Areas of parallelograms and triangles · 4 marks

In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (see the figure). Prove that Area (∆BPQ) = 12 Area (∆ABC).

Answer: Join CD. △DPQ and △DPC have the same base DP and lie between the parallels DP and QC, so they have equal area. Then ar(BPQ) = ar(BPD) + ar(DPQ) = ar(BPD) + ar(DPC) = ar(BDC) = 12ar(ABC) (CD is a median).

Step-by-step solution

Given: D is the midpoint of AB; P on BC; Q on AB with CQ ∥ PD
To find: Prove ar(△BPQ) = 12ar(△ABC)

Idea: Triangles on the same base and between the same parallels have equal areas; a median halves a triangle.

Join CD. Since D is the midpoint of AB, CD is a median of △ABC: ar(△BDC) = 12ar(△ABC).1 mark
△DPQ and △DPC have the same base DP, and their third vertices Q and C lie on a line parallel to DP (CQ ∥ PD). So they have equal heights and ar(△DPQ) = ar(△DPC).1 mark
ar(△BPQ) = ar(△BPD) + ar(△DPQ) (Q lies beyond D on BA, so DP splits △BPQ).1 mark
= ar(△BPD) + ar(△DPC) = ar(△BDC) = 12ar(△ABC). Proved.1 mark

Area(ΔBPQ) = area(ΔBPD) + area(ΔDPQ) = area(ΔBPD) + area(ΔDPC) = area(ΔBDC) = ½ area(ΔABC).

Check: Coordinates B(0, 0), C(6, 0), A(2, 6): D = (1, 3). Take P = (4, 0); CQ ∥ PD gives Q = (1.5, 4.5) on AB. ar(△BPQ) = 12 × 4 × 4.5 = 9 and ar(△ABC) = 12 × 6 × 6 = 18 ✓.

Answer to write in the exam

Join CD. ar(△BDC) = 12ar(△ABC) (CD is a median)

ar(△DPQ) = ar(△DPC) (same base DP, between parallels DP and QC)

ar(△BPQ) = ar(△BPD) + ar(△DPQ)

= ar(△BPD) + ar(△DPC) = ar(△BDC)

∴ ar(△BPQ) = 12ar(△ABC)

Common mistakes that cost marks

Forgetting to join CD; without it there is no triangle to compare △DPQ with.
Saying △DPQ ≅ △DPC. They have equal area, not equal shape.
Writing ar(△BPQ) = ar(△BPD) − ar(△DPQ). Q is beyond D, so the areas add.

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): If XY ∥ PQ, then ar(△PQX) = ar(△PQY).
Reason (R): Triangles on the same base and between the same parallels are equal in area.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Show answer

(A) Both A and R are true, and R is the correct explanation of A.
X and Y are at the same distance from PQ, so the triangles on base PQ have equal heights. R explains A.

Try one yourself

In △ABC, D is the midpoint of AB and ar(△ABC) = 30 cm². With P, Q as in this question, find ar(△BPQ) and ar(quadrilateral AQPC).