CBSE Class 8 Mathematics Congruence of Triangles MCQs Set 28

Answer: (c) CM = ½ AB

Explanation:
1. In triangles AMC and BMD: AM = BM, ∠AMC = ∠BMD (vertically opposite) and CM = DM, so △AMC ≅ △BMD (SAS). Hence BD = AC and ∠MBD = ∠MAC, so BD ∥ AC and ∠DBC = 90°. Then △DBC ≅ △ACB (SAS: BD = AC, BC common, both angles 90°), so CD = AB. As CM = ½ CD, CM = ½ AB.

Teacher's Note:
1. The mid-point of the hypotenuse of a right triangle is equidistant from all three vertices.
2. Prove the two triangles congruent first, then use corresponding parts.

Answer: (d) 30°

Explanation:
1. The angles above AB add up to 180°: 3x + 3x = 180, so x = 30°. Check below the line: x + 150° = 180°.

Teacher's Note:
1. Angles on a straight line add up to 180°.
2. Verify with the lower pair of angles.

Answer: (c) 130°

Explanation:
1. The marked 130° angle is vertically opposite to the angle below AB on the right of the transversal. That angle and x are corresponding angles for the parallel lines AB and CD, so x = 130°.

Teacher's Note:
1. Look for corresponding angles after using vertically opposite angles.
2. Alternate exterior angles are equal as well.

Answer: (a) 20°

Explanation:
1. Since PQ ∥ RS, ∠QRS = ∠PQR = 70° (alternate angles). Since UV ∥ RS, ∠URS = 180° − 130° = 50° (interior angles on the same side). So x = ∠QRS − ∠URS = 70° − 50° = 20°.

Teacher's Note:
1. Use alternate angles at Q and R, and co-interior angles at U and R.
2. The required angle is a difference of two angles at R.

Answer: (d) 85°

Explanation:
1. ∠A = 180° − 45° − 55° = 80°, so ∠BAD = 40°. ∠ADC is an exterior angle of △ABD, so ∠ADC = ∠B + ∠BAD = 45° + 40° = 85°.

Teacher's Note:
1. Use the exterior angle of triangle ABD.
2. Check with triangle ADC: 40° + 55° + 85° = 180°.

Answer: (c) 126°

Explanation:
1. ∠BOC = 90° + ½ ∠A = 90° + 36° = 126°. (∠B + ∠C = 108°, half is 54°, and 180° − 54° = 126°.)

Teacher's Note:
1. Half of ∠B + ∠C is taken for the triangle BOC.
2. The result is 90° plus half of ∠A.

Answer: (d) None of these

Explanation:
1. The angle sum gives (x − 40) + (x − 20) + (½x − 10) = 180, so 2.5x − 70 = 180 and x = 100, which is not among the other options. (The angles are 60°, 80° and 40°.)

Teacher's Note:
1. Add the three expressions and equate to 180°.
2. Always solve first, then compare with the options.

Answer: (d) 121°

Explanation:
1. ∠XZY = 180° − 62° − 54° = 64°. Then ∠OYZ = 27° and ∠OZY = 32°, so ∠YOZ = 180° − 27° − 32° = 121°. (Formula: 90° + ½∠X = 121°.)

Teacher's Note:
1. Find the third angle, then halve the two base angles.
2. Use the angle sum in triangle YOZ.

Answer: (a) 38°

Explanation:
1. AD ∥ BC, so ∠ACB = ∠DAC = 32° (alternate angles). ∠AOB is an exterior angle of △OBC, so ∠AOB = ∠OBC + ∠OCB, giving ∠DBC = 70° − 32° = 38°.

Teacher's Note:
1. Alternate angles on the parallel sides AD and BC are equal.
2. Use the exterior angle property in triangle OBC.

Answer: (d) 70°

Explanation:
1. The other two angles add up to 115° and differ by 25°, so they are (115 + 25)/2 = 70° and (115 − 25)/2 = 45°. The largest angle is 70°.

Teacher's Note:
1. Use sum and difference to find two numbers.
2. Compare with the given 65° to pick the largest.

Answer: (d) equal

Explanation:
1. By the converse of the isosceles triangle theorem, sides opposite equal angles are equal.

Teacher's Note:
1. Recall the isosceles triangle theorem and its converse.
2. Equal angles go with equal opposite sides.

Answer: (c) equal

Explanation:
1. This is the isosceles triangle theorem: angles opposite equal sides are equal.

Teacher's Note:
1. This is proved using SSS or SAS congruence after drawing the bisector.
2. The converse is also true.

Answer: (a) BC

Explanation:
1. The side opposite ∠A is BC and the side opposite ∠C is AB. Equal angles have equal opposite sides, so AB = BC.

Teacher's Note:
1. Name the side opposite each angle carefully.
2. The side opposite ∠C is AB, not AC.

Answer: (a) AC

Explanation:
1. Area = ½ × AB × CE = ½ × AC × BF. Since CE = BF, AB = AC, so the triangle is isosceles.

Teacher's Note:
1. Write the area of the triangle in two ways.
2. Equal altitudes imply equal corresponding bases.

Answer: (d) CF

Explanation:
1. In right triangles BDC and CFB, BC is common and ∠B = ∠C (as AB = AC), so △BDC ≅ △CFB (AAS). Hence BD = CF.

Teacher's Note:
1. The altitudes to the equal sides of an isosceles triangle are equal.
2. Use AAS congruence on the two right triangles.

Answer: (b) less than

Explanation:
1. The perpendicular is the shortest segment from a point to a line, so AD < AB, BE < BC and CF < CA. Adding the three inequalities, the sum of the altitudes is less than the perimeter.

Teacher's Note:
1. The perpendicular is the shortest distance from a point to a line.
2. Compare each altitude with a side.

Answer: (b) included angle

Explanation:
1. This is the SAS criterion: two sides and the angle between them.

Teacher's Note:
1. SAS stands for side, angle, side.
2. The angle must lie between the two sides.

Answer: (a) ASA

Explanation:
1. Two angles and the side between them match: this is the ASA criterion. AAA is not a congruence criterion.

Teacher's Note:
1. ASA means angle, side, angle.
2. AAA gives similar triangles but not congruent ones.

Answer: (d) congruent

Explanation:
1. The third angles are equal too (angle sum), so the two angles and the included side are equal: ASA holds. This is the AAS criterion, and the triangles are congruent.

Teacher's Note:
1. Find the third angle using the angle sum.
2. Then reduce AAS to ASA.

Answer: (c) three sides

Explanation:
1. This is the SSS criterion.

Teacher's Note:
1. SSS needs no angle at all.
2. Three equal angles are not enough.

Answer: (c) hypotenuse

Explanation:
1. This is the RHS criterion: right angle, hypotenuse, side.

Teacher's Note:
1. RHS applies only to right-angled triangles.
2. The hypotenuse must be matched with the hypotenuse.

Answer: (b) perpendicular segment

Explanation:
1. Of all segments from the point to the line, the perpendicular is the shortest, and its length is the distance.

Teacher's Note:
1. The perpendicular is the shortest segment from a point to a line.
2. This fact is used to prove inequalities in triangles.