$A B C D$ is a parallelogram with side $A B = 10 c m$ . Its diagonals $A C$ and $B D$ are of length $12 c m$ and $16 c m$ respectively. Find the area of a parallelogram $A B C D$

Open in App

Solution

$\begin{matrix} A C = 12 c m B D = 16 c m A r e a o f Δ P A B = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{12 (12 - 6) (12 - 8) (12 - 10)} & P A = \frac{1}{2} A C = 6 = \sqrt{12 \times 6 \times 4 \times 2} & P B = \frac{1}{2} B D = 8 = \sqrt{576} = 24 c m^{2} & S = \frac{6 + 8 + 10}{2} = 12 A r e a o f Δ P A B = \frac{1}{2} \times P Q \times A B P Q = \frac{24 \times 2}{10} = 4.8 c m ∴ A E = 2 P Q = 2 \times 4.8 = 9.6 c m A r e a o f r e c tan g l e A B C D = 10 \times 9.6 = 96 c m^{2} \end{matrix}$

Suggest Corrections

Similar questions

A B = 12 c m

B C = 10 c m

A C = 16 c m

A B C D, A B = 16 c m, B C = 12 c m

A C = 20 c m

10 c m

12 c m

16 c m

In parallelogram ABCD, AB = 16 cm, BC = 12 cm and diagonal AC = 20 cm. Find the area of the parallelogram ABCD.

Join BYJU'S Learning Program