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

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Bunuel
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In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   11 Jan 2019, 01:43
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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

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2019-01-11_1242.png
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Ana4001
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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   11 Jan 2019, 02:16
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ACD I'd half of rectangle

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

Answer is 16

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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   11 Jan 2019, 03:55
Bunuel wrote:

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

Show Spoiler
Attachment:
2019-01-11_1242.png


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
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Re: In the figure above,ABCD is a rectangle. If the area of triangle AEB [#permalink] New post   11 Mar 2020, 04:59
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.
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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   03 Sep 2020, 05:31
Archit3110 wrote:
Bunuel wrote:

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

Show Spoiler
Attachment:
2019-01-11_1242.png


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


How is AEB a 45/45/90 ?
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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   16 Sep 2020, 19:17
Bunuel wrote:

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

Show Spoiler
Attachment:
2019-01-11_1242.png


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)
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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   03 Jan 2022, 06:32
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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)
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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink] New post   14 Jan 2022, 11:09
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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Re: In the figure above, ABCD is a rectangle. If the area of ΔAEB is 8, wh [#permalink]
14 Jan 2022, 11:09
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