O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Question

Homi · Accepted Answer

Triangles with the same base and equal heights have equal areas.

Step 1 — Identify common base

Let's look at $\triangle PSO$ and $\triangle PQO$ . Point O lies on the diagonal PR. Both triangles share the side PO. So, PO is their common base.

Diagram 1

Step 2 — Compare heights

PQRS is a parallelogram. The diagonal PR divides it. This means $\triangle PSR$ and $\triangle PQR$ have equal areas. They share the same base PR. So, their corresponding heights must be equal. Let $h$ be this common height. This $h$ is the perpendicular distance from S to PR. It is also the perpendicular distance from Q to PR. For $\triangle PSO$ , its base is PO. Its height is the perpendicular distance from S to PR, which is $h$ . For $\triangle PQO$ , its base is PO. Its height is the perpendicular distance from Q to PR, which is $h$ .

Step 3 — Conclude areas are equal

Both $\triangle PSO$ and $\triangle PQO$ have the same base PO. They also have the same height $h$ . The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$ . $\text{Area}(\triangle PSO) = \frac{1}{2} \times \text{PO} \times h$ $\text{Area}(\triangle PQO) = \frac{1}{2} \times \text{PO} \times h$ Therefore, we can say:

$\boxed{\text{Area}(\triangle PSO) = \text{Area}(\triangle PQO)}$

Answer

The areas of triangles PSO and PQO are equal.

Question 7

Answer

More questions in Exercise 6.2