Right answer is **(D)** **16 : 81**

Given, Sides of two similar triangles are in the ratio 4 : 9.

Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And \( \dfrac{AB}{DE}\) = \( \dfrac{AC}{DF}\) = \( \dfrac{BC}{EF}\) = \( \dfrac{4}{9}\)

As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

\(
\dfrac{ Area(ΔABC)}{Area(ΔDEF)}\) = \( \dfrac{AB^2}{DE^2}\)^{ }

\(
\dfrac{ Area(ΔABC)}{Area(ΔDEF)}\) = (\( \dfrac{4}{9}\))^{2 }= \( \dfrac{16}{81}\) = 16:81

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm^{2}. Find ar(∆COD)

Find the length of the altitude of an equilateral triangle of side 2a cm.

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.