NCERT Solutions for Class 9th Maths Chapter 6 Think and Reflect — Area of a Triangle

Book page 134 Updated on2026-09-08

Q1.

Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g., 1. A square and non-square rectangle with equal area, 2. Two triangles with different shapes but equal area, 3. A triangle and a square with equal area. Formulate a conjecture of your own about this.

Answer

Conjecture: yes — equal area is enough. Any two polygons of the same area can be cut into finitely many pieces that reassemble into each other. Here is how the three suggested cases go.

1. Square ↔ non-square rectangle. Take a 4 × 4 square and a 8 × 2 rectangle, both of area 16. Cut the rectangle in half across its length to get two 4 × 2 pieces and stack them: you get the 4 × 4 square. Two pieces.

For a harder pair, such as 16 × 1 and 4 × 4, use the "staircase" cut: cut a step pattern along the rectangle and slide the upper part along. In general a rectangle can be turned into a square of the same area by the P-slide — and note that Baudhāyana's construction in Section 6.9 does exactly this job with straightedge and compass.

2. Two triangles of equal area. First turn each triangle into a parallelogram: cut along the midline (joining the midpoints of two sides) and rotate the small top triangle through 180°. Now both are parallelograms of the same area. Shear each into a rectangle of a common base, and the two rectangles are then equal in both dimensions.

3. Triangle ↔ square. Chain the two previous ideas: triangle → parallelogram → rectangle → square. Each step costs only a few pieces.

Why it happens: the whole method is a chain of three moves, each of which preserves area — cut and reflect, shear, and the rectangle-to-square P-slide. Because every polygon can be cut into triangles, and every triangle can be pushed through the chain to a square of the same area, any polygon can be turned into "the square of its area". Do this to P, do it to Q, and since the two squares are identical you can go P → square → Q.

Did you know? Your conjecture is a real theorem: the Wallace–Bolyai–Gerwien theorem (1807–1835). Hilbert asked the same question for solids in 1900, and Max Dehn answered it within the year — no: a regular tetrahedron and a cube of equal volume cannot be cut into each other.

Q2.

Think of various rectangles with perimeter 40 units (the sides do not have to be integers). 1. How many such rectangles are there? 2. Among them, is there one whose area is the largest? What are its dimensions? 3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?

Answer

1. Infinitely many. 2. Yes — the 10 × 10 square, area 100 sq. units. 3. No — there is no smallest.

Perimeter 40 ⟹ 2(x + y) = 40 ⟹ x + y = 20 ⟹ y = 20 − x
Any x with 0 < x < 20 gives a rectangle, so there are infinitely many.

Area A(x) = x(20 − x) = 20x − x²

Largest area. Complete the square:

A = 20x − x²
= −(x² − 20x)
= −(x² − 20x + 100) + 100
= 100 − (x − 10)²

(x − 10)² ≥ 0 always, and equals 0 only when x = 10
∴ A ≤ 100, with A = 100 exactly when x = 10, y = 10

So the largest area is 100 sq. units, from the 10 × 10 square.

x	y = 20 − x	Area
1	19	19
5	15	75
9	11	99
10	10	100
15	5	75
19.9	0.1	1.99
19.99	0.01	0.1999

Smallest area. There is none. As x creeps up towards 20, the rectangle becomes a longer and longer splinter and its area drops towards 0 — but never reaches it, because x = 20 would leave no rectangle at all. For any rectangle you name, x = 19.999… gives one with smaller area.

Why it happens: the surprise cuts both ways. Fixing the perimeter does not fix the area — that alone is worth remembering, and it is the same lesson as "the sides do not determine a parallelogram's area". And while the maximum is attained at the most symmetric shape, the minimum is only approached, never reached: the set of allowed values of x is the open interval (0, 20), and an open interval has no endpoint to sit at.

Tip: "among all shapes with a given perimeter, the most symmetric one has the greatest area" is a genuine principle. Among all rectangles it picks the square; among all plane shapes it picks the circle. That is the isoperimetric property — and it explains why a soap bubble is round.

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