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NCERT Solutions for Class 7th Maths Chapter 1 Figure it Out — Angles of Isosceles and Equilateral Triangles

Book page 20–21 Updated on2026-09-19

Q1.
ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles.
Answer

Read the letters in the same order on both sides of the ≅ sign.

Corresponding verticesCorresponding sidesCorresponding angles
A and FAI and FL∠A = ∠F
I and LIR and LY∠I = ∠L
R and YRA and YF∠R = ∠Y
Why it happens: a side is named by the two vertices at its ends. Since A goes with F and I goes with L, the side AI must go with the side FL. An angle is named by the vertex it sits at, so ∠A goes with ∠F.
Tip: the same congruence could be written ∆IRA ≅ ∆LYF or ∆RAI ≅ ∆YFL. The pairs A–F, I–L, R–Y never change.
Q2.
Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent. (a) AB = DE, BC = EF, CA = DF (b) AB = EF, ∠A = ∠E, AC = ED (c) AB = DF, ∠B = ∠D = 90°, AC = FE (d) ∠A = ∠D, ∠B = ∠E, AC = DF (e) AB = DF, ∠B = ∠F, AC = DE
Answer

Four of the five cases give congruent triangles.

CaseCongruent?ConditionCongruence
(a)YesSSS — all three sides match∆ABC ≅ ∆DEF
(b)YesSAS — ∠A lies between AB and AC, ∠E lies between EF and ED∆ABC ≅ ∆EFD
(c)YesRHS — a right angle, the hypotenuse and one more side∆ABC ≅ ∆FDE
(d)YesAAS — two angles and a non-included side∆ABC ≅ ∆DEF
(e)NoSSA — the angle is not between the two sides—

(a) AB = DE, BC = EF, CA = FD. Three pairs of sides, so SSS gives ∆ABC ≅ ∆DEF.

(b) ∠A is the angle between the sides AB and AC. ∠E is the angle between EF and ED. Since AB = EF and AC = ED, the equal angle is the included one in both — SAS gives ∆ABC ≅ ∆EFD.

(c) ∠B = ∠D = 90°. In the first triangle the side opposite the right angle is AC; in the second it is FE. These are the hypotenuses, and AC = FE. The extra pair AB = DF is one more side. RHS gives ∆ABC ≅ ∆FDE.

(d) ∠A = ∠D and ∠B = ∠E, with AC = DF. The side AC is not between ∠A and ∠B, so this is AAS. Find the third angle: ∠C = 180° − ∠A − ∠B = 180° − ∠D − ∠E = ∠F. Now ∠A = ∠D, AC = DF, ∠C = ∠F is ASA, so ∆ABC ≅ ∆DEF.

(e) AB = DF, AC = DE and ∠B = ∠F. The angle ∠B sits at the end of AB, opposite the side AC — it is not between the two given sides. This is the SSA case, which we have seen can give two different triangles. The triangles need not be congruent.

Why it happens: the safe conditions all pin the third vertex down completely. SSA does not, because an arc can cut a line at two points. Always check where the equal angle sits before naming the condition.
Q3.
It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a transversal for these two lines. Are there any equal alternate angles?]
Answer

The segments AD and BC cross at O. Look at ∆AOB and ∆DOC.

OA = OD (given)
∠AOB = ∠DOC (vertically opposite angles)
OB = OC (given)
→ SAS condition → ∆AOB ≅ ∆DOC

Corresponding angles of congruent triangles are equal, so

∠OAB = ∠ODC, that is ∠DAB = ∠ADC

AD is a transversal cutting the two lines AB and DC. The angles ∠DAB and ∠ADC are alternate angles for that transversal, and they are equal. Hence

AB ∥ CD
Why it happens: equal alternate angles is the test for parallel lines. The congruence supplies exactly that pair of equal angles. Notice how congruence is being used here as a tool — we wanted a fact about parallel lines, and congruent triangles delivered it.
Tip: the same congruence also gives AB = DC. So ABDC has one pair of sides that are both equal and parallel.
Q4.
ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent to ΔCDA? Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways?
Answer

Part 1 — ∆ABC ≅ ∆ADC. The diagonal AC splits the square into two triangles.

AB = AD (sides of a square)
BC = DC (sides of a square)
AC = AC (common diagonal)
→ SSS condition → ∆ABC ≅ ∆ADC

Part 2 — is ∆ABC ≅ ∆CDA as well? Yes. Check the three pairs of that correspondence:

∆ABC∆CDASides matchedEqual?
ACAB with CDYes — sides of the square
BDBC with DAYes — sides of the square
CACA with ACYes — same segment

So ∆ABC ≅ ∆CDA also holds, again by SSS. The same pair of triangles is congruent in two different ways.

Why it happens: ∆ABC is isosceles — AB = BC, since both are sides of the square. An isosceles triangle can be laid on its partner in two ways: the ordinary way, and after being flipped over its line of symmetry. Each way gives one correct congruence statement.

More examples of two ways: take any isosceles triangle ∆PQR with PQ = PR and a copy of it, ∆XYZ with XY = XZ. Then both ∆PQR ≅ ∆XYZ and ∆PQR ≅ ∆XZY are correct.

Six ways: take an equilateral triangle and a copy of it — say ∆PQR and ∆XYZ, each with all sides 5 cm. Every one of the six matchings works:

∆PQR ≅ ∆XYZ, ∆PQR ≅ ∆XZY, ∆PQR ≅ ∆YXZ
∆PQR ≅ ∆YZX, ∆PQR ≅ ∆ZXY, ∆PQR ≅ ∆ZYX
Tip: the number of ways is the number of ways of matching equal sides. Scalene → 1 way, isosceles → 2 ways, equilateral → 6 ways.
Q5.
Find ∠B and ∠C, if A is the centre of the circle. (∠A = 120°)
Answer

∠B = ∠C = 30°.

B and C lie on the circle and A is its centre, so AB and AC are both radii:

AB = AC (radii of the same circle)
→ ∆ABC is isosceles
→ ∠B = ∠C (angles opposite equal sides)

Now use the angle sum, writing x for each of the equal angles:

∠A + ∠B + ∠C = 180°
120° + x + x = 180°
2x = 60°
x = 30°
∠B = ∠C = 30°
Why it happens: every radius of a circle has the same length, so any triangle formed by the centre and two points of the circle is automatically isosceles. That single fact does all the work here.
Check it yourself: 120° + 30° + 30° = 180° ✓
Q6.
Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘|’ are equal to each other and those marked with a double ‘|’ are equal to each other, etc.
Answer

The figure is a rectangle ACDB cut into 16 triangles. Four points inside it are not named in the book, so let us call them P, Q, S and T, as marked here.

CRVD AKLB UF STQP
The rectangle with the four inner points named S, T, Q and P.

The marked equal sides. Single ‘|’: CR = RV = VD = CU = UA = US = UT = VQ. Triple ‘|||’: RQ = AT = KL = LB. Double ‘||’: PF = PB = FB.

Working, step by step.

  1. ∆CRU has CU = CR and ∠C = 90°, so it is isosceles: ∠CRU = ∠CUR = 45°.
  2. ∆UTA has UA = UT and ∠UAT = 56° (given at A), so ∠UTA = 56° and ∠AUT = 180° − 112° = 68°.
  3. ∆RVQ has VR = VQ and ∠RVQ = 68°, so ∠VRQ = ∠VQR = (180° − 68°) ÷ 2 = 56°.
  4. C, R, V, D lie on one line, so at R: 45° + ∠URS + 34° + 56° = 180°, giving ∠URS = 45°.
  5. C, U, A lie on one line, so at U: 45° + ∠RUS + 34° + 68° = 180°, giving ∠RUS = 33°. Then in ∆RUS, ∠RSU = 180° − 45° − 33° = 102°.
  6. In ∆RSQ: ∠RSQ = 180° − 34° − 44° = 102°.
  7. ∆UST has US = UT and ∠SUT = 34°, so ∠UST = ∠UTS = 73°.
  8. The four angles at S fill a full turn: ∠TSQ = 360° − 102° − 73° − 102° = 83°. Then in ∆STQ, ∠STQ = 180° − 83° − 46° = 51°.
  9. In ∆TQP: ∠TQP = 90° and ∠QPT = 56°, so ∠QTP = 34°.
  10. In ∆ATK: ∠AKT = 180° − 34° − 44° = 102°.
  11. The six angles at T fill a full turn: ∠KTP = 360° − (34° + 51° + 73° + 56° + 44°) = 102°. Then in ∆TPK, ∠TKP = 180° − 102° − 30° = 48°.
  12. A, K, L, B lie on one line, so at K: ∠PKL = 180° − 102° − 48° = 30°. In ∆KPL, ∠PLK = 90°, so ∠KPL = 60°.
  13. ∆PLK and ∆PLB have LK = LB, ∠PLK = ∠PLB = 90° and PL common, so by SAS they are congruent. Hence ∠PBL = 30°, ∠LPB = 60° and PK = PB.
  14. ∆PBF has PF = PB = FB (all double marks), so it is equilateral: ∠BPF = ∠PFB = ∠PBF = 60°.
  15. The six angles at P fill a full turn: ∠FPQ = 360° − (56° + 30° + 60° + 60° + 60°) = 94°.
  16. D, F, B lie on one line, so at F: ∠QFP = 180° − 98° − 60° = 22°. Then in ∆PFQ, ∠PQF = 180° − 94° − 22° = 64°.
  17. At V, ∠QVD = 180° − 68° = 112°. ∆QVD has VQ = VD, so ∠VQD = ∠VDQ = (180° − 112°) ÷ 2 = 34°.
  18. At the corner D: ∠QDF = 90° − 34° = 56°. Then in ∆QFD, ∠DQF = 180° − 98° − 56° = 26°.

All the angles, triangle by triangle.

TriangleAnglesTriangleAngles
∆CRU90°, 45°, 45°∆TPK102° (at T), 30° (at P), 48° (at K)
∆RUS45° (at R), 33° (at U), 102° (at S)∆KPL30°, 60°, 90°
∆UST34° (at U), 73°, 73°∆PLB90° (at L), 60° (at P), 30° (at B)
∆UTA68° (at U), 56°, 56°∆PBF60°, 60°, 60°
∆ATK34° (at A), 44° (at T), 102° (at K)∆PFQ94° (at P), 22° (at F), 64° (at Q)
∆RSQ34° (at R), 102° (at S), 44° (at Q)∆QFD98° (at F), 56° (at D), 26° (at Q)
∆STQ83° (at S), 51° (at T), 46° (at Q)∆QDV112° (at V), 34°, 34°
∆TQP34° (at T), 90° (at Q), 56° (at P)∆QVR68° (at V), 56°, 56°
Why it happens: only three tools are used again and again — angles opposite equal sides are equal, the three angles of a triangle add to 180°, and the angles at a point add to 360° (or 180° along a straight line). Each answer unlocks the next triangle, like links in a chain.
Check it yourself: the seven angles around Q must total 360°. Adding them: 44° + 46° + 90° + 64° + 26° + 34° + 56° = 360° ✓ Notice also that ∆RSQ and ∆ATK have the same three angles 34°, 44°, 102° and the equal sides RQ = AT, so ∆RSQ ≅ ∆AKT by ASA.
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