CBSE Class 10 Mathematics Constructions MCQs Set 20

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

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Multiple Choice Questions

Question 1: To divide a line segment of 12 cm internally in the ratio 3 : 2 by the ray method, the number of equal parts marked on the ray is:
(a) 5
(b) 3
(c) 2
(d) 6
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. m + n = 3 + 2 = 5 equal points A1, ..., A5 are marked on the ray AX.

Teacher's Note:
1. To divide a segment AB internally in the ratio m : n draw a ray AX at an acute angle, mark m + n equal parts A1, A2, ..., join B to A(m+n) and draw a parallel through A(m) to meet AB at P.

Question 2: In that construction, after marking the points, which point is joined to B?
(a) A3, and a parallel is drawn through A5
(b) A2, and a parallel is drawn through A3
(c) A4, and a parallel is drawn through A1
(d) A5, and a parallel to A5B is drawn through A3
Show Answer & Explanation

Answer: (d) A5, and a parallel to A5B is drawn through A3

Explanation:
1. B is joined to A(m + n) = A5, and the line through A(m) = A3 parallel to A5B meets AB at P with AP : PB = 3 : 2.

Teacher's Note:
1. To divide a segment AB internally in the ratio m : n draw a ray AX at an acute angle, mark m + n equal parts A1, A2, ..., join B to A(m+n) and draw a parallel through A(m) to meet AB at P.

Question 3: By the alternative method, a segment is divided in the ratio 3 : 4. The number of points marked on AX and BY (parallel to AX on the other side) is:
(a) 4 on AX and 3 on BY
(b) 7 on AX and none on BY
(c) 3 on AX and 4 on BY
(d) 3 on each
Show Answer & Explanation

Answer: (c) 3 on AX and 4 on BY

Explanation:
1. Mark m = 3 equal parts on AX and n = 4 equal parts on BY; the line A3B4 meets AB at P with AP : PB = 3 : 4.

Teacher's Note:
1. To divide a segment AB internally in the ratio m : n draw a ray AX at an acute angle, mark m + n equal parts A1, A2, ..., join B to A(m+n) and draw a parallel through A(m) to meet AB at P.

Question 4: To construct a triangle whose sides are 2/3 of the corresponding sides of a given triangle ABC, the number of equal points marked on the ray AX is:
(a) 2
(b) 3
(c) 5
(d) 6
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The larger of 2 and 3 is 3, so three equal points A1, A2, A3 are marked. A3 is joined to B and the parallel is drawn through A2.

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 5: For scale factor 4/3 (greater than 1), the points marked on the ray at B and the point joined to C are:
(a) 3 points, and B4 is joined to C
(b) 4 points, and the third point B3 is joined to C
(c) 4 points, and B4 is joined to C
(d) 7 points, and B3 is joined to C
Show Answer & Explanation

Answer: (b) 4 points, and the third point B3 is joined to C

Explanation:
1. The larger of 4 and 3 is 4, so B1 to B4 are marked. The smaller number 3 is used for the joining point: B3 is joined to C and a parallel to B3C is drawn through B4.

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 6: A triangle ABC has AB = 5 cm, BC = 6 cm and AC = 7 cm. A triangle with sides 2/3 of these has sides:
(a) \( \frac{10}{3},\ 4,\ \frac{14}{3}\ \text{cm} \)
(b) \( \frac{15}{2},\ 9,\ \frac{21}{2}\ \text{cm} \)
(c) \( \frac{5}{3},\ 2,\ \frac{7}{3}\ \text{cm} \)
(d) \( \frac{10}{3},\ 6,\ \frac{14}{3}\ \text{cm} \)
Show Answer & Explanation

Answer: (a) \( \frac{10}{3},\ 4,\ \frac{14}{3}\ \text{cm} \)

Explanation:
1. Multiply each side by 2/3: 5 x 2/3 = 10/3, 6 x 2/3 = 4, 7 x 2/3 = 14/3.

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 7: A triangle similar to an equilateral triangle of side 5 cm has each side 6/7 of the corresponding side. Its side is:
(a) \( \frac{35}{6}\ \text{cm} \)
(b) \( \frac{6}{7}\ \text{cm} \)
(c) \( 5\ \text{cm} \)
(d) \( \frac{30}{7}\ \text{cm} \)
Show Answer & Explanation

Answer: (d) \( \frac{30}{7}\ \text{cm} \)

Explanation:
1. Side = 5 x 6/7 = 30/7 cm, about 4.29 cm. The ray AX needs 7 equal points (the larger of 6 and 7).

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 8: A triangle similar to an isosceles triangle with base QR = 5 cm has each side 5/3 of the corresponding side. Its base is:
(a) \( 3\ \text{cm} \)
(b) \( \frac{15}{5}\ \text{cm} \)
(c) \( \frac{25}{3}\ \text{cm} \)
(d) \( \frac{5}{3}\ \text{cm} \)
Show Answer & Explanation

Answer: (c) \( \frac{25}{3}\ \text{cm} \)

Explanation:
1. Base = 5 x 5/3 = 25/3 cm.

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 9: The tangent to a circle at a point P on it is drawn by:
(a) drawing a line through P parallel to OP
(b) joining P to the centre and extending it
(c) drawing a line through P perpendicular to the radius OP
(d) drawing the chord through P of greatest length
Show Answer & Explanation

Answer: (c) drawing a line through P perpendicular to the radius OP

Explanation:
1. The tangent at P is perpendicular to the radius OP, so draw OP and then the perpendicular to OP at P.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 10: To draw the tangent at a point P of a circle without using the centre, the angle made by the tangent with chord PQ is made equal to:
(a) half of angle PRQ
(b) the angle PRQ in the alternate segment
(c) the angle POQ
(d) 90 degrees
Show Answer & Explanation

Answer: (b) the angle PRQ in the alternate segment

Explanation:
1. The angle between the tangent and a chord equals the angle in the alternate segment (angle PRQ for R on the major arc). The line BPA so drawn is the tangent.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 11: To draw tangents from an external point P (centre O known), a circle is drawn on OP as diameter because:
(a) the angle in a semicircle is 90 degrees, so OA is perpendicular to AP
(b) it passes through the centre of the given circle
(c) its radius equals the given radius
(d) it is always tangent to the given circle
Show Answer & Explanation

Answer: (a) the angle in a semicircle is 90 degrees, so OA is perpendicular to AP

Explanation:
1. If A lies on both circles, angle OAP is an angle in a semicircle and equals 90 degrees. So PA is perpendicular to the radius OA, hence PA is a tangent.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 12: A point P is 6 cm from the centre of a circle of radius 2.5 cm. The length of each tangent from P is:
(a) \( \frac{\sqrt{119}}{4}\ \text{cm} \)
(b) \( 6\ \text{cm} \)
(c) \( \sqrt{29}\ \text{cm} \)
(d) \( \frac{\sqrt{119}}{2}\ \text{cm}\ (\approx 5.45\ \text{cm}) \)
Show Answer & Explanation

Answer: (d) \( \frac{\sqrt{119}}{2}\ \text{cm}\ (\approx 5.45\ \text{cm}) \)

Explanation:
1. Length = sqrt(6^2 - 2.5^2) = sqrt(36 - 6.25) = sqrt29.75 = sqrt119 / 2, about 5.45 cm.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 13: A point P is 5.6 cm from the centre of a circle of radius 3 cm. The length of the tangent from P is about:
(a) 4.2 cm
(b) 5.6 cm
(c) 6.35 cm
(d) 4.73 cm
Show Answer & Explanation

Answer: (d) 4.73 cm

Explanation:
1. Length = sqrt(5.6^2 - 3^2) = sqrt(31.36 - 9) = sqrt22.36, about 4.73 cm.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 14: In constructing tangents from P to a circle with centre O and OP = 6 cm, the auxiliary circle on OP as diameter has radius:
(a) 6 cm
(b) 12 cm
(c) 3 cm
(d) 2.5 cm
Show Answer & Explanation

Answer: (c) 3 cm

Explanation:
1. The diameter is OP = 6 cm, so the radius is 3 cm and its centre is the mid-point of OP.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 15: Without using the centre, tangents are drawn from P by taking a secant PAB, PC = PA on the other side, a semicircle on CB, and PD perpendicular to CB. If PA = 4 cm and PB = 9 cm, the tangent length PT equals:
(a) 5 cm
(b) 6 cm
(c) 6.5 cm
(d) 36 cm
Show Answer & Explanation

Answer: (b) 6 cm

Explanation:
1. PD^2 = CP x PB = PA x PB = 36 (angle CDB = 90 degrees), and PT = PD, so the tangent length is 6 cm.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 16: A segment AB = 7 cm is divided externally in the ratio 3 : 5. The dividing point P lies on the line AB:
(a) beyond A, with PA = 10.5 cm
(b) between A and B, 2.625 cm from A
(c) beyond B, with PB = 10.5 cm
(d) beyond A, with PA = 4.2 cm
Show Answer & Explanation

Answer: (a) beyond A, with PA = 10.5 cm

Explanation:
1. For external division PA : PB = 3 : 5, the smaller part is nearer A, so P is beyond A. With PA = 3x, PB = 5x, AB = PB - PA = 2x = 7, so x = 3.5 and PA = 10.5 cm.

Teacher's Note:
1. To divide a segment AB internally in the ratio m : n draw a ray AX at an acute angle, mark m + n equal parts A1, A2, ..., join B to A(m+n) and draw a parallel through A(m) to meet AB at P.

Question 17: A segment AB = 7 cm is divided externally in the ratio 5 : 3. The dividing point P lies:
(a) beyond B, with PB = 10.5 cm
(b) beyond A, with PA = 10.5 cm
(c) between A and B, 4.375 cm from A
(d) beyond B, with PB = 4.2 cm
Show Answer & Explanation

Answer: (a) beyond B, with PB = 10.5 cm

Explanation:
1. For PA : PB = 5 : 3 the smaller part is at B, so P is beyond B. With PA = 5x, PB = 3x, AB = 2x = 7, so x = 3.5 and PB = 10.5 cm.

Teacher's Note:
1. To divide a segment AB internally in the ratio m : n draw a ray AX at an acute angle, mark m + n equal parts A1, A2, ..., join B to A(m+n) and draw a parallel through A(m) to meet AB at P.

Question 18: A triangle ABC has BC = 7 cm, angle B = 45 degrees and angle A = 105 degrees. A triangle with sides 4/3 of the corresponding sides has its base along BC extended to:
(a) \( \frac{21}{4}\ \text{cm} \)
(b) \( \frac{14}{3}\ \text{cm} \)
(c) \( 9\ \text{cm} \)
(d) \( \frac{28}{3}\ \text{cm} \)
Show Answer & Explanation

Answer: (d) \( \frac{28}{3}\ \text{cm} \)

Explanation:
1. Angle C = 180 - 45 - 105 = 30 degrees. The corresponding base BC' = 7 x 4/3 = 28/3 cm.

Teacher's Note:
1. Similar triangle by scale factor m/n: draw the given triangle, make an acute angle at one end of a base on the side opposite the vertex, mark the larger of m and n equal points, join the n-th point to the other end of the base, draw a parallel through the m-th point, then a parallel to the third side.

Question 19: A circle of diameter 6 cm is drawn and a tangent is to be drawn at the end B of the diameter AOB. The tangent is:
(a) a line through B parallel to AB
(b) a line through O perpendicular to AB
(c) a line through B perpendicular to AB
(d) the chord through B and a point on the circle
Show Answer & Explanation

Answer: (c) a line through B perpendicular to AB

Explanation:
1. The tangent at B is perpendicular to the radius OB, which lies along AB, so it is the perpendicular to AB at B.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

Question 20: A circle of radius 4 cm with centre O and a point P at distance 10 cm from O. The tangents PA and PB make with OP an angle whose sine is:
(a) 0.6
(b) 0.4
(c) 0.8
(d) 0.25
Show Answer & Explanation

Answer: (b) 0.4

Explanation:
1. In the right triangle OAP, OA = 4 and OP = 10, so sin(angle OPA) = 4/10 = 0.4.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact. From an external point P the tangents are drawn using the circle on OP as diameter (angle in a semicircle is 90 degrees), or, without the centre, using a secant and a semicircle on CB.

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