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NCERT Solutions for Class 7th Maths Chapter 1 Measuring the Angles · Two Sides and the Included Angle (SAS) · A Non-included Angle (SSA) · Two Angles and the Included Side (ASA) — In-text Questions

Book page 9–13 Updated on2026-09-19

Q1.
Instead of measuring the three sidelengths of the triangular frame, if Meera and Rabia measure the three angles, can they recreate the triangle exactly?
Answer

No. The three angles fix the shape but not the size.

Two triangles with the same three angles look alike, but one may be a big version of the other. They can be placed one over the other only if they also happen to be the same size, and the angles do not tell us that.

Why it happens: the angles say how steeply the sides lean, not how long they are. You can keep the leaning the same and stretch every side to twice its length — the angles do not change at all.
Q2.
Suppose the angles are 30°, 70°, and 80°. Can we create an exact copy of the frame with this?
Answer

No. Many different triangles have these three angles.

30° + 70° + 80° = 180° ✓ — so such triangles do exist
But nothing in these numbers says how long a side must be

Draw a base of 4 cm and put 30° at one end and 70° at the other; the third angle comes out as 80°. Now do it again with a base of 8 cm. Both triangles have angles 30°, 70°, 80°, but the second is twice as big as the first.

Why it happens: the two rays from the ends of the base meet at one point whatever the base length is. Lengthening the base simply pushes that meeting point further away, giving a larger triangle with the same three angles. So two triangles with the same set of angles need not be congruent — AAA is not a congruence condition.
Q3.
ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 5 cm, and ∠A = ∠X = 30°. Are they congruent?
Answer

Yes, they are congruent.

Here the equal angle lies between the two equal sides — ∠A is the angle between AB and AC, and ∠X is the angle between XY and XZ.

AB = XY = 6 cm
∠A = ∠X = 30° (included angle)
AC = XZ = 5 cm
→ SAS condition → ∆ABC ≅ ∆XYZ
Why it happens: start at the vertex A. Draw the 30° angle, mark 6 cm along one arm to get B and 5 cm along the other to get C. Both B and C are now decided, so the third side BC is decided too. Nobody following these instructions can end up with a different triangle.
Q4.
Construct a triangle having the above measurements. Compare it with the triangles constructed by your classmates. Are the triangles all congruent? Explain why all such triangles with these measurements are congruent.
Answer

Yes — every triangle in the class will be congruent (some may be flipped over).

The construction leaves no free choice:

  1. Draw AB = 6 cm.
  2. At A, draw a ray making 30° with AB. There is only one such ray on a given side of AB.
  3. On that ray mark AC = 5 cm. There is only one such point C.
  4. Join BC — there is only one segment joining two fixed points.
One choice for the angle + one choice for C → one triangle
Why it happens: each step is forced by the step before it. Because the given angle is squeezed between the two given sides, both ends of the third side are pinned down before that side is drawn. This is exactly what the SAS condition claims.
Check it yourself: cut out two classmates' triangles and place one on the other. If one was drawn with the 30° opening to the left, turn it over first.
Q5.
ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B = ∠Y = 30°. Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out.
Answer

Not necessarily. Two different triangles can be drawn from these measurements.

Here the equal angle is not between the two equal sides — ∠B lies at the end of AB, while AC starts at A. This is the SSA case. Follow the construction:

  1. Draw the base PQ = 6 cm.
  2. From P draw a line l making 30° with PQ.
  3. From Q draw a long arc of radius 4 cm. It cuts the line l at two points, R and S.

Both ∆PQR and ∆PQS have a 6 cm side, a 4 cm side and a 30° angle at P — yet they are clearly different triangles.

P Q R S 6 cm 30° l
One arc, two crossing points — two triangles that are not congruent.
Why it happens: an arc can cut a straight line twice. Because the given angle is not squeezed between the two given sides, the third vertex is not pinned down — it may sit at either crossing. So the SSA condition does not guarantee congruence.
Q6.
How do we find the required triangle from this figure?
30°6 cmPQRSl
Step 3 of the construction on page 11 — base PQ = 6 cm, the line l at 30° to PQ, and an arc of radius 4 cm drawn from Q cutting l at R and S.
Answer

A point where the arc meets the line l gives the third vertex. But the arc meets l at two points, R and S — so the figure gives two triangles, ∆PQR and ∆PQS.

Both satisfy: PQ = 6 cm, QR = QS = 4 cm, ∠P = 30°
But PR is longer than PS
→ the two triangles are not congruent
Why it happens: the measurements do not tell us which crossing point to pick, so the data is incomplete. This is what makes SSA unsafe: with SSS, SAS or ASA the construction never offers such a choice.
Tip: a rough diagram drawn before the actual construction shows at once whether the given angle lies between the two given sides. That decides whether you are in the safe SAS case or the risky SSA case.
Q7.
∆ABC and ∆XYZ are two triangles with, BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z = 30°. Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out.
Answer

Yes, they are congruent, and no other triangle is possible with these measurements.

The equal side lies between the two equal angles, so this is the ASA case.

  1. Draw BC = 5 cm.
  2. At B draw a ray making 50° with BC.
  3. At C draw a ray making 30° with CB, on the same side.
  4. The two rays meet at exactly one point — that is A.
∠B = ∠Y = 50°
BC = YZ = 5 cm (included side)
∠C = ∠Z = 30°
→ ASA condition → ∆ABC ≅ ∆XYZ
Why it happens: two rays that are not parallel meet at exactly one point. Since both ends of the base and both directions are fixed, the meeting point cannot move. Everyone in the class gets the same triangle.
Tip: the third angle is decided too — 180° − 50° − 30° = 100°. That is a useful check on your drawing.
Q8.
In the figure, Point O is the midpoint of AD and BC. What can one say about the lengths AB and CD?
ACBDO
The figure on page 12 — AD and BC cross at O.
Answer

AB = CD. The two lengths are equal.

Look at ∆AOB and ∆DOC:

AO = OD (O is the midpoint of AD)
∠AOB = ∠DOC (vertically opposite angles)
BO = OC (O is the midpoint of BC)
→ SAS condition → ∆AOB ≅ ∆DOC

The correspondence is A ↔ D, O ↔ O, B ↔ C. So AB and DC are corresponding sides, and

AB = DC
Why it happens: the two triangles are joined at O like a bow tie. Turning ∆AOB half a turn about O sends A to D and B to C, because O is the midpoint of both segments. The side AB is carried to the side DC, so the two must have the same length.
Q9.
We have, AO = OD (as O is the midpoint of AD) and BO = OC (as O is the midpoint of BC). Are there any other equal sides or angles?
Answer

Yes — ∠AOB = ∠DOC, because they are vertically opposite angles formed where the segments AD and BC cross at O.

Two known sides: AO = OD and BO = OC
One known angle: ∠AOB = ∠DOC
The angle lies between the two sides → SAS
Why it happens: this is the piece of information that completes the condition. Without it we would only have two pairs of equal sides, which is never enough. The crossing point O supplies the third fact free of cost, just as a common side does.
Tip: whenever two straight lines cross in a figure, look at once for vertically opposite angles. They are equal without being marked.
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