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# In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of

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In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of  [#permalink]

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20 Sep 2017, 02:02
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In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of EB. What is the area of ∆ ADE?

(A) 8
(B) 9
(C) 12
(D) 18
(E) 24

Attachment:

2017-09-20_1021.png [ 3.94 KiB | Viewed 757 times ]

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Re: In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of  [#permalink]

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20 Sep 2017, 02:25
Hello,
• The area of rectangle ABCD is 72 -> AB*AD=72
• AE = 1/3 EB -> EB=3AE
• AE+EB=AB -> 4AE=AB
• Formula area of triangle: (base*height)/2 --> (AE*AD)/2=18/2=9

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Re: In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of  [#permalink]

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20 Sep 2017, 03:02
Given : In the figure above, the area of rectangle ABCD is 72 => AB * AD =72
AE = 1/3 of EB

=> AB = AE+EB = AE + 3AE = 4AE
=> AE =1/4 * AB
What is the area of ∆ ADE => Area = 1/2 * AE * AD = 1/2 * (1/4 * AB) * AD = 1/8 * AB* AD = 1/8 * 72 = 9

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Re: In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of  [#permalink]

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20 Sep 2017, 08:52
1
Bunuel wrote:

In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of EB. What is the area of ∆ ADE?

(A) 8
(B) 9
(C) 12
(D) 18
(E) 24

Attachment:
The attachment 2017-09-20_1021.png is no longer available

Attachment:

2017-09-20_1021ed.png [ 9.67 KiB | Viewed 555 times ]

Too many side and segment names are hard for me to follow.

Let the base (AE) of ∆ ADE = $$x$$

$$x$$ is $$\frac{1}{3}$$ of $$EB$$

$$\frac{1}{3} = \frac{x}{EB}$$

$$EB = 3x$$

Let the height of ∆ ABE and short side of rectangle ABCD, (AD) = $$y$$

Area of rectangle ABCD is

$$4x * y =$$
$$4(xy) = 72$$
$$xy = 18$$

$$\frac{(x*y)}{2}$$
$$xy = 18$$
$$\frac{18}{2} = 9$$

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Re: In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of  [#permalink]

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25 Sep 2017, 18:02
Bunuel wrote:

In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of EB. What is the area of ∆ ADE?

(A) 8
(B) 9
(C) 12
(D) 18
(E) 24

Attachment:
2017-09-20_1021.png

Since the area of a rectangle is base times height, we see that the area of rectangle ABCD is AD x AB, which is given to be 72.

Since the area of a triangle is one-half times base times height, we see that the area of triangle ADE is ½(AD x AE). Since AE = ⅓(EB), 3(AE) = EB. Since AB = AE + EB, AB = AE + 3(AE) = 4(AE) or ¼(AB) = AE.

Thus, the area of triangle ADE is ½(AD x AE) = ½(AD x ¼(AB)) = (½) x (AD) x (¼) x AB = ⅛(AD x AB). Since we know AD x AB = 72, the area of triangle ADE is ⅛(72) = 9.

Alternate Solution:

Let’s denote the length of the segment AE by x. Since AE = 1/3 of EB, the length of EB is 3x. Thus, the length of AB is x + 3x = 4x. Let’s express the length of AD in terms of x. Since the area of the rectangle is 72, |AD|(4x) = 72; thus |AD| = 72/(4x) = 18/x. Now, the area of the triangle ADE is (1/2)(x)(18/x) = 18/2 = 9.

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Re: In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of &nbs [#permalink] 25 Sep 2017, 18:02
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