In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh

Question

https://gmatclub.com/forum/download/file.php?id=45945 In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, what is the area of ΔACD? A. 8 B. 12 C. 16 D. 24 E. 32 2019-01-11_1242.png

Ana4001 · Answer

ACD I'd half of rectangle

While as AEB=ACD
BOTH Eyal to half of rectangle= 16

Answer is 16

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Archit3110 · Answer

triangel AEB = 45:45:90 
ae=eb= 4 and ab = 4 sqrt 2
triangle adc ac= 8 , dc= 4 swrt 2
ad= 4sqrt2 
area ; 0.5*4 sqrt2*4 sqrt2 
16 IMO C

roneygeorge · Answer

can we say the area of ACD will be double of AEB?
If we apply 'area = 1/2* base * height' formula. For both triangles, the base remains the same, height is twice for ACD compared to AEB. Hence the area is twice.

chrtpmdr · Answer

How is AEB a 45/45/90 ?

adstudy · Answer

From area of parallelogram's we know, that it's diagonals bisect each other.

Using this property we can deduce, 
AE = EC and DE = EB

Also we know, AB = DC, as they are opposite sides of a rectangle.
As AE = EC, DE = EB and AB = DC
ΔAEB = ΔCED,
As they are equal, it means - 
Perpendicular from E on AB = Perpendicular from E on CD --- (1)

Now comparing ΔAEB and ΔACD, we get

AB (From ΔAEB) = CD (From ΔACD),
and AD (From ΔACD) = 2 x Perpendicular from E on AB (From ΔAEB)

That means area of ΔACD = 2 x ΔAEB = 16,
  Hence (C)

bv8562 · Answer

Let,
Length of rectangle = L
Width of rectangle = W

Area of △AEB = 1/2*Base*Height = 8

Base = L
Height = W/2

∴ 1/2*L*W/2= 8 => L*W = 32

Area of △ACD = (Area of rectangle)/2 = L*W/2 = 32/2 = 16 (C)

trentgmatclub123 · Answer

C)16 nothing stops me from converting it to a square. I can divide AEB into half and assume 4 cms and multiply that by 4.

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bumpbot · Answer

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In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh

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In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh

Prep Toolkit

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