EN हिं

NCERT Solutions for Class 7th Maths Chapter 1 Measuring Two Angles and a Non-Included Side (AAS) · Measuring Two Sides in a Right Triangle (RHS) — In-text Questions

Book page 14–17 Updated on2026-09-19

Q1.
The following triangles ∆ABC and ∆XYZ are such that ∠A = ∠X = 35°, ∠C = ∠Z = 75°, and BC = YZ = 4 cm. Are the triangles congruent? Give a reason.
Answer

Yes, they are congruent.

The given side BC is not between the two given angles ∠A and ∠C, so ASA cannot be used straight away. But the third angle can be found first.

∠B + 35° + 75° = 180°
∠B + 110° = 180°
∠B = 70°, and in the same way ∠Y = 70°

Now use ∠B, BC and ∠C, which do fit the ASA pattern:

∠B = ∠Y = 70°
BC = YZ = 4 cm (included side)
∠C = ∠Z = 75°
→ ASA condition → ∆ABC ≅ ∆XYZ
Why it happens: in a triangle the three angles always add to 180°, so knowing two of them gives the third for free. That is why two angles and any one side are enough. This shortcut is called the AAS condition.
Q2.
What are the measures of ∠B and ∠Y?
Answer

Both are 70°.

In ∆ABC: ∠A + ∠B + ∠C = 180°
35° + ∠B + 75° = 180°
∠B + 110° = 180°
∠B = 70°

In ∆XYZ: 35° + ∠Y + 75° = 180°
∠Y = 70°

So ∠B = ∠Y.

Why it happens: the two triangles were given the same pair of angles, 35° and 75°. Since all three angles must total 180°, whatever is left over is the same in both — 70°.
Q3.
Does this help in showing that ΔABC and ΔXYZ are congruent?
Answer

Yes. Finding the third angle turns the problem into an ordinary ASA problem.

∠B = ∠Y = 70°
BC = YZ = 4 cm
∠C = ∠Z = 75°

Now the equal side BC lies between the two equal angles ∠B and ∠C, so ASA applies and

∆ABC ≅ ∆XYZ
Why it happens: the side did not have to be the included one from the start. Once the third angle is worked out, every side of the triangle is included between some pair of known angles. So AAS always guarantees congruence.
Tip: AAS is not really a new rule. It is ASA with one extra line of arithmetic.
Q4.
ΔABC and ΔXYZ are right-angled triangles such that BC = YZ = 4 cm, ∠B = ∠Y = 90° and AC = XZ = 5 cm. Are they congruent? Can there exist non-congruent triangles having these measurements? Construct and find out.
Answer

Yes, they are congruent, and no different triangle can be drawn with these measurements.

Follow the construction with a rough diagram in front of you:

  1. Draw the base QR = 4 cm.
  2. At Q draw a line l perpendicular to QR.
  3. From R cut an arc of radius 5 cm on the line l. Call the crossing point P.
  4. Join PR. ∆PQR is the required triangle.
∠Q = 90°
PR = 5 cm (the hypotenuse, opposite the right angle)
QR = 4 cm (one other side)
→ RHS condition → ∆ABC ≅ ∆XYZ
Why it happens: this looks like the risky SSA case — two sides and an angle that is not between them. The difference is that here the known angle is a right angle and the known side opposite it is the hypotenuse, the longest side. The arc from R can meet the perpendicular line above Q or below Q, and those two triangles are mirror images of each other, so they are congruent. There is no third possibility.
Did you know? The third side comes out as 3 cm, since 3, 4, 5 is the smallest whole-number right triangle.
Q5.
Consider the downward extension of line l below QR. Would the arc from R meet this line downwards as well (as in the case of triangle construction when the sidelengths are given)? If so, would this lead to a triangle whose size and shape are different from ΔPQR, and yet has the given measurements?
Answer

Yes, the arc does meet the line below QR — but no, the new triangle is not different.

Call the lower crossing point P′. Then ∆P′QR also has ∠Q = 90°, QR = 4 cm and P′R = 5 cm, exactly the required measurements.

P and P′ lie on the same line l, one above Q and one below
QP = QP′ (both are cut by the same 5 cm arc from R)
→ ∆PQR and ∆P′QR are mirror images in the line QR
Why it happens: fold the paper along QR. The perpendicular line l falls back on itself and the arc from R falls back on itself, so P lands exactly on P′. A figure and its mirror image are always congruent, so the second triangle is not a new triangle at all.
Q6.
It can be seen that the other triangle we get below is also congruent to ΔPQR. Why? Therefore, all triangles having these measurements will be congruent to each other.
Answer

Because the two triangles share a side and match in two more parts:

QR = QR (common side)
∠PQR = ∠P′QR = 90° (both are right angles on the same perpendicular)
QP = QP′ (the same arc from R cuts off equal lengths on either side of Q)
→ SAS condition → ∆PQR ≅ ∆P′QR

So both possible triangles are congruent to each other, and therefore every triangle with these measurements is congruent to ∆PQR.

Why it happens: QP = QP′ can also be seen from the right angle itself. In each triangle the third side is decided by the other two, so equal hypotenuses and equal bases force equal heights. This is what makes RHS safe while ordinary SSA is not.
Was this helpful?