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In the given figure, DE || BC and DE : BC = 3 : 5 the ratio of the areas of Δ ADE and the trapezium BCED.
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9 25 -
12 25 -
3 4 -
9 16 - None of these
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Correct Option: D
From above given figure , we can see that
Given , DE : BC = 3 : 5
ΔADE ∼ ΔABC (AA Similarity)
we know that ,
| ∴ | ar(∆ADE) | = | DE2 | = | 9 |
| ar(∆ABC) | BC2 | 25 |
Let ar(ΔADE) = 9x sq units.
Then, ar(ΔABC) = 25x sq. units
Now ar(trap. BCED) = ar(ΔABC) − ar(ΔADE) = 25x - 9x = 16x sq. units
| Hence | ar(∆ADE) | = | 9x | = | 9 |
| ar(∆BCED) | 16x | 16 |
