CBSE Class 10 Mathematics Lines and Angles MCQs Set 07

Practice MCQs for Class 10 Mathematics Lines and Angles

Access targeted multiple-choice questions for Lines and Angles designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: If x° is the measure of an angle equal to its complement and y° is the measure of an angle equal to its supplement, then x°/y° is
(a) 1
(b) 2
(c) 0.5
(d) 2.5
Show Answer & Explanation

Answer: (c) 0.5

Explanation:
1. x = 90 − x, so x = 45
2. y = 180 − y, so y = 90
3. x/y = 45/90 = 0.5.

Teacher's Note:
1. An angle equal to its complement is half of 90°.
2. An angle equal to its supplement is half of 180°.

Question 2: Lines EF and GH are parallel and GI is a transversal, with I on EF and G on GH; F and H lie on the same side of GI. If ∠IGH = y° and ∠FIG = 3y°, then the ratio of the supplement to the complement of y° is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 3
(d) 3 : 2
Show Answer & Explanation

Answer: (b) 3 : 1

Explanation:
1. ∠FIG and ∠IGH are interior angles on the same side of the transversal, so 3y + y = 180°
2. y = 45°
3. Supplement = 135° and complement = 45°, ratio 3 : 1.

Teacher's Note:
1. Interior angles on the same side of a transversal are supplementary.
2. Then find supplement and complement separately.

Question 3: PQ ∥ RS and PR ∥ QS, so PQSR is a parallelogram with angles a, b, c, d at P, Q, R, S respectively. If c = 120°, then a is
(a) 80°
(b) 70°
(c) 60°
(d) 90°
Show Answer & Explanation

Answer: (c) 60°

Explanation:
1. Angles at adjacent vertices R and P of a parallelogram are supplementary (PR is a transversal of PQ and RS)
2. a = 180° − 120° = 60°.

Teacher's Note:
1. Co-interior angles between parallel lines add up to 180°.
2. P and R are joined by the side PR.

Question 4: In parallelogram PQSR (PQ ∥ RS, PR ∥ QS) the angles at P, Q, R, S are ∠A, ∠B, ∠C, ∠D respectively. If ∠C = 110°, then ∠B =
(a) 90°
(b) 110°
(c) 100°
(d) 120°
Show Answer & Explanation

Answer: (b) 110°

Explanation:
1. Q and R are opposite vertices of the parallelogram
2. Opposite angles of a parallelogram are equal, so ∠B = ∠C = 110°.

Teacher's Note:
1. Opposite angles of a parallelogram are equal.
2. Adjacent angles are supplementary.

Question 5: If a and b are complementary angles, b and c are supplementary angles, and the average of a and c is 85°, then a + b + c =
(a) 200°
(b) 220°
(c) 210°
(d) 240°
Show Answer & Explanation

Answer: (b) 220°

Explanation:
1. a + b = 90° and b + c = 180°, so c = 90° + a
2. a + c = 170°, so 2a + 90° = 170° and a = 40°
3. b = 50°, c = 130°, so a + b + c = 220°.

Teacher's Note:
1. Express a and c in terms of b.
2. Use the average to form an equation.

Question 6: The ratio between the complement and the supplement of an angle is 1 : 2. The angle is
(a) 90°
(b) 180°
(c) 135°
(d) 0°
Show Answer & Explanation

Answer: (d) 0°

Explanation:
1. (90 − x) : (180 − x) = 1 : 2, so 2(90 − x) = 180 − x
2. 180 − 2x = 180 − x, so x = 0°.

Teacher's Note:
1. Form the ratio equation with both measures in terms of x.
2. Here the equation gives the degenerate angle 0°.

Question 7: AFB is a straight line. Three angles x°, y°, z° are formed one after another on one side of AB at point F, with x = 2y = z. Then (x + y) : (y + 2z) is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 3 : 5
Show Answer & Explanation

Answer: (d) 3 : 5

Explanation:
1. x + y + z = 180° on a straight line, with x = z = 2y
2. 5y = 180°, so y = 36°, x = z = 72°
3. (x + y) : (y + 2z) = 108 : 180 = 3 : 5.

Teacher's Note:
1. Angles on a straight line add up to 180°.
2. Write all three angles in terms of y.

Question 8: MN and OP are two straight lines intersecting at R, and ∠ORN = (5/2) of ∠ORM. Then ∠NRP : ∠MRP is
(a) 2 : 5
(b) 5 : 2
(c) 1 : 1
(d) 3 : 7
Show Answer & Explanation

Answer: (a) 2 : 5

Explanation:
1. ∠ORM and ∠ORN form a linear pair, so ∠ORM = 360°/7 and ∠ORN = 900°/7
2. ∠NRP is vertically opposite to ∠MRO, and ∠MRP is vertically opposite to ∠ORN
3. So ∠NRP : ∠MRP = ∠ORM : ∠ORN = 2 : 5.

Teacher's Note:
1. Vertically opposite angles are equal.
2. Linear pair angles add up to 180°.

Question 9: Lines ℓ ∥ m ∥ n are cut by two transversals p and q. Points A, B, C lie on p (on ℓ, m, n respectively) and D, E, F lie on q (on ℓ, m, n respectively). If AB : BC = 2 : 1 and DE = 5 cm, then EF =
(a) 3 cm
(b) 2 cm
(c) 5 cm
(d) 2.5 cm
Show Answer & Explanation

Answer: (d) 2.5 cm

Explanation:
1. Three parallel lines cut transversals in the same ratio: DE : EF = AB : BC = 2 : 1
2. EF = 5/2 = 2.5 cm.

Teacher's Note:
1. Parallel lines make proportional intercepts on any two transversals.
2. This is the equal-ratio (intercept) theorem.

Question 10: Lines ℓ ∥ m ∥ n are cut by two transversals p and q. Transversal p meets them at P, Q, R and q meets them at L, M, N. If PQ : QR = 4 : 3 and LN = 21 cm, the lengths of LM and MN are
(a) 9 cm, 15 cm
(b) 9 cm, 12 cm
(c) 12 cm, 9 cm
(d) 15 cm, 9 cm
Show Answer & Explanation

Answer: (c) 12 cm, 9 cm

Explanation:
1. LM : MN = PQ : QR = 4 : 3
2. LM = 4/7 × 21 = 12 cm and MN = 3/7 × 21 = 9 cm.

Teacher's Note:
1. The ratio of intercepts is the same on both transversals.
2. Divide the total length 21 in the ratio 4 : 3.

Question 11: Lines AB, CD and EF are parallel, and lines GH and IJ are parallel to each other and cut AB, CD, EF at K, L, M (on GH) and N, O, P (on IJ). If KL : LM = 2 : 3, then NO : OP =
(a) 2 : 3
(b) 3 : 2
(c) 2 : 5
(d) 3 : 5
Show Answer & Explanation

Answer: (a) 2 : 3

Explanation:
1. The three parallel lines AB, CD, EF cut the transversals GH and IJ in the same ratio
2. NO : OP = KL : LM = 2 : 3.

Teacher's Note:
1. Parallel lines cut any two transversals in the same ratio.
2. The ratio does not change if the transversals are parallel.

Question 12: CD and GH are two straight lines intersecting at O. If the supplement and the complement of ∠GOC are in the ratio 31 : 13, then ∠HOC (the angle adjacent to ∠GOC on line GH) =
(a) 155°
(b) 125°
(c) 25°
(d) 75°
Show Answer & Explanation

Answer: (a) 155°

Explanation:
1. (180 − x) : (90 − x) = 31 : 13, so 13(180 − x) = 31(90 − x)
2. 2340 − 13x = 2790 − 31x, so x = 25°
3. ∠HOC = 180° − 25° = 155°.

Teacher's Note:
1. Let the angle be x and form the ratio equation.
2. Adjacent angles on a straight line are supplementary.

Question 13: There are two angles a and b such that a = b + 15°. If the supplement of 3b is twice the complement of (b + 15°), the value of a is
(a) 30
(b) 45
(c) 50
(d) 55
Show Answer & Explanation

Answer: (b) 45

Explanation:
1. 180 − 3b = 2(90 − b − 15) = 150 − 2b
2. b = 30°, so a = 45°.

Teacher's Note:
1. Write each condition as an equation.
2. Solve for b, then find a.

Question 14: In quadrilateral ABCD, AD ∥ BC, ∠DCB = 100° and ∠DAB = 50°. Then ∠ABC − ∠ADC =
(a) 50°
(b) 180°
(c) 200°
(d) 210°
Show Answer & Explanation

Answer: (a) 50°

Explanation:
1. AB is a transversal of AD and BC, so ∠ABC = 180° − 50° = 130°
2. DC is a transversal, so ∠ADC = 180° − 100° = 80°
3. Difference = 130° − 80° = 50°.

Teacher's Note:
1. Co-interior angles between parallel lines are supplementary.
2. Use each leg of the trapezium as a transversal.

Question 15: MN ∥ OP and EF is a transversal meeting MN at A and OP at B (E above A, F below B). If ∠FBO = (3x + 5)° and ∠MAE = (2x)°, then (supplement of ∠PBA) : (complement of ∠NAB) =
(a) 2 : 7
(b) 7 : 2
(c) 1 : 1
(d) 1 : 6
Show Answer & Explanation

Answer: (b) 7 : 2

Explanation:
1. ∠MAE equals the corresponding angle ∠OBA, and ∠FBO + ∠OBA = 180°, so 3x + 5 + 2x = 180°, x = 35°
2. ∠MAE = 70°, ∠PBA = 110°, so its supplement is 70°
3. ∠NAB is vertically opposite ∠MAE, so ∠NAB = 70° and its complement is 20°
4. Ratio = 70 : 20 = 7 : 2.

Teacher's Note:
1. Identify corresponding and vertically opposite angles first.
2. Then compute the supplement and the complement.

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Practice MCQs for Class 10 Mathematics Lines and Angles

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FAQs

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