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What is the area of quadrilateral ABCD?

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

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New post 23 Aug 2018, 03:15
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

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

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Director
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Re: What is the area of quadrilateral ABCD?  [#permalink]

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New post 23 Aug 2018, 03:48
Bunuel wrote:
Image
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
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Re: What is the area of quadrilateral ABCD?  [#permalink]

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New post 23 Aug 2018, 03:50
Bunuel wrote:
Image
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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Re: What is the area of quadrilateral ABCD?  [#permalink]

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New post 23 Aug 2018, 04:00
Attachment:
image031.jpg
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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