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Q.

In ΔABC point D is on side AB and point E on side AC such that quadrilateral. BCED is a trapezium. If DE:BC=3:5 then find the ratio of area of ΔADE and Quadrilateral.BCED.

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a

7:17

b

8:18

c

8:10

d

9:16

answer is D.

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Detailed Solution

https://www.vedantu.com/question-sets/addbaf23-82a5-4ca7-9273-030c61e21e3c818925047300488851.png

Since, the quadrilateral BCED is a trapezium.
Therefore, line DE is parallel to BC.
Now, in triangle ADE and triangle ABC. We have
∠ADE=∠ABC(corresponding angles)
∠AED=ACB (corresponding angles)
Therefore, ΔADE∼ΔABC
Now, by area theorem. We have a ratio of area of two similar triangles equal to the ratio of squares or their corresponding sides.
Therefore, we have,

\frac{ar (ΔADE)}{ar (ΔABC)} = {(\frac{DE}{BC})}^{2}

But it is given that ratio of sides DE to BC is 3:5 or we can write it as:

\frac{DE}{BC} = \frac{3}{5}

Using this value of

\frac{DE}{BC}

in the above formed equation. We have

\frac{ar (ΔADE)}{ar (ΔABC)} = {(\frac{3}{5})}^{2}

\frac{ar (ΔADE)}{ar (ΔABC)} = \frac{9}{25}

\frac{ar (ΔABC)}{ar (ΔADE)} = \frac{25}{9}

ΔABC=ΔADE+Quadrilateral (BCDE) (Given)
Using this in the above formed equation. We have,

\frac{ΔADE + Quadrilateral . (BCDE)}{ar (ΔADE)} = \frac{25}{9}

\Rightarrow \frac{ΔADE}{ΔADE} + \frac{Quadrilateral . (BCDE)}{ΔADE} = \frac{25}{9}

\Rightarrow 1 + \frac{Quadrilateral (BCDE)}{ΔADE} = \frac{25}{9}

\Rightarrow \frac{Quadrilateral (BCDE)}{ΔADE} = \frac{25}{9} - 1

\Rightarrow \frac{Quadrilateral (BCDE)}{ΔADE} = \frac{25 - 9}{9}

\Rightarrow \frac{Quadrilateral (BCDE)}{ΔADE} = \frac{16}{9}

\Rightarrow

\frac{ΔADE}{Quadrilateral (BCDE)} = \frac{9}{16}

Therefore, from above we see that the ratio of area of triangle ADE to quadrilateral or trapezium is

\frac{9}{16}

.
So, the correct answer is “

\frac{9}{16}

”.

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