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In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...
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Updated on: 30 Jan 2019, 06:45
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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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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...
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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...
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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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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...
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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 kiaericThank 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...
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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 345 right triangle properties. (1) ∆ ADB is a right 3x4x5x 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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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...
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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 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...
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04 Feb 2019, 09:23
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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Re: In the figure below, ABCD is a rectangle with AB = 20, DA = 15. AE...
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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 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 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...
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06 Feb 2019, 08:15






