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605-655 (Medium)|   Geometry|      
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jpfg259
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... Updated on: 30 Jan 2019, 07:45
Originally posted by jpfg259 on 29 Jan 2019, 08:18.
Last edited by Sajjad1994 on 30 Jan 2019, 07:45, edited 1 time in total.
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45% (medium)

Question Stats:

73% (02:55) correct 27% (02:48) wrong based on 62 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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D

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KanishkM
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 29 Jan 2019, 08:38
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jpfg259
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
Show more

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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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 30 Jan 2019, 01:53
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KanishkM
jpfg259
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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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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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 31 Jan 2019, 08:16
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kiaeric


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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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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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 03 Feb 2019, 10:36
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jpfg259
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
Show more
Attachment:
rectangleABCD 2019.02.03.jpg
rectangleABCD 2019.02.03.jpg [ 39.99 KiB | Viewed 3489 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 =\((\frac{16}{25} * 150) = 96\)

Answer D
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 04 Feb 2019, 10:07
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jpfg259
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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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 ADB similar triangle AEB
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... 04 Feb 2019, 10:23
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Once you realise:-

ADB is 15:20:25
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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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 06 Feb 2019, 09:15
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jpfg259
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 ADB similar triangle AEB
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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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!
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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE... 09 Jun 2020, 05:57
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