ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = 1/8 of the area of parallelogram ABCD.

Question

KnowledgeBoat · Accepted Answer

From figure, ABCD is the parallelogram and BD is the diagonal of the parallelogram. ∴ BD divides || gm ABCD into two triangles of equal area. ∴ Area of Δ ABD = Area of Δ DBC = Area of || gm ABCD ..........(1) Since, P is the mid-point of AB, ∴ DP is the median of Δ ABD. ∴ Area of Δ APD = Area of Δ DPB = Area of Δ ABD (Median divides triangle into two triangles of equal area) ⇒ Area of Δ APD = Area of Δ ABD ⇒ Area of Δ APD = Area of || gm ABCD [From equation (1)] ⇒ Area of Δ APD = Area of || gm ABCD .......(2) In Δ APD, Q is the mid-point of AD. ∴ PQ is the median. ∴ Area of Δ APQ = Area of Δ DPQ = Area of Δ APD ⇒ Area of Δ APQ = Area of Δ APD ⇒ Area of Δ APQ = Area of || gm ABCD [From equation (2)] ⇒ Area of Δ APQ = Area of || gm ABCD. Hence, proved that area of Δ APQ = area of || gm ABCD.

Mathematics

ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = 1/8 of the area of parallelogram ABCD.

Theorems on Area

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