Answer :
Triangles APB and BPC share the same base and height, resulting in their areas being equal. Therefore, the area of triangle BPC is also 24.5 cm².
In parallelogram ABCD, O is the intersection point of its diagonals, making O the midpoint of both diagonals. P is a point on DO. We are given that the area of triangle APB is 24.5 cm² and need to find the area of triangle BPC.
Since diagonals of a parallelogram bisect each other, O is the midpoint, meaning triangles AOB, BOC, COD, and DOA each have equal areas as they are formed by the halving of diagonals.
Step-by-Step Solution:
- Triangles AOB and COD are equal in area because diagonals split the parallelogram into two equal areas.
- Area of triangle AOB (or BOC, COD, DOA) is half of the total area of the parallelogram; thus, they are equal in area.
- Since triangles APB and BPC share the same base PB and height from point O (intersection), they form equal areas within the parallelogram's divided sections.
Given area of triangle APB is 24.5 cm², the area of triangle BPC must also be 24.5 cm².
