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NCERT Solutions for Class 7th Maths Chapter 1 In-text Questions — Congruence of Triangles · Measuring the Sidelengths · Conventions to Express Congruence

Book page 5–8 Updated on2026-09-19

Q1.
Meera and Rabia have been asked to make a cardboard cutout identical to a triangular frame they have in school. They see that the frame is too big to be traced on a paper and replicated. What do you think they can do?
Answer

They should measure the frame and then build the cutout from those measurements.

With a measuring tape they can take the three sidelengths of the triangular frame. Those three numbers can be carried back to the desk, and a triangle with exactly those sides can be drawn on cardboard with a ruler and a compass.

Why it happens: tracing needs the original and the copy to touch each other. Measurements do not — a number written in a notebook carries the shape from the playground to the desk. This is the whole idea behind the congruence conditions in this chapter.
Q2.
Using a measuring tape, the girls measure the sides of the triangle to be 40 cm, 60 cm, and 80 cm. Then, Rabia takes out her protractor to measure the angles. She is stopped by Meera. Meera: The angles of the triangle are not required! With the side lengths we have measured, we can create a triangle congruent to this one. Do you agree with Meera?
Answer

Yes, Meera is right. The three sidelengths are enough.

Try to draw a triangle with sides 40 cm, 60 cm and 80 cm in more than one way. Draw the 60 cm side first, then swing an arc of 40 cm from one end and an arc of 80 cm from the other. The arcs meet at only two points, one above the line and one below — and those two triangles are mirror images of each other, so they are congruent.

Three sides fixed → only one triangle possible (up to a flip)
This is the SSS condition
Why it happens: once the longest side is drawn, the third corner has to be at a fixed distance from each of its ends. Two distances from two fixed points leave only two possible positions, and those two give congruent triangles. Nothing is left to choose, so the angles cannot come out differently — measuring them would only repeat information we already have.
Q3.
Instead of the lengths being 40 cm, 60 cm, and 80 cm, suppose the sidelengths had been 4 cm, 6 cm, 8 cm (this triangle can fit on our page). Is this information sufficient to replicate the triangle with the same size and shape? If yes, can you do so?
Answer

Yes, it is sufficient. Here is the construction, drawn small enough to fit on a page.

  1. Draw AB = 6 cm with a ruler.
  2. With A as centre and radius 4 cm, draw an arc.
  3. With B as centre and radius 8 cm, draw a second arc cutting the first at E.
  4. Join AE and BE. ∆ABE has sides 6 cm, 4 cm and 8 cm.
A B E 6 cm 4 cm 8 cm
The two arcs decide the third corner E. Only 4 cm, 6 cm and 8 cm were used.
Check it yourself: compare your triangle with your neighbour's. Cut them out and place one on the other — they fit, perhaps after a flip.
Q4.
Rabia: If I were to construct this triangle, I would first draw a line segment having one of the given lengths, say 6 cm, and then draw circles from each of its end points with radii 4 cm and 8 cm. But the circles would intersect at two points, forming two triangles: ∆ABE and ∆ABF. Rabia: Do these two triangles have the same shape and size? If not, then we will not be sure which of these would actually be congruent to the original triangle we are trying to replicate.
ABEFAB = 6 cm
The construction on page 5 — a 6 cm segment AB, a circle of radius 4 cm about A and a circle of radius 8 cm about B, meeting at E and F.
Answer

Yes, they do have the same shape and size. ∆ABE and ∆ABF are congruent, so Rabia's worry does not arise.

Both triangles are built from the same three lengths:

AE = AF = 4 cm (both on the circle centred at A)
BE = BF = 8 cm (both on the circle centred at B)
AB is the same side in both
→ same three sidelengths
Why it happens: the whole construction above the line AB is repeated below it in exactly the same way. So AB acts as a line of symmetry: fold the paper along AB and E falls on F. Whichever of the two triangles Rabia draws, it will be congruent to the frame.
Q5.
Examine whether ∆ABE and ∆ABF are congruent. For this, you could use one or more of the following methods — tracing and comparing, taking a cutout and superimposing, or observing that AB acts as a line of symmetry due to the ‘sameness’ of the act of construction above and below this line.
Answer

All three methods give the same answer: ∆ABE ≅ ∆ABF.

  • Tracing: trace ∆ABE, flip the tracing paper over about the line AB, and it lands exactly on ∆ABF.
  • Cutout: cut both triangles out. Turn one over and place it on the other — the three sides match.
  • Symmetry: the arcs below AB were drawn with the same centres and the same radii as the arcs above it. So the lower half is the mirror image of the upper half in the line AB.
Why it happens: a reflection does not change any length or any angle. It only turns the figure over. Since congruence allows a figure to be flipped before superimposing, a figure and its mirror image are always congruent.
Did you know? This is exactly why the SSS condition works. Any two triangles with the same three sidelengths can be constructed by the same arcs, so they must be congruent.
Q6.
The two triangles given below are congruent. How can these two triangles be superimposed? Which vertices of ∆XYZ and ∆ABC should we overlap? This has to be done so that the equal sides overlap. Figure out how.
Answer

Place A over X, B over Y and C over Z.

AB fits over XY
BC fits over YZ
AC fits over XZ
→ the triangles fit exactly, one over the other

Since the triangles now fit exactly, their angles also match:

∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z
Why it happens: superimposing is only allowed to put equal sides on equal sides. If you tried to put A over Y, the side AB would land on YX; but AB and YX need not be equal, so the two triangles would stick out beyond each other. Matching the equal sides first automatically matches the vertices and the angles.
Tip: This matching is written as ∆ABC ≅ ∆XYZ. The order of the letters is the message — first with first, second with second, third with third.
Q7.
Are there other ways of overlapping the vertices so that the triangles fit exactly over each other?
Answer

For these triangles, no — there is only one way.

The three sides of the triangle have three different lengths, so each side has exactly one partner of the same length in the other triangle. That fixes the matching completely.

Why it happens: a second way of overlapping would need two sides of the same triangle to be equal. That happens only in special triangles:
  • Isosceles triangle — two equal sides, so two ways of overlapping.
  • Equilateral triangle — all three sides equal, so six ways of overlapping.
Q8.
Can you identify a pair of congruent triangles below? Why are they congruent?
ABCD
Fig. 1.1 (page 7) — rectangle ABCD with the diagonal BD drawn.
Answer

The diagonal BD cuts the rectangle into ∆ABD and ∆CDB, and these two are congruent.

AB = CD (opposite sides of a rectangle)
AD = CB (opposite sides of a rectangle)
BD = DB (the same side, common to both)
→ all three sides match → SSS condition

So ∆ABD and ∆CDB are congruent.

Why it happens: the third side of each triangle is the diagonal itself. A side is always equal to itself, so it costs nothing — that is why a common side is so useful in congruence proofs. Two given equalities plus one free one complete the SSS condition.
Q9.
Can they be the following? ∆ABD ↔ ∆CDB with A ↔ C, B ↔ B, D ↔ D. Verify this by superimposing paper cutouts of the triangles obtained from the rectangle ABCD (Fig. 1.1).
ABCD
Fig. 1.1 (page 7) — rectangle ABCD with the diagonal BD drawn.
Answer

No, this matching does not work. Cut the two triangles out of the rectangle and try it — they will not settle on each other.

A over C, B over B, D over D
puts the side AB over the side CB
But AB and CB are the length and the breadth — they need not be equal
Why it happens: a correspondence is only correct if every pair of matched sides is a pair of equal sides. Here AB is matched with CB, and in a rectangle those two are next to each other, not opposite, so they are usually different. The cutouts overlap only partly, and congruence is not established.
Tip: before writing a congruence, list the matched sides. If any pair on your list is not equal, the order of the letters is wrong.
Q10.
Identify the correct correspondence of vertices and express the congruence between the two triangles.
Answer

The correct matching is A ↔ C, B ↔ D, D ↔ B.

∆ABD ≅ ∆CDB

AB ↔ CD (equal, opposite sides)
BD ↔ DB (the common diagonal)
DA ↔ BC (equal, opposite sides)

Every matched pair is a pair of equal sides, so this correspondence is correct.

Why it happens: think of it as a physical move — lift ∆ABD, turn it half a turn about the centre of the rectangle, and set it down. A lands on C, B lands on D and D lands on B. That is exactly what the letters ∆ABD ≅ ∆CDB record.
Tip: ∆ADB ≅ ∆CBD says the same thing, because the pairs A↔C, D↔B, B↔D are unchanged. Only the order of listing has changed.
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