CBSE Class 10 Mathematics Constructions MCQs Set 18

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 12 Surface Areas and Volumes

Review structured MCQ sets for Class 10 Mathematics Chapter 12 Surface Areas and Volumes. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 12 Surface Areas and Volumes MCQs for Class 10 Mathematics

View or download the dedicated Chapter 12 Surface Areas and Volumes MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: A line segment of length 7 cm is divided in the ratio 2 : 3. The lengths of the two parts are:
(a) 2.8 and 4.2 cm
(b) 3 and 4 cm
(c) 3.5 and 3.5 cm
(d) 2 and 5 cm
Show Answer & Explanation

Answer: (a) 2.8 and 4.2 cm

Explanation:
1. The parts are 7 x 2/5 = 2.8 cm and 7 x 3/5 = 4.2 cm.

Teacher's Note:
1. If a segment is divided in the ratio m : n, the parts are m/(m + n) and n/(m + n) of its length.

Question 2: A line segment of length 7.8 cm is divided in the ratio 5 : 8. The lengths of the two parts are:
(a) 3.9 and 3.9 cm
(b) 2.6 and 5.2 cm
(c) 4.8 and 3 cm
(d) 3 and 4.8 cm
Show Answer & Explanation

Answer: (d) 3 and 4.8 cm

Explanation:
1. The parts are 7.8 x 5/13 = 3 cm and 7.8 x 8/13 = 4.8 cm (in that order).

Teacher's Note:
1. If a segment is divided in the ratio m : n, the parts are m/(m + n) and n/(m + n) of its length.

Question 3: To divide a line segment AB in the ratio 3 : 2 by drawing a ray AX making an acute angle with AB, the number of equal-distance points marked on AX is:
(a) 3
(b) 2
(c) 5
(d) 6
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. The number of points marked is m + n = 3 + 2 = 5. Join the last point A5 to B and draw a line through A3 parallel to A5B to meet AB at C.

Teacher's Note:
1. To divide a segment in the ratio m : n, mark m + n equal steps on a ray from one end and join the last point to the other end.

Question 4: In the construction dividing AB in the ratio 3 : 2, a line through A3 parallel to A5B meets AB at C. The reason AC : CB = 3 : 2 is:
(a) the Pythagoras theorem
(b) the Basic Proportionality Theorem in triangle AA5B
(c) the angle sum property of a triangle
(d) the SSS congruence criterion
Show Answer & Explanation

Answer: (b) the Basic Proportionality Theorem in triangle AA5B

Explanation:
1. CA3 is parallel to BA5, so by the Basic Proportionality Theorem AC/CB = AA3/A3A5 = 3/2.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 5: A triangle with sides 5 cm, 6 cm and 7 cm is to be enlarged or reduced to a similar triangle with sides 2/3 of the corresponding sides. The sides of the new triangle are:
(a) 5, 6 and 7 cm
(b) 10/3, 4 and 14/3 cm
(c) 3, 4 and 5 cm
(d) 7.5, 9 and 10.5 cm
Show Answer & Explanation

Answer: (b) 10/3, 4 and 14/3 cm

Explanation:
1. Multiply each side by 2/3: 10/3 cm, 4 cm and 14/3 cm.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 6: A triangle with sides 5 cm, 6.5 cm and 7.6 cm is enlarged to a similar triangle with sides 7/5 of the corresponding sides. The sides of the new triangle are:
(a) 7, 9.1 and 10.64 cm
(b) 3.57, 4.64 and 5.43 cm
(c) 6.4, 7.9 and 9.0 cm
(d) 7, 9.1 and 10.4 cm
Show Answer & Explanation

Answer: (a) 7, 9.1 and 10.64 cm

Explanation:
1. Multiply each side by 7/5: 5 x 1.4 = 7, 6.5 x 1.4 = 9.1 and 7.6 x 1.4 = 10.64 cm.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 7: A triangle ABC with sides 5 cm, 12 cm and 13 cm is to be reduced to a similar triangle with sides 3/5 of the corresponding sides. The sides of the new triangle are:
(a) 2, 4.8 and 5.2 cm
(b) 3, 7.2 and 8.0 cm
(c) 5, 12 and 13 cm
(d) 3, 7.2 and 7.8 cm
Show Answer & Explanation

Answer: (d) 3, 7.2 and 7.8 cm

Explanation:
1. Multiply each side by 3/5: 3 cm, 7.2 cm and 7.8 cm. Since 5² + 12² = 13², ABC (and the new triangle) is right-angled.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 8: A triangle with sides 6 cm, 7 cm and 8 cm is enlarged to a similar triangle whose sides are 1.4 times the corresponding sides. The new sides are:
(a) 7.4, 8.4 and 9.4 cm
(b) 4.3, 5, and 5.7 cm
(c) 8.4, 9.8 and 11.2 cm
(d) 8.4, 9.8 and 12.2 cm
Show Answer & Explanation

Answer: (c) 8.4, 9.8 and 11.2 cm

Explanation:
1. Multiply each side by 1.4: 8.4 cm, 9.8 cm and 11.2 cm.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 9: A right triangle has the two sides (other than the hypotenuse) 3 cm and 4 cm. A similar triangle is constructed with sides 5/3 of the corresponding sides. Its hypotenuse is:
(a) 5 cm
(b) 20/3 cm
(c) 25/3 cm
(d) 10 cm
Show Answer & Explanation

Answer: (c) 25/3 cm

Explanation:
1. The hypotenuse of the given triangle is 5 cm. The new hypotenuse is 5 x 5/3 = 25/3 cm. The other sides are 5 cm and 20/3 cm.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 10: An isosceles triangle has base 7 cm and altitude 4 cm. A similar triangle is constructed with sides 3/2 times the corresponding sides. Its base is:
(a) 4.67 cm
(b) 10.5 cm
(c) 7 cm
(d) 14 cm
Show Answer & Explanation

Answer: (b) 10.5 cm

Explanation:
1. The base of the new triangle is 7 x 3/2 = 10.5 cm and its altitude is 6 cm.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 11: Triangle ABC has BC = 6 cm, angle B = 45 degrees and angle A = 105 degrees. A triangle is constructed similar to it with sides 4/3 of the corresponding sides. The side corresponding to BC is:
(a) 8 cm
(b) 4.5 cm
(c) 6 cm
(d) 7.5 cm
Show Answer & Explanation

Answer: (a) 8 cm

Explanation:
1. The side corresponding to BC is 6 x 4/3 = 8 cm. The angles of the new triangle are the same: 45, 105 and 30 degrees.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 12: Triangle ABC has BC = 6 cm, angle B = 45 degrees and angle A = 105 degrees. The measure of angle C is:
(a) 45 degrees
(b) 60 degrees
(c) 105 degrees
(d) 30 degrees
Show Answer & Explanation

Answer: (d) 30 degrees

Explanation:
1. The angle sum is 180 degrees, so angle C = 180 - 45 - 105 = 30 degrees.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 13: To construct a triangle similar to a given triangle with scale factor 3/5 (smaller triangle), the number of equal parts marked on the ray from the vertex is:
(a) 3
(b) 2
(c) 8
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. The greater of 3 and 5 is 5, so five points are marked. The 5th point is joined to the other end of the base and a parallel is drawn through the 3rd point.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 14: To construct a triangle similar to a given triangle with scale factor 7/5 (larger triangle), the number of equal parts marked on the ray is:
(a) 5
(b) 2
(c) 7
(d) 12
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. For a scale factor m/n with m greater than n, mark m = 7 equal parts. Join the 5th point to the end of the base and draw a parallel through the 7th point to the extended base.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 15: In the construction of a triangle similar to triangle ABC with scale factor 4/3, the criterion that proves the new triangle A'BC' is similar to ABC is:
(a) SSS congruence
(b) AA similarity
(c) SAS congruence
(d) RHS congruence
Show Answer & Explanation

Answer: (b) AA similarity

Explanation:
1. A'C' is parallel to AC, so angle BA'C' = angle BAC and angle B is common. The triangles are similar by the AA criterion.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 16: Two triangles ABC and A'B'C' are similar with each side of A'B'C' equal to 1.5 times the corresponding side of ABC. The ratio of their perimeters is:
(a) 1.5 : 1
(b) 2.25 : 1
(c) 1 : 1.5
(d) 1 : 2.25
Show Answer & Explanation

Answer: (a) 1.5 : 1

Explanation:
1. The perimeter of similar triangles is in the same ratio as the corresponding sides, 1.5 : 1. The areas would be in the ratio 2.25 : 1.

Teacher's Note:
1. A line parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem). Triangles with equal corresponding angles are similar (AA criterion), and corresponding sides of similar triangles are in the same ratio.

Question 17: To draw a tangent to a circle at a point P on it using the centre O, the tangent is drawn:
(a) perpendicular to the radius OP at P
(b) parallel to the radius OP at P
(c) along the line OP
(d) at 45 degrees to the radius OP
Show Answer & Explanation

Answer: (a) perpendicular to the radius OP at P

Explanation:
1. A tangent at a point of contact is perpendicular to the radius through that point.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius at the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is the square root of (d^2 - r^2).
2. The two tangents from an external point are equal in length.

Question 18: A point P is at a distance of 5 cm from the centre of a circle of radius 4 cm. The length of the tangent from P to the circle is:
(a) 1 cm
(b) 4 cm
(c) √41 cm
(d) 3 cm
Show Answer & Explanation

Answer: (d) 3 cm

Explanation:
1. The tangent length = √(5² - 4²) = √9 = 3 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius at the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is the square root of (d^2 - r^2).
2. The two tangents from an external point are equal in length.

Question 19: A point P is 7 cm from the centre of a circle of diameter 8 cm. The length of each tangent from P to the circle is, to one decimal place:
(a) 3.0 cm
(b) 6.0 cm
(c) 5.7 cm
(d) 8.1 cm
Show Answer & Explanation

Answer: (c) 5.7 cm

Explanation:
1. The radius is 4 cm. Tangent length = √(49 - 16) = √33 = 5.7 cm (the two tangents are equal).

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius at the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is the square root of (d^2 - r^2).
2. The two tangents from an external point are equal in length.

Free study material for Mathematics

Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

Class 10 Mathematics Chapter 12 Surface Areas and Volumes Objective Test Questions

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FAQs

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