Question

(i)Write a property of diagonals of parallelogram.

(ii)ABCD is a parallelogram in which CD is produced to P.DC =6 cm and height BQ = 4 cm. Find the area of \(\Delta APB\)

4cm 

Answer 1

(i) Diagonals of parallelograms bisects each other.

(ii)Theorem used:

Area of triangle is half the area of parallelogram, if both are on same base and between the same parallels.

From the given figure:

AB||CD (ABCD is a parallelogram)

AB=CD=6cm (given)

Altitude BQ=4cm (given)

$\Delta ABP$ and parallelogram ABCD are on same base and between the same parallels.

Area of $\Delta APB=\frac{1}{2}\times$ Area of ABCD

Area of ABCD = Base$\times$Altitude

Area of ABCD = 6$\times$4=24 cm${}^2$

Area of $\Delta APB=\frac{1}{2}\times$Area of ABCD

Area of $\Delta APB=\frac{1}{2}\times$24

Area of $\Delta APB=12{\text{cm}}^2$