In the given figure, PQR is a triangle whose area is 75 cm². S and T are mid-points of sides PQ and PR respectively. Calculate the area of quadrilateral SQRT.

56.25 cm²

Reason

Using midpoint theorem, which states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half the length of the third side.

Given,

S and T are the midpoints of PQ and PR.

ST || QR and ST = $\dfrac{1}{2}QR$

$\Rightarrow \dfrac{ST}{QR} = \dfrac{1}{2}$

Draw PM perpendicular to QR. Let PM meets ST at N.

As, ST || QR

∴ ∠SNP = ∠QMN = 90°

As, S is mid-point of PQ and SN || QM,

∴ N is mid-point of PM.

Area of triangle = $\dfrac{1}{2} \times \text{ base} \times \text{ height}$

$\Rightarrow \text{Area of △PST }= \dfrac{1}{2} \times ST \times PN \\[1em] \Rightarrow\text{Area of △PST }= \dfrac{1}{2} \times \dfrac{1}{2} QR \times \dfrac{1}{2}PM \\[1em] \Rightarrow\text{Area of △PST }= \dfrac{1}{4} \times \dfrac{1}{2} \times QR \times PM \\[1em] \Rightarrow\text{Area of △PST }= \dfrac{1}{4} \times \text{ Area of △PQR} = \dfrac{1}{4} \times 75 = 18.75 \text{ cm}^2.$

Area of trapezium STRQ = Area of △PQR - Area of △PST = 75 - 18.75 = 56.25 cm².