In Fig, DE || BC, If DE: BC = 3: 5. Calculate the ratio of the areas

1.	In Fig, DE \|\| BC, If DE: BC = 3: 5. Calculate the ratio of the areas of ΔADE and the trapezium BCED.
Answer» Given, DE ∥ BC. In ΔADE and ΔABC We know that, ∠ADE = ∠B [Corresponding angles] ∠DAE = ∠BAC [Common] Hence, ΔADE ~ ΔABC (AA Similarity) Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have, \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{DE^2}{BC^2}\) \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{3^2}{5^2}\) \(\frac{Ar(ΔADE)}{Ar(ΔABC)}\) = \(\frac{9}{25}\) Assume that the area of ΔADE = 9 x sq units And, area of ΔABC = 25 x sq units So, Area of trapezium BCED = Area of ΔABC – Area of ΔADE = 25x – 9x = 16x Now, \(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9x}{16x}\) \(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9}{16}\)

In Fig, DE || BC, If DE: BC = 3: 5. Calculate the ratio of the areas of ΔADE and the trapezium BCED.

Answer»

Given,

DE ∥ BC.

In ΔADE and ΔABC

We know that,

∠ADE = ∠B [Corresponding angles]

∠DAE = ∠BAC [Common]

Hence, ΔADE ~ ΔABC (AA Similarity)

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,

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

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

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

Assume that the area of ΔADE = 9 x sq units

And, area of ΔABC = 25 x sq units

So, Area of trapezium BCED = Area of ΔABC – Area of ΔADE

= 25x – 9x

= 16x

Now, \(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9x}{16x}\)

\(\frac{Ar(ΔADE)}{Ar(trap\, BCED)}\) = \(\frac{9}{16}\)