and if we add the quadrilateral region CBED to each of these triangles, we have quadrilateral ABCD and triangle CFD having the same area. So instead of finding the area of quadrilateral ABCD, we can just find the area of triangle CFD, which is easier since it’s an isosceles right triangle. Of course, we need to know one side of this triangle and we will make another argument that FD = 6 since it’s equal to AB. The argument is as follows:

We have mentioned above that BE = DE and since triangles AED and BEF are congruent, AE = FE. However, since AB = BE + AE and FD = FE + DE, AB = FD. Lastly, since AB = 6, FD = 6.

Since triangle CFD is an isosceles right triangle, its area is ½ x (FD)^2 = ½ x 6^2 = 18.

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What is the area of quadrilateral ABCD shown above?   [#permalink] 20 Jun 2020, 13:16

# What is the area of quadrilateral ABCD shown above?

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