NCERT Solutions Ganita Prakash (Part 2) Chapter 7 Triangles between Parallel Lines with a Common Base — In-text Questions

Book page 157 Updated on2026-09-05

Q1.

What can we say about the lengths of AB and its reflection AB´?

Answer

They are equal: AB = AB′.

X is the point where BB′ crosses l, and the mirror sends B to B′ with BX = B′X
∠AXB = ∠AXB′ = 90°, and AX is common
So ∆AXB ≅ ∆AXB′ (SAS)
Hence AB = AB′, and in the same way AC = AC′

So the bent path B → A → C and the bent path B → A → C′ have exactly the same length.

Why it happens: a reflection is a rigid motion — it flips the plane over without stretching it. Every length in the figure survives the flip unchanged, which is why the mirror can be used to replace an awkward path by an equal one that is easier to shorten.

Q2.

Analyse whether A lies on the perpendicular bisector of BC. [Math Talk]

Answer

Yes, it does — so the triangle of least perimeter is the isosceles one.

Set BC along a line and let l be parallel to it at distance d. Reflecting C in l puts C′ directly above C at height 2d, while B is at height 0.

The segment BC′ rises from height 0 to height 2d
It crosses l at exactly half that rise, i.e. at the midpoint of BC′
That crossing point sits horizontally halfway between B and C
So A is directly above the midpoint of BC → AB = AC

Why it happens: because l ‖ BC, the mirror image C′ is exactly as far above l as C is below it. The straight line BC′ therefore meets l at its own midpoint, and that point lies on the perpendicular bisector of BC. Intuition and proof agree here — but notice that the proof, not the picture, is what settles it.

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