NCERT Solutions for Class 9th Maths Chapter 6 In-text Questions — Area of a Triangle

Book page 133 Updated on2026-09-08

Q1.

Do you see why the two triangles fit together to make a parallelogram? (If you study the angles in the figure (e.g., ∠B'C'A' and ∠BCA), you will see why this is so. Keep in mind the criterion by which we check whether two lines are parallel.)

Answer

Yes — because a pair of equal alternate angles forces the opposite sides to be parallel.

Take ΔABC and a congruent copy ΔA′B′C′, and place the copy so that side B′C′ falls exactly on side CB (with B′ on C and C′ on B). Call the new figure ABDC, where D is the image of A.

Congruence gives ∠A′C′B′ = ∠ACB, i.e. ∠DBC = ∠ACB.
These are alternate angles for the lines AC and BD with transversal BC.
Equal alternate angles ⟹ AC ∥ BD.

Congruence also gives A′C′ = AC, i.e. BD = AC.
A 4-gon with one pair of sides both equal and parallel is a parallelogram.
∴ ABDC is a parallelogram.

Hence area of parallelogram = 2 × area of triangle, and since the parallelogram has the same base BC = b and the same height h as the triangle,

2 × ar(ΔABC) = bh ⟹ ar(ΔABC) = ½ bh

Why it happens: a half-turn (rotation through 180°) about the midpoint of BC carries the triangle onto its copy. A half-turn sends every line to a parallel line, which is precisely why the two outer sides end up parallel. This is a far cleaner argument than the rectangle one, because a half-turn does not care whether any angle of the triangle is obtuse.

Q2.

We shall spoil the surprise by revealing that it is possible. But we will not tell you the least number of pieces required. Try to find the answer!

Answer

ΔABD can be cut into just two pieces that reassemble to cover ΔACD — one straight cut is enough.

Recall the set-up: AD is a median of ΔABC, so BD = DC and ar(ΔABD) = ar(ΔACD). The two triangles share the side AD and have equal bases BD, DC on the same line.

Let M be the midpoint of AB, and let the cut run from M to D.
This splits ΔABD into ΔMBD and ΔAMD.

Now rotate ΔMBD through 180° about M.
B ↦ A (since M is the midpoint of AB), and D ↦ D′ where M is the midpoint of DD′.
MD ∥ AC (midpoint theorem in ΔABC: M and D are midpoints of AB and BC),
and MD = ½AC, so the image of BD lands along AC and D′ is on it.

The rotated piece together with ΔAMD covers ΔACD exactly. So two pieces, one straight cut MD, plus a half-turn.

Why it happens: a half-turn about the midpoint of a segment is a rigid motion that swaps the endpoints. Choosing the midpoint of AB as the pivot makes B land on A, and the midpoint theorem guarantees that the cut MD points along AC so that the piece lands flush. If you cut anywhere else, the pieces overlap or leave a gap.

Try This: cut the two triangles out of paper and check. Then ask the harder question that the next Think and Reflect box asks: can any two polygons of equal area be cut up into each other? (Yes — that is the Wallace–Bolyai–Gerwien theorem.)

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