Question

ABCD is a parallelogram and BPC is a triangle with P falling on AD. If the area of parallelogram ABCD=26 cm square, find the area of triangle BPC.

Answer 1

the area of BPC will be 13 i.e. half of 26

according to the theorem that the area of a triangle lying between two parallel lines of a parallelogram is having area half the parallelogram..


hope it helps u..

Answer 2

The area of triangle BPC is 13$(cm^{2})$.

Step-by-step explanation:

1. Area of parallelogram ABCD = 26 $(cm^{2})$.

2. $\Delta BCP$ which base(BC) is a side of parallelogram. It vertices P lie on opposite side of parallelogram AD.

3. If a triangle and parallelogram with have common side and both triangle and parallelogram lie between two parallel line, relation between area of triangle and area of parallelogram is given below.

   $area of triangle=\frac{1}{2}\times area of parallelogram$    ...1)

4. So  from equation 1)

  $area of \Delta BCP=\frac{1}{2}\times area of ABCD$

   $area of \Delta BCP=\frac{1}{2}\times 26=13(cm^{2})$