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# Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe

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Math Expert
Joined: 02 Sep 2009
Posts: 58142
Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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13 May 2019, 01:03
00:00

Difficulty:

85% (hard)

Question Stats:

43% (02:46) correct 58% (02:58) wrong based on 40 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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Director
Joined: 19 Oct 2018
Posts: 879
Location: India
Re: Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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14 May 2019, 03:55
1
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

Bunuel wrote:
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Director
Joined: 27 May 2012
Posts: 852
Re: Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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16 Jul 2019, 12:41
nick1816 wrote:
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

Bunuel wrote:
________________________
BUMPING FOR DISCUSSION.

Hi nick1816,
A little elaboration could help. How did you calculate BE?
_________________
- Stne
Director
Joined: 19 Oct 2018
Posts: 879
Location: India
Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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Updated on: 17 Jul 2019, 12:37
1
Area of triangle ABC= 1/2*AB*BC=1/2*AC*BE
$$BE= \frac{AB*BC}{AC}$$=$$\frac{4*3}{5}$$=2.4

stne wrote:
nick1816 wrote:
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

Bunuel wrote:
________________________
BUMPING FOR DISCUSSION.

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

Attachments

y.png [ 8.22 KiB | Viewed 437 times ]

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.
Manager
Joined: 25 Jul 2018
Posts: 211
Re: Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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16 Jul 2019, 13:38
1
Please find the attached file

Posted from my mobile device
Attachments

File comment: PFA

C72E6766-7B43-48A8-BE91-FD9A745D1660.jpeg [ 2.13 MiB | Viewed 431 times ]

Director
Joined: 27 May 2012
Posts: 852
Re: Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe  [#permalink]

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17 Jul 2019, 12:34
nick1816 wrote:
Area of triangle ABC= 1/2*AB*BC=1/2*AC*BE
$$BE= \frac{AB*AC}{AC}$$=$$\frac{4*3}{5}$$=2.4

stne wrote:
nick1816 wrote:
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?

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.
_________________
- Stne
Re: Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe   [#permalink] 17 Jul 2019, 12:34
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# Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpe

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