Question 17 of 42advanced🔍 AnalyzeLong Answer5 marks

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

Correct Answer

We are given a parallelogram PQRS and O is any point on its diagonal PR. We need to prove that the areas of triangles PSO and PQO are equal.

First, a diagonal of a parallelogram divides it into two congruent triangles. Therefore, diagonal PR divides PQRS into Δ\Delta PQR and Δ\Delta PSR, which have equal areas.

Since Area(Δ\Delta PQR) = Area(Δ\Delta PSR), and both share the same base PR, their corresponding heights must be equal. This means the perpendicular distance from Q to PR equals the perpendicular distance from S to PR. Let this height be hh.

Now, consider Δ\Delta PQO and Δ\Delta PSO. Since O is on diagonal PR, both triangles share the same base PO. Their corresponding heights are the perpendicular distances from Q and S to PR, which are both hh.

Using the area of a triangle formula:

12×base×height\frac{1}{2} \times \text{base} \times \text{height}

we get:

Area(Δ PQO)=12×PO×h\text{Area}(\Delta \text{ PQO}) = \frac{1}{2} \times \text{PO} \times h

and

Area(Δ PSO)=12×PO×h\text{Area}(\Delta \text{ PSO}) = \frac{1}{2} \times \text{PO} \times h

Thus, Area(Δ\Delta PQO) = Area(Δ\Delta PSO). Hence proved.

Exercise: Exercise Set 6.2 | Q: 7 | (Page 25)
For More Understanding

Explanation

The proof relies on the property that a diagonal of a parallelogram bisects it into two triangles of equal area. Because these two triangles share the same base (the diagonal), their heights must be equal.

When a point O is chosen on the diagonal, the two new triangles formed (PSO and PQO) share a common base (a segment of the diagonal) and have the same heights as the original large triangles, making their areas equal.

Solution Steps

  1. Step 1: State that diagonal PR divides parallelogram PQRS into two congruent triangles, Δ\Delta PQR and Δ\Delta PSR, so Area(Δ\Delta PQR) = Area(Δ\Delta PSR).

  2. Step 2: Conclude that since Δ\Delta PQR and Δ\Delta PSR share the same base PR, their corresponding heights (perpendicular distances from Q and S to PR) are equal, say hh.

  3. Step 3: Consider Δ\Delta PQO and Δ\Delta PSO. Since O is on PR, both triangles share the same base PO.

  4. Step 4: The height of Δ\Delta PQO is the perpendicular distance from Q to PR, which is hh. The height of Δ\Delta PSO is the perpendicular distance from S to PR, which is also hh.

  5. Step 5: Calculate areas:

    Area(Δ PQO)=12×PO×h\text{Area}(\Delta \text{ PQO}) = \frac{1}{2} \times \text{PO} \times h

    and

    Area(Δ PSO)=12×PO×h\text{Area}(\Delta \text{ PSO}) = \frac{1}{2} \times \text{PO} \times h

    Therefore, Area(Δ\Delta PQO) = Area(Δ\Delta PSO).

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