ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (∆PCD): area (∆QCD)?
Answer: Both triangles have base CD and the same height (the distance between AB and CD), so their areas are equal: ratio 1 : 1 (each is half the parallelogram).
Step-by-step solution
Idea: Triangles on the same base and between the same parallels have equal area.
- △PCD and △QCD both have base CD. P and Q lie on AB, and AB ∥ CD, so each apex is at the same distance h from CD.1 mark
- area(△PCD) = 12 × CD × h = area(△QCD). So the ratio is 1 : 1. (Each is half of the parallelogram, whose area is CD × h.)1 mark
area(ΔPCD) : area(ΔQCD) = 1 : 1.
Answer to write in the exam
Common base CD; AB ∥ CD ⇒ equal heights h
ar(△PCD) = 12 × CD × h = ar(△QCD)
∴ ar(△PCD) : ar(△QCD) = 1 : 1
Common mistakes that cost marks
- Thinking the ratio depends on where P and Q are. Their distance from CD is the same anywhere on AB.
- Measuring the slanting sides PC or PD instead of the perpendicular height.
How this can come in the exam
MCQ (1 mark)
ABCD is a parallelogram of area 60 cm2 and X is any point on AB. The area of △XCD is
- 15 cm2
- 20 cm2
- 30 cm2
- 60 cm2
Show answer
(C) 30 cm2
Same base and same height as the parallelogram, so half of it: 30 cm2.
Try one yourself
PQRS is a parallelogram with SR = 9 cm and height 4 cm. M is any point on PQ. Find area(△MSR).
Show answer
12 × 9 × 4 = 18 cm2.
More questions like this
- O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
- If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered later, when quadrilaterals are studied.)
- In ∆ABC, the midpoint of BC is D (see the figure). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).
- Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (see the figure). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?
- 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).