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# In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...

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Joined: 22 Feb 2018
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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Updated on: 30 Jan 2019, 06:45
4
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Difficulty:

35% (medium)

Question Stats:

75% (03:00) correct 25% (02:54) wrong based on 24 sessions

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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Source: Manhattan Review

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Screen Shot 2019-01-29 at 4.16.29 PM.png [ 100.45 KiB | Viewed 479 times ]

Originally posted by jpfg259 on 29 Jan 2019, 07:18.
Last edited by SajjadAhmad on 30 Jan 2019, 06:45, edited 1 time in total.
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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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29 Jan 2019, 07:38
jpfg259 wrote:
In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Hear me out on this

Total Area of the rectangle= 20 *15 = 300 unit^2

triangle ADE, will be an isosceles triangle with area = 1/4 * 15 *15 = 56.25 unit^2

Area of triangle DCB = 300/2 = 150 unit^2

Area of AEB = 300 - [150 + 56.25]
= 300 - 206.25
~ 94.75 unit^2

D
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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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30 Jan 2019, 00:53
KanishkM wrote:
jpfg259 wrote:
In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Hear me out on this

Total Area of the rectangle= 20 *15 = 300 unit^2

triangle ADE, will be an isosceles triangle with area = 1/4 * 15 *15 = 56.25 unit^2

Area of triangle DCB = 300/2 = 150 unit^2

Area of AEB = 300 - [150 + 56.25]
= 300 - 206.25
~ 94.75 unit^2

D

Hello KanishkM,

Why is ADE an isosceles triangle? And could you please explain how you came up with 1/4*15*15 in order to calculate ADE?

Thank you very much,
kiaeric
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Posts: 914
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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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31 Jan 2019, 07:16
kiaeric wrote:

Hello KanishkM,

Why is ADE an isosceles triangle? And could you please explain how you came up with 1/4*15*15 in order to calculate ADE?

Thank you very much,
kiaeric

Hi kiaeric

Thank you for the question. After reworking on the question, there was a flaw in my approach.

I solved it this way now, Join diagonal AC, since in rectangle both diagonals are equal and they bisect each other.

diagonal BD will be 25 ( $$\sqrt{625}$$), take O as the point of intersection.

Area of AOD, 1/2 * 12.5 * 12.5 ~= 72

Again use Pythagoras theorem, to calculate the height of AEO
Area of AEO = 1/2 * 10* 6 ~ = 30

Total area ~ 102,

which is nearer to D.

Let me know if this brings some clarity.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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03 Feb 2019, 09:36
jpfg259 wrote:
In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Source: Manhattan Review

Attachment:

rectangleABCD 2019.02.03.jpg [ 39.99 KiB | Viewed 283 times ]

Use properties of right triangles divided by altitude
and properties of similar triangles to find the area of ∆ AEB

All three triangles are similar. (see rule below)

Find side length of DB from 3-4-5 right triangle properties.

(1) ∆ ADB is a right 3x-4x-5x triangle.
-- $$4x = 20$$
-- $$x = 5$$
-- side DB = $$5x = 25$$

Rule: A right triangle divided by an altitude creates two right triangles
similar to the original triangle and similar to each other.

Find area of ∆ AEB from area of ∆ ADB

(2) ∆ AEB is similar to ∆ ADB - from rule above

(3) Rule: If two triangles are similar and their sides are in ratio $$\frac{a}{b}$$, then their areas will be in ratio $$\frac{a^2}{b^2}$$

• ratio of sides is $$\frac{20}{25} = \frac{4}{5}$$

• ratio of areas is $$\frac{4^2}{5^2} = \frac{16}{25}$$

(4) Area of large ∆ ADB = $$\frac{1}{2}$$ of rectangle ABCD

• rectangle area = $$(l * w) = (15* 20) = 300$$

• area of ∆ ADB =$$(\frac{1}{2} * 300) = 150$$

(5) Area of ∆ AEB = $$\frac{16}{25}$$ of area of ∆ ADB

Area of ∆ AEB =$$(\frac{16}{25} * 150) = 96$$

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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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04 Feb 2019, 09:07
jpfg259 wrote:
In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Source: Manhattan Review

use triangle similarity concept to solve
area of the rectangle = 15* 20 = 300
and area of triangle ADB = 150
since the triangle ADB has an altitude E , its divided into two triangles ADE and AEB
whose angles angle ADE= angle EAB and angle DAE= angle EBA
Triangle DAB side = AB:DB = 20/25 = 4:5 = square of the sides = 16:25 which is 16:25 of area of triangle ADB
16* 150 /25 = 96
IMO D
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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04 Feb 2019, 09:23
Once you realise:-

Then using similarity rule: AEB is 12:16:20
Hence the area is 1/2*12*16=96

Option D is the correct ans.
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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...  [#permalink]

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06 Feb 2019, 08:15
Archit3110 wrote:
jpfg259 wrote:
In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE is the altitude on diagonal BD. What is the area of the △AEB?

(A) 24
(B) 48
(C) 64
(D) 96
(E) 120

Source: Manhattan Review

use triangle similarity concept to solve
area of the rectangle = 15* 20 = 300
and area of triangle ADB = 150
since the triangle ADB has an altitude E , its divided into two triangles ADE and AEB
whose angles angle ADE= angle EAB and angle DAE= angle EBA
Triangle DAB side = AB:DB = 20/25 = 4:5 = square of the sides = 16:25 which is 16:25 of area of triangle ADB
16* 150 /25 = 96
IMO D

Hi,

Could you please explain why you said:
"since the triangle ADB has an altitude E , it's divided into two triangles ADE and AEB, whose angles angle ADE= angle EAB and angle DAE= angle EBA"

I don't understand... why are <ADE = <EAB and <DAE = <EBA?

Thank you!
Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...   [#permalink] 06 Feb 2019, 08:15
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