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Bunuel
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Attachment:
image031.jpg
image031.jpg [ 9.79 KiB | Viewed 1554 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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Bunuel

What is the area of quadrilateral ABCD?


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


Attachment:
image031.jpg

Quadrilateral ABCD can be split into lower rectangle and upper triangle.

Height of upper triangle = \(\sqrt{25^2 - 24^2 } = 7\)

Area of quadrilateral ABCD = 24*7 + 1/2 * 24 * 7 = 24*7 * 3/2 = 252

IMO D
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