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.)
Step-by-step solution
To find: Prove ar(PQRS) = 12 ar(ABCD)
Idea: A median halves a triangle’s area. Using it twice, the corner triangle APS is a quarter of △ABD.
- In △ABD, P is the midpoint of AB, so DP is a median: ar(△APD) = 12ar(△ABD). In △APD, S is the midpoint of AD, so PS is a median: ar(△APS) = 12ar(△APD) = 14ar(△ABD).1 mark
- In the same way ar(△CQR) = 14ar(△CBD). Adding: ar(△APS) + ar(△CQR) = 14[ar(△ABD) + ar(△CBD)] = 14ar(ABCD).1 mark
- Using the other diagonal AC: ar(△BPQ) + ar(△DRS) = 14[ar(△BAC) + ar(△DAC)] = 14ar(ABCD).1 mark
- The four corner triangles together = 12ar(ABCD). PQRS is what is left: ar(PQRS) = ar(ABCD) − 12ar(ABCD) = 12ar(ABCD). Proved (for a 4-gon whose corners all point outwards).1 mark
Check: Square of side 2: the midpoints form a square of side √2, area 2 = 12 × 4 ✓.
Answer to write in the exam
In △ABD: ar(△APD) = 12ar(△ABD) (DP median)
In △APD: ar(△APS) = 12ar(△APD) = 14ar(△ABD) (PS median)
Similarly ar(△CQR) = 14ar(△CBD) ⇒ ar(△APS) + ar(△CQR) = 14ar(ABCD)
Similarly ar(△BPQ) + ar(△DRS) = 14ar(ABCD)
ar(PQRS) = ar(ABCD) − 12ar(ABCD)
∴ ar(PQRS) = 12ar(ABCD)
Common mistakes that cost marks
- Saying ar(△APS) = 12ar(△ABD) after one median step; it takes two halvings.
- Using the same diagonal for all four corners; corners A and C go with BD, corners B and D with AC.
- Assuming ABCD is a parallelogram or a square; the result holds for any such 4-gon.
How this can come in the exam
The midpoints of the sides of a 4-gon of area 50 cm2 are joined in order. The area of the figure formed is
- 12.5 cm2
- 20 cm2
- 25 cm2
- 40 cm2
Show answer
(C) 25 cm2
It is half of the 4-gon: 25 cm2.
Try one yourself
A rectangle is 12 cm by 8 cm. Its midpoints are joined in order. Find the area of the rhombus formed, in two ways.
Show answer
Half the rectangle: 12 × 96 = 48. Diagonals 12 and 8: 12 × 12 × 8 = 48 cm2 ✓.
More questions like this
- 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).
- Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?
- Such reasoning suggests that for a circle too, if C = circumference and A = area, the ratio C2 : A must be some fixed constant. But what is this constant?