RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Two triangles are congruent when one could be picked up and placed exactly on top of the other, and this chapter gives you four reliable tests for deciding that without any cutting out. The real skill is choosing the right criterion and listing the corresponding parts in the correct order. From congruence flow the elegant results about isosceles triangles and about which side of a triangle is longest.
Which of the following is NOT a valid criterion for the congruence of two triangles?
Answer
AAA is not a valid congruence criterion. If all three angles of one triangle equal those of another, the triangles have the same shape but not necessarily the same size — one could be an enlargement of the other. Such triangles are similar, not congruent. SAS, RHS and SSS all fix the size as well as the shape, so each of those does prove congruence.
In triangle DEF, DE = DF and ∠E = 52°. The measure of ∠D is:
Answer
∠D = 76°. Since DE = DF, the triangle is isosceles and the angles opposite the equal sides are equal, so ∠F = ∠E = 52°. By the angle sum property, ∠D = 180° - 52° - 52° = 76°.
Which of the following sets of lengths can form a triangle?
Answer
5 cm, 9 cm, 6 cm can form a triangle. The triangle inequality requires the sum of any two sides to exceed the third. Here 5 + 6 = 11 > 9, 5 + 9 = 14 > 6, and 6 + 9 = 15 > 5, so all three checks pass. The others fail: 3 + 4 = 7 < 8; 2 + 2 = 4 < 5; and 7 + 1 = 8 < 9.
In triangle PQR, ∠P = 40° and ∠Q = 95°. The longest side of the triangle is:
Answer
The longest side is PR. First find ∠R = 180° - 40° - 95° = 45°. The greatest angle is ∠Q = 95°, and the longest side is the one opposite the greatest angle. The side opposite ∠Q is PR, so PR is the longest side.
Assertion (A): Two triangles having equal areas are always congruent. Reason (R): Congruent triangles always have equal areas.
Answer
A is false but R is true. Reason R is true: congruent triangles are identical copies, so their areas must be equal. Assertion A reverses this wrongly. A triangle with base 8 cm and height 3 cm has area 12 cm², and so does one with base 12 cm and height 2 cm, yet the two are clearly not congruent. Equal area does not force equal shape and size.
In triangles ABC and PQR, AB = PQ, ∠B = ∠Q and BC = QR. Name the congruence criterion that applies and write the congruence statement correctly.
Answer
The equal angle ∠B = ∠Q lies between the two pairs of equal sides AB = PQ and BC = QR, so it is the included angle. The applicable criterion is therefore SAS (side-angle-side). The congruence is written as triangle ABC ≅ triangle PQR, with the vertices listed so that A corresponds to P, B to Q and C to R.
In triangle LMN, LM = LN and the bisector of ∠L meets MN at O. Prove that O is the midpoint of MN.
Answer
In triangles LMO and LNO: LM = LN (given) ∠MLO = ∠NLO (LO bisects ∠L) LO = LO (common side) By the SAS criterion, triangle LMO ≅ triangle LNO. By CPCT (corresponding parts of congruent triangles), MO = NO. Since O lies on MN and divides it into two equal parts, O is the midpoint of MN.
Two sides of a triangle measure 9 cm and 14 cm. Between what values must the third side lie?
Answer
By the triangle inequality, the third side must be less than the sum of the other two and greater than their difference. Sum = 9 + 14 = 23 cm, and difference = 14 - 9 = 5 cm. So the third side x must satisfy 5 cm < x < 23 cm. Any value strictly between 5 cm and 23 cm will form a valid triangle.
In quadrilateral ABCD, AB = AD and AC bisects ∠BAD. (a) Prove that triangle ABC ≅ triangle ADC. (b) Deduce that BC = DC. (c) Deduce that ∠ABC = ∠ADC. (d) State whether AC is perpendicular to BD, giving a reason.
Answer
(a) In triangles ABC and ADC: AB = AD (given) ∠BAC = ∠DAC (AC bisects ∠BAD) AC = AC (common side) The equal angle is included between the two pairs of equal sides, so by the SAS criterion, triangle ABC ≅ triangle ADC. (b) Since the triangles are congruent, corresponding parts are equal, so BC = DC by CPCT. (c) Again by CPCT, the corresponding angles ∠ABC and ∠ADC are equal. (d) Yes, AC is perpendicular to BD. Since AB = AD and CB = CD, both A and C are equidistant from B and D, so line AC is the perpendicular bisector of BD. Hence AC meets BD at right angles.
(a) State and prove the theorem that angles opposite to equal sides of a triangle are equal. (b) In triangle STU, ST = SU and the angle ∠S is 40° more than each base angle. Find all three angles.
Answer
(a) Theorem: If two sides of a triangle are equal, then the angles opposite them are equal. Proof: Let triangle ABC have AB = AC. Draw AD, the bisector of ∠BAC, meeting BC at D. In triangles ABD and ACD: AB = AC (given) ∠BAD = ∠CAD (AD is the bisector) AD = AD (common) By SAS, triangle ABD ≅ triangle ACD. Hence by CPCT, ∠ABD = ∠ACD, that is ∠B = ∠C, which are the angles opposite the equal sides AC and AB respectively. Proved. (b) Since ST = SU, the base angles opposite them are equal; let each be x°. Then ∠S = (x + 40)°. By the angle sum property, x + x + (x + 40) = 180. So 3x = 140, giving x = 140/3 = 46 2/3° approximately 46.67°. Hence the two base angles are each 46 2/3° and ∠S = 86 2/3°. Checking: 46 2/3 + 46 2/3 + 86 2/3 = 180°.
Read the following and answer the questions that follow: An engineer designs a symmetric roof truss shaped like triangle ABC, where AB = AC. A vertical support AM runs from the apex A down to the midpoint M of the horizontal beam BC. The apex angle ∠BAC measures 50°. (a) Find the measures of ∠ABC and ∠ACB. (b) Prove that triangle ABM ≅ triangle ACM. (c) Hence find ∠AMB. (d) If AB = 6.5 m and BC = 5 m, find BM and state which side of triangle ABM is the longest, with a reason.
Answer
(a) Since AB = AC, the base angles opposite them are equal. Let each be x. Then x + x + 50° = 180°, so 2x = 130° and x = 65°. Thus ∠ABC = ∠ACB = 65°. (b) In triangles ABM and ACM: AB = AC (given) BM = CM (M is the midpoint of BC) AM = AM (common side) By the SSS criterion, triangle ABM ≅ triangle ACM. (c) By CPCT, ∠AMB = ∠AMC. These two angles form a linear pair along BC, so they add to 180°. Hence each is 90°, giving ∠AMB = 90°, which confirms the support is vertical. (d) BM = (1/2)BC = (1/2)(5) = 2.5 m. In triangle ABM the largest angle is ∠AMB = 90°, and the longest side is the one opposite the largest angle, which is AB (the hypotenuse), measuring 6.5 m.
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