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# What is the area of quadrilateral ABCD?

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Math Expert
Joined: 02 Sep 2009
Posts: 51071

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23 Aug 2018, 03:15
00:00

Difficulty:

25% (medium)

Question Stats:

100% (01:19) correct 0% (00:00) wrong based on 10 sessions

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What is the area of quadrilateral ABCD?

A. 84
B. 168
C. 172
D. 252
E. 336

Attachment:

image031.jpg [ 2.45 KiB | Viewed 285 times ]

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Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management

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23 Aug 2018, 03:48
Bunuel wrote:

What is the area of quadrilateral ABCD?

A. 84
B. 168
C. 172
D. 252
E. 336

Attachment:
image031.jpg

drop a perpendicular from D to AB at P
now,
BP = 7
PD = 24
AP = 7

total area = area of DPBC + area of triangle ADP
=(24*7) + $$\frac{1}{2}$$* 24*7 = 24*7 ( $$\frac{3}{2}$$) = 252
Director
Status: Learning stage
Joined: 01 Oct 2017
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23 Aug 2018, 03:50
Bunuel wrote:

What is the area of quadrilateral ABCD?

A. 84
B. 168
C. 172
D. 252
E. 336

Attachment:
image031.jpg

Draw DE perpendicular to AB
Area of quadrilateral= Area of right angled triangle AED+Area of rectangle EBCD=1/2*7*24+ (24*7)=84+168=252

Ans. (D)
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23 Aug 2018, 04:00
Attachment:

image031.jpg [ 9.79 KiB | Viewed 179 times ]

In the figure, triangle AXD is a right-angled triangle with hypotenuse 25 and one of the sides 24
This is because BCDX is a rectangle and BC = XD(since we have drawn DX to be a perpendicular line)

Using this, we can find the length of the third sides using Pythagoras theorem - $$AD^2 = AX^2 + DX^2$$
Re-writing, we get $$AX^2 = AD^2 - DX^2$$ -> $$AX^2 = 25^2 - 24^2 = 625 - 576 = 49$$ -> AX = $$\sqrt{49} = 7$$

The area of the triangle is $$\frac{1}{2} * 24 * 7 = 12 * 7 = 84$$ | Area of the rectangle = $$24 * 7 = 168$$

Therefore, the area of the quadrilateral ABCD is the sum of the individual areas, which is 252(Option D)

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Re: What is the area of quadrilateral ABCD? &nbs [#permalink] 23 Aug 2018, 04:00
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