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# In the figure, ABCD is a rectangle, and the area of triangle ACE is 10

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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Dec 2010, 05:45
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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10, what is the area of rectangle ?

A. 18
B. 22.5
C. 36
D. 54
E. 60

[Reveal] Spoiler:
Attachment:

Untitled.png [ 4.15 KiB | Viewed 325 times ]
[Reveal] Spoiler: OA

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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Dec 2010, 06:07
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hirendhanak wrote:

In the figure, ABCD is a rectangle, and the area of triangle ACE is 10, what is the area of rectangle ?

A. 18
B. 22.5
C. 36
D. 54
E. 60

$$area_{ACE}=\frac{1}{2}*base*height=\frac{1}{2}*CE*AB=10$$;

$$\frac{1}{2}*5*AB=10$$;

$$AB=4$$.

$$area_{ABCD}=AB*CB=AB*(CE+EB)=4*9=36$$.

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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Dec 2010, 07:23
thanks bunuel

I was not clear why we took AB as the height of triangle ACE . because AB is out of scope of the triangle ACE.

I thought AE becomes height of the triangle ACE.
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Dec 2010, 07:39
hirendhanak wrote:
thanks bunuel

I was not clear why we took AB as the height of triangle ACE . because AB is out of scope of the triangle ACE.

I thought AE becomes height of the triangle ACE.

The height of a triangle is the perpendicular from the base to the opposite vertex, sometimes the base may need to be extended. Since there are three possible bases, there are also three possible heights, so if we consider the base to be CE then the height will be the perpendicular from CE (CB) to the opposite vertex A, thus height is AB.
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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Sep 2016, 13:04
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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10. What is the area of the rectangle?
Attachment:

24.PNG [ 18.34 KiB | Viewed 2017 times ]

A. 18
B. 22.5
C. 36
D. 44
E. 45
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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28 Sep 2016, 19:11
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Expert's post
iMyself wrote:
Hi Bunuel,
What's the procedure to solve this problem?
Thanks Expert...

Area of a triangle is height*base/2...
In ∆ACE, CE is the base and height is CD...
So CE*CD/2 = 10.....5*CD=20.....CD is 4...

So the two sides of the rectangle ABCD are 4 and (5+4) or 9..
Thus Area of rectangle is 4*9=36
C
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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29 Sep 2016, 00:00
chetan2u wrote:
iMyself wrote:
Hi Bunuel,
What's the procedure to solve this problem?
Thanks Expert...

Area of a triangle is height*base/2...
In ∆ACE, CE is the base and height is CD...
So CE*CD/2 = 10.....5*CD=20.....CD is 4...

So the two sides of the rectangle ABCD are 4 and (5+4) or 9..
Thus Area of rectangle is 4*9=36
C

Thank you chetan2u
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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30 Sep 2016, 02:26
I have a small doubt regarding this question, how can we consider CD an altitude for the triangle ACE?
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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30 Sep 2016, 02:58
pushpitkc wrote:
I have a small doubt regarding this question, how can we consider CD an altitude for the triangle ACE?

That was also my question but end of the day we have to take it as altitude !

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In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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30 Sep 2016, 05:02
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iMyself wrote:
pushpitkc wrote:
I have a small doubt regarding this question, how can we consider CD an altitude for the triangle ACE?

That was also my question but end of the day we have to take it as altitude !

Sent from my iPhone using GMAT Club Forum mobile app

Height of a triangle is defined as the perpendicular distance from one of the vertices to its base.

So, If we are considering CE a base of the triangle. The perpendicular distance from C to its base is CD only. Notice, Base has been extended to B to get the height.
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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17 Oct 2016, 20:10
This question tests your understanding of similar triangles.

You should derive the following equation using this concept:

(1/2)(9)(AB) - (1/2)(4)(AB) = 10
AB(5/2) = 10
AB = 4

Therefore, we know the rectangle is a 4x9. Area = 36
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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17 Jan 2017, 07:43
Hi,,,lawiniecke,,,I have doubt that you subtract only ABE from whole,,,but there is another triangle ACD has which you didn't consider,,,so,,what's the reason??.?I mean ,,Area of ACE=10=1/2*9*AB-1/2*4*AB-area of ACE,,,why is it wrong?..Pls,,,,
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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20 Jan 2017, 07:09
Area of a triangle ACE = 10
Area = bh/2
b = 5
10 = 5xh/2
h = 4
Now, Rectangle length = 4, width = 9(4+5)
Area = 4x9 = 36
C
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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27 Feb 2017, 05:51
Area of triangle ACE = ½ *EC *AB
or 10 = ½ * 5*AB
or AB =4 BC =9.
Area of rectangle = 9*4
= 36. Option C
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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10 [#permalink]

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Re: In the figure, ABCD is a rectangle, and the area of triangle ACE is 10   [#permalink] 14 Dec 2017, 05:02
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