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Question
In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED
Solution
∆ADE ~ ∆ABC.
`\therefore \frac{ar(\Delta ADE)}{ar(\Delta ABC)}=(DE^2)/(BC^2)=((DE)/(BC))^2=(3/5)^2=9/25`
Let ar (∆ADE) = 9x sq units
Then, ar (∆ABC) = 25x sq units
ar (trap. BCED) = ar (∆ABC) – ar (∆ADE)
= (25x – 9x) = (16x) sq units
`\therefore \frac{ar(\Delta ADE)}{ar(trap.BCED)}=\frac{9x}{16x}=\frac{9}{16}`
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