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

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.

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

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

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.

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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