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In a triangle ABC, Point D is on side AB and point E is on side AC,

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In a triangle ABC, Point D is on side AB and point E is on side AC,  [#permalink]

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New post 12 May 2016, 03:55
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In a triangle ABC, Point D is on side AB and point E is on side AC, such that BCDE is a trapezium. If DE: BC=3:5 then Calculate the ratio of the area of ΔADE and that of trapezium BCED.

A.3:4
B.9:16
C.3:5
D.9:25
E.5:9
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Posts: 7199
Re: In a triangle ABC, Point D is on side AB and point E is on side AC,  [#permalink]

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New post 12 May 2016, 05:54
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techiesam wrote:
In a triangle ABC, Point D is on side AB and point E is on side AC, such that BCDE is a trapezium. If DE: BC=3:5 then Calculate the ratio of the area of ΔADE and that of trapezium BCED.

A.3:4
B.9:16
C.3:5
D.9:25
E.5:9


Hi,
what you get is Triangle ABC and triangle ADE as two similar triangle..where ABC is the larger triangle..
RATIO of areas is SQUARE of sides of two triangles..

SO Area of ADE/Area of ABC = \((\frac{3}{5})^2 = \frac{9}{25}\) = 9x/25x...

But area of BCDE = area of ABC - Area of ADE = 25x-9x = 16x..
so Area of ADE/Area of BCDE = \(\frac{9x}{16x}\) = 9/16...
B
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Re: In a triangle ABC, Point D is on side AB and point E is on side AC,  [#permalink]

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New post 12 May 2016, 06:17
I solved this in the following manner: (I am assuming the folks reading this have drawn the appropriate figure)

Let the height of triangle ADE be x and the height of the trapezium be y
Now, let's assume that 3 and 5 are the actual lengths of sides DE and BC respectively

The area of triangle ADE is 1/2*3x
The area of trapezium DECB is 1/2*(3+5)y i.e., 1/2*8y

The ratio of their areas is thus: 3x/2 divided by 8y/2

This gives us: 3x/8y.

Now look through the options and see which one of the values given can possibly fit this expression. 'B' is the only one!
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Re: In a triangle ABC, Point D is on side AB and point E is on side AC,  [#permalink]

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New post 03 Jan 2018, 03:06
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Re: In a triangle ABC, Point D is on side AB and point E is on side AC, &nbs [#permalink] 03 Jan 2018, 03:06
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