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805+ (Hard)|   Geometry|      
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Bunuel
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 13 May 2019, 01:03
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46% (02:46) correct 54% (02:59) wrong based on 74 sessions

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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpendicular from B to diagonal AC. What is the area of triangle AED?

A. 1
B. 42/25
C. 28/15
D. 2
E. 54/25
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E
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 14 May 2019, 03:55
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BE=3*4/5=2.4
EC= BC^2/AC
AC^2=3^2+4^2
AC=5
EC= 4^2/5=3.2

Perpendicular distance of E from BC= 2.4*3.2/4= 1.92
Perpendicular distance of E from AD=3-1.92=1.08

Area of AED= 0.5*1.08*4=2.16 or 54/25

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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 16 Jul 2019, 12:41
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nick1816
BE=3*4/5=2.4
EC= BC^2/AC
AC^2=3^2+4^2
AC=5
EC= 4^2/5=3.2

Perpendicular distance of E from BC= 2.4*3.2/4= 1.92
Perpendicular distance of E from AD=3-1.92=1.08

Area of AED= 0.5*1.08*4=2.16 or 54/25

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Hi nick1816,
A little elaboration could help. How did you calculate BE?
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe Updated on: 17 Jul 2019, 12:37
Originally posted by nick1816 on 16 Jul 2019, 13:08.
Last edited by nick1816 on 17 Jul 2019, 12:37, edited 1 time in total.
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Area of triangle ABC= 1/2*AB*BC=1/2*AC*BE
\(BE= \frac{AB*BC}{AC}\)=\(\frac{4*3}{5}\)=2.4


stne
nick1816
BE=3*4/5=2.4
EC= BC^2/AC
AC^2=3^2+4^2
AC=5
EC= 4^2/5=3.2

Perpendicular distance of E from BC= 2.4*3.2/4= 1.92
Perpendicular distance of E from AD=3-1.92=1.08

Area of AED= 0.5*1.08*4=2.16 or 54/25

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Hi nick1816,
A little elaboration could help. How did you calculate BE?
Show more

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lacktutor
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 16 Jul 2019, 13:38
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Please find the attached file

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File comment: PFA
C72E6766-7B43-48A8-BE91-FD9A745D1660.jpeg
C72E6766-7B43-48A8-BE91-FD9A745D1660.jpeg [ 2.13 MiB | Viewed 7952 times ]

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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 17 Jul 2019, 12:34
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nick1816
Area of triangle ABC= 1/2*AB*BC=1/2*AC*BE
\(BE= \frac{AB*AC}{AC}\)=\(\frac{4*3}{5}\)=2.4


stne
nick1816
BE=3*4/5=2.4
EC= BC^2/AC
AC^2=3^2+4^2
AC=5
EC= 4^2/5=3.2

Perpendicular distance of E from BC= 2.4*3.2/4= 1.92
Perpendicular distance of E from AD=3-1.92=1.08

Area of AED= 0.5*1.08*4=2.16 or 54/25



Hi nick1816,
A little elaboration could help. How did you calculate BE?
Show more

Hi nick1816,

Thanks man, but there is a small typo above , it should be BE=\(\frac{AB*BC}{AC}\), Hope you'll correct it. Thanks.
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 13 Jul 2023, 22:29
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1) use proportions of similar triangles.

AB:BC:AC = 3 : 4 : 5
=BE:EC:CB = x : y : 4

X = 3*4/5 = 12/5,
Y = 4*4/5 = 16/5


2) use formula of same area (1/2 * base * height)

X*Y = (12*16)/25
= 4*h1 (hight OF BEC)
---> h1 = 48/25 *at this point it's very likely that the answer is a fraction with the denominator 25 ...

h2 = 3 - 48/25 = 27/25

area of AED = 1/2 * 4 * 27/25 = 54/25
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Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe 23 Jul 2023, 23:05
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Concept of Similar Triangles, Rectangles and Pythagoras Theorem is tested in this question.

AB = 3
BC = 4
Therefore, AC = sqrt(4^2+3^2) = 5

Angle BEC = Angle ABC
Angle ECB = Angle BCA
Therefore the third angles of both the triangle will be same.
This proves that the Triangles ABC and BEC are similar triangles.

The sides of similar triangles are in proportion.
BE/AB = EC/BC = BC/AC
BE/3 = EC/4 = 4/5
Therefore BE = 12/5 and EC = 16/5
AE = AC - EC = 5 - 16/5 = 9/5

Triangle DAC and Triangle BCA
This is because the diagonal of a rectangle divides the rectangle into 2 congruent triangles since length of all the sides of both the triangles are equal.
This means the height of triangle BCA (which is BE since perpendicular) and DAC will be the same which implies the same height for triangle ADE.

Area of triangle ADE = 1/2 * BE(height) * AE(base) = 1/2 * 12/5 * 9/5 = 54/25
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