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)?
Answer: With side a, the heights of △PAB and △PCD from P add up to a, so red = 12a × a = 12a2. Then green = 12a2 too. Ratio 1 : 1, wherever P is.
Step-by-step solution
Idea: AB and CD are opposite (parallel) sides, so P’s distances to them add up to the side of the square.
- Let the side be a, and let P be at distance x from AB and y from CD. AB ∥ CD and they are a apart, so x + y = a.1 mark
- Red = ar(△PAB) + ar(△PCD) = 12ax + 12ay = 12a(x + y) = 12a2.1 mark
- Green = whole square − red = a2 − 12a2 = 12a2. So red : green = 1 : 1.1 mark
The red and green regions have equal areas: the ratio is 1 : 1, for every point P inside the square.
Check: If P is the centre, all four triangles are equal (14a2 each), so red = green ✓.
Answer to write in the exam
Let side = a; distances of P from AB and CD = x, y; x + y = a
Red = 12ax + 12ay = 12a(x + y) = 12a2
Green = a2 − 12a2 = 12a2
∴ Red : Green = 1 : 1
Common mistakes that cost marks
- Assuming P is the centre. The result holds for any P inside.
- Pairing triangles on adjacent sides (PAB with PBC); the proof needs the opposite sides AB and CD.
How this can come in the exam
MCQ (1 mark)
P is any point inside a rectangle of area 48 cm2, joined to the four corners. The sum of the areas of the two triangles on opposite sides is
- 12 cm2
- 16 cm2
- 24 cm2
- Depends on P
Show answer
(C) 24 cm2
The two heights add up to the distance between the opposite sides, so the sum is half the rectangle: 24 cm2.
Try one yourself
A square has side 10 cm. P inside it is 3 cm from AB. Find ar(△PAB), ar(△PCD) and their sum.
Show answer
12 × 10 × 3 = 15; 12 × 10 × 7 = 35; sum 50 cm2 = half of 100.
More questions like this
- 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?
- As the slices become smaller and smaller, the arcs in the figure (B) become more and more closer to a line. This makes the figure more and more closer to a parallelogram with base = half the circumference (why?) = πr, height = radius r.
- Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.