Triangles [Congruency in Triangles]
ICSE Class 9 · Mathematics · Chapter 9
What this chapter covers
- Congruent figures
- Congruent triangles and corresponding parts
- Illustration of congruency through cut-outs
- The SSS case
- The SAS case
- The AAS case
- The RHS case
- Conditions that do not prove congruency
- Writing a congruency proof
- Simple applications of congruency
Key points
- Congruent figures have exactly the same shape and size; the symbol cong means "is congruent to".
- In △ ABC cong △ PQR, the order of letters gives the correspondence A leftrightarrow P, B leftrightarrow Q, C leftrightarrow R.
- Congruent triangles have all six corresponding parts equal; using this in a proof is written CPCT.
- SSS: three sides of one triangle equal to three sides of the other.
- SAS: two sides and the angle included between them.
- AAS: two angles and a corresponding side; ASA, with the side between the angles, is a form of it.
- RHS: in right-angled triangles, the hypotenuse and one side.
- AAA and SSA are not conditions of congruence.
- The four conditions are verified with cut-outs and superposition; their proofs are not required.
- A congruency proof names the triangles, lists three equal pairs with reasons, states the congruence in the correct order with its condition, and then uses CPCT.
- Congruency is used to prove equal lengths and angles, midpoints, angle bisectors and parallel lines.
Questions answered
Define congruent triangles.
Congruent triangles are two triangles in which the three sides and three angles of one are equal, respectively, to the three sides and three angles of the other.
State the four conditions for the congruency of two triangles.
SSS (three sides), SAS (two sides and the included angle), AAS (two angles and a corresponding side; ASA is a form of it), and RHS (right angle, hypotenuse and one side, for right-angled triangles only).
If △ ABC cong △ RPQ, write the three pairs of equal sides and the three pairs of equal angles.
The correspondence is A leftrightarrow R, B leftrightarrow P, C leftrightarrow Q. So AB = RP, BC = PQ, CA = QR, and ∠ A = ∠ R, ∠ B = ∠ P, ∠ C = ∠ Q.
△ DEF cong △ KLM, with ∠ D = 48° and ∠ M = 77°. Find ∠ E.
∠ F = ∠ M = 77°. So ∠ E = 180° - 48° - 77° = 55°.
△ PQR cong △ STU, with PQ = (2y + 3) cm and ST = (4y - 5) cm. Find y and the length PQ.
PQ corresponds to ST, so 2y + 3 = 4y - 5, giving y = 4 and PQ = 11 cm.
In △ ABC and △ XYZ, AB = XY, BC = YZ and ∠ A = ∠ X. Are the triangles necessarily congruent? Give a reason.
No. The angle ∠ A is not included between AB and BC (the included angle is ∠ B). This is the SSA arrangement, which can give two different triangles.
Why is AAA not a condition for congruence?
Three equal angles fix the shape of a triangle but not its size, so two triangles with the same angles can be of different sizes. They are similar but need not be congruent.
Why does the RHS condition work even though the right angle is not between the hypotenuse and the given side?
Because the angle is 90° and the hypotenuse is the longest side, an arc of the hypotenuse's length from the end of the given side meets the other arm of the right angle at only one point, so only one triangle is possible.
Past board exam questions
ICSE 2019 sample paper · 3 marks
Without using tables, evaluate: 4tan 60° sec 30° + (sin 31° sec 59° + cot 59° cot 31°)/(8sin2 30° - tan2 45°)
Show answer
4tan 60° sec 30° = 4 × √(3) × (2)/(√(3)) = 8
sin 31° sec 59° = (sin 31°)/(cos(90°-31°)) = (sin 31°)/(sin 31°) = 1; cot 59° cot 31° = tan 31° cot 31° = 1
Denominator = 8 × (1)/(4) - 1 = 1
Value = 8 + (1+1)/(1) = 10
ICSE 2019 sample paper · 4 marks
ABCD is a parallelogram. △APD and △BQC are equilateral triangles drawn outside the parallelogram on sides AD and BC respectively. Prove that: (i) ∠PAB = ∠QCD (ii) PB = QD
Show answer
(i) ∠PAD = ∠QCB = 60° (angles of equilateral triangles) and ∠DAB = ∠BCD (opposite angles of a parallelogram).
∠PAB = ∠PAD + ∠DAB = 60° + ∠DAB
∠QCD = ∠QCB + ∠BCD = 60° + ∠BCD
∴ ∠PAB = ∠QCD.
(ii) In △PAB and △QCD:
PA = AD = BC = QC (sides of equilateral triangles, AD = BC opposite sides of parallelogram)
AB = CD (opposite sides of parallelogram)
∠PAB = ∠QCD (proved)
∴ △PAB ≅ △QCD (SAS), so PB = QD (c.p.c.t.).
ICSE 2019 sample paper · 3 marks
Solve for x; where 0° ≤ x ≤ 90° sin2 x + cos2 30° = (5)/(4)
Show answer
cos2 30° = ((√(3))/(2))2 = (3)/(4)
sin2 x = (5)/(4) - (3)/(4) = (1)/(2) ⇒ sin x = (1)/(√(2))
x = 45°
ICSE 2019 sample paper · 3 marks
In △ABC, ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC (E on AB, D on BC), and AD and CE intersect at P. Prove that: CP > AP
Show answer
In △ABD, ∠BAD = 90° − 66° = 24°, so ∠BAC = 24° + 38° = 62°.
In △ABC, ∠ACB = 180° − 66° − 62° = 52°.
In △BEC, ∠BCE = 90° − 66° = 24°, so ∠ACP = ∠ACB − ∠BCE = 52° − 24° = 28°.
In △APC, ∠PAC = 38° > ∠ACP = 28°.
The side opposite the greater angle is greater, so CP > AP.
ICSE 2019 sample paper · 4 marks
In △ABC, CD is perpendicular to AB, with D on AB. AC = 10 m, CB = 5√(2) m, ∠ACD = 60°, CD = x and ∠CBA = y°. Evaluate x and y.
Show answer
In right △ADC: cos 60° = (CD)/(AC) ⇒ x = 10 × (1)/(2) = 5 m.
In right △BDC: sin y° = (CD)/(CB) = (5)/(5√(2)) = (1)/(√(2)) ⇒ y = 45.
x = 5 m, y = 45°.
ICSE 2019 sample paper · 3 marks
△ABC is isosceles triangle such that AB = AC. D is a point on side AB such that BC = CD. Given ∠BAC = 28°. Find the value of ∠DCA.
Show answer
AB = AC ⇒ ∠ABC = ∠ACB = (180° - 28°)/(2) = 76°.
In △BCD, BC = CD ⇒ ∠CDB = ∠CBD = 76°, so ∠BCD = 180° − 76° − 76° = 28°.
∠DCA = ∠ACB − ∠BCD = 76° − 28° = 48°.