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

Question

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

KanishkM · Answer

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 -  
= 300 - 206.25
~ 94.75 unit^2

D

kiaeric · Answer

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

KanishkM · Answer

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.

generis · Answer

rectangleABCD 2019.02.03.jpg 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

Archit3110 · Answer

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

CAMANISHPARMAR · Answer

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.

jpfg259 · Answer

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!

bumpbot · Answer

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

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