CBSE Class 10 Mathematics Question Bank Set 05

CONSTRUCTION

IMPORTANT CONCEPTS:-
TAKE A LOOK

1. Division of line segment in the given ratio.
2. Construction of triangles:-
a. When three sides are given.
b. When two sides and included angle given.
c. When two angles and one side given.
d. Construction of right angled triangle.
3. Construction of triangle similar to a given triangle as per given scale.
4. Construction of tangents to a circle.

LEVEL - I

Question 1. Divide a line segment in given ratio.
Answer: To divide a given line segment AB in a given ratio \(m:n\) (where both \(m\) and \(n\) are positive integers):
1. Draw a line segment AB of the given length.
2. Draw any ray AX making an acute angle with AB.
3. Locate \(m + n\) points (i.e., \(A_1, A_2, \dots, A_{m+n}\)) on AX at equal distances.
4. Join \(B\) to \(A_{m+n}\).
5. Through the point \(A_m\), draw a line parallel to \(BA_{m+n}\) intersecting AB at point C.
Point C divides AB internally in the ratio \(m:n\).
In simple words: A line is divided proportionally by drawing an acute ray, marking equal segments, connecting the last mark to the end of the line, and drawing a parallel line.

Exam Tip: Ensure the acute angle made by ray AX with AB is sharp (around \(60^\circ\) or \(45^\circ\)) for accurate compass work.

Question 2. Draw a line segment AB=8cm and divide it in the ratio 4:3.
Answer: Steps of construction:
1. Draw a line segment \(AB = 8\) cm.
2. Draw a ray AX making an acute angle with AB.
3. Locate \(4 + 3 = 7\) points \(A_1, A_2, A_3, A_4, A_5, A_6, A_7\) on ray AX at equal intervals.
4. Join \(A_7\) to B.
5. Through \(A_4\), draw a line parallel to \(A_7B\) intersecting AB at point C.
6. AC : CB = 4 : 3.
In simple words: Divide an 8 cm line into 7 equal parts using a ray and parallel lines, such that 4 parts are on one side of point C and 3 on the other.

Exam Tip: Measure AC and CB at the end to verify that \(AC = \frac{4}{7} \times 8\) cm and \(CB = \frac{3}{7} \times 8\) cm.

Question 3. Divide a line segment of 7cm internally in the ratio 2:3.
Answer: Steps of construction:
1. Draw a line segment \(AB = 7\) cm.
2. Make an acute angle at A by drawing ray AX.
3. Mark \(2 + 3 = 5\) equidistant points \(A_1, A_2, A_3, A_4, A_5\) on AX.
4. Join \(A_5\) to B.
5. Through \(A_2\), draw a line parallel to \(A_5B\) meeting AB at point C.
6. Thus, point C divides AB internally in the ratio 2:3.
In simple words: Use a ray with 5 equal divisions to split a 7 cm segment into a 2:3 ratio.

Question 4. Draw a circle of radius 4 cm. Take a point P on it. Draw tangent to the given circle at p.
Answer: Steps of construction:
1. Draw a circle of radius \(r = 4\) cm with centre O.
2. Take a point P on the circumference of the circle.
3. Join radius OP.
4. Construct an angle of \(90^\circ\) at point P using a compass or set square, or draw a perpendicular line to radius OP at point P.
5. Extend the line to get the required tangent XY to the circle at point P.
In simple words: Draw a radius to the given point on the circle and construct a perpendicular line at that point to form the tangent.

Exam Tip: Remember that the radius is always perpendicular to the tangent at the point of contact.

Question 5. Construct an isosceles triangle whose base 7.5 cm and altitude is 4.2 cm.
Answer: Steps of construction:
1. Draw a line segment \(BC = 7.5\) cm as the base.
2. Draw the perpendicular bisector of BC intersecting BC at point D.
3. With D as centre, cut an arc of length \(AD = 4.2\) cm along the perpendicular bisector to locate vertex A.
4. Join AB and AC.
5. Triangle ABC is the required isosceles triangle.
In simple words: Draw the base, find its midpoint, erect a perpendicular line equal to the given height, and connect the top to the base endpoints.

Exam Tip: The perpendicular bisector of the base in an isosceles triangle passes through the opposite vertex and bisects the base.

LEVEL - II

Question 1. Construct a triangle of sides 4cm , 5cm and 6cm and then triangle similar to it whose side are 2/3 of corresponding sides of the first triangle.
Answer: Steps of construction:
1. Draw a line segment \(BC = 6\) cm.
2. With B as centre and radius 5 cm, draw an arc. With C as centre and radius 4 cm, draw another arc intersecting the first arc at point A.
3. Join AB and AC to form triangle ABC.
4. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
5. Locate 3 points \(B_1, B_2, B_3\) on BX at equal distances such that \(BB_1 = B_1B_2 = B_2B_3\).
6. Join \(B_3\) to C.
7. Through \(B_2\), draw a line parallel to \(B_3C\) intersecting BC at C'.
8. From C', draw a line parallel to CA intersecting AB at A'.
9. Triangle A'BC' is the required similar triangle.
In simple words: Construct the original triangle, draw an acute ray with 3 divisions, join the end, and draw parallel lines to scale down the triangle by a factor of 2/3.

Exam Tip: Since the scaling factor is less than 1 (\(2/3 < 1\)), the new triangle will lie inside the original triangle.

Question 2. Construct a triangle similar to a given triangle ABC such that each of its sides is 2/3rd of the corresponding sides of ABC. It is given that AB=4cm BC=5cm and AC=6cm also write the steps of construction.
Answer: Steps of construction:
1. Draw base \(BC = 5\) cm.
2. With B as centre and radius 4 cm, draw an arc. With C as centre and radius 6 cm, draw an arc intersecting at A.
3. Join AB and AC to form triangle ABC.
4. Draw an acute ray BX making an angle below BC.
5. Mark 3 equal points \(B_1, B_2, B_3\) on BX.
6. Join \(B_3C\).
7. Draw a line through \(B_2\) parallel to \(B_3C\) meeting BC at C'.
8. Draw a line through C' parallel to CA meeting AB at A'.
9. Triangle A'BC' is the required similar triangle.
In simple words: Draw triangle ABC with given side lengths, then use a ratio scale of 2/3 via a side ray to construct a smaller similar triangle inside it.

Exam Tip: Clearly label all points A, B, C, A', C', and the division points on the ray to ensure full marks.

Question 3. Draw a right triangle ABC in which \(\angle B = 90^\circ\), AB=5cm, BC=4cm then construct another triangle ABC whose sides are 5/3 times the corresponding sides of \(\Delta ABC\).
Answer: Steps of construction:
1. Draw a line segment \(BC = 4\) cm.
2. At point B, construct a right angle (\(90^\circ\)) using a compass.
3. With B as centre and radius 5 cm, cut an arc on the perpendicular line to locate vertex A.
4. Join AC to form right-angled triangle ABC.
5. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
6. Locate 5 points \(B_1, B_2, B_3, B_4, B_5\) on BX at equal distances.
7. Join \(B_3\) to C.
8. Through \(B_5\), draw a line parallel to \(B_3C\) meeting the extended line BC at C'.
9. Through C', draw a line parallel to CA meeting the extended line AB at A'.
10. Triangle A'BC' is the required similar triangle.
In simple words: Construct the right triangle, extend its base and side, and use a 5-division ray to construct a larger similar triangle outside it by a factor of 5/3.

Exam Tip: Since the scaling factor (\(5/3 > 1\)) is greater than 1, the new triangle will be larger than the original and will enclose it.

Question 4. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other at an angle of \(60^\circ\).
Answer: Steps of construction:
1. Draw a circle of radius \(r = 5\) cm with centre O.
2. Choose a point A on the circumference and draw radius OA.
3. Draw another radius OB such that angle between radii \(\angle AOB = 180^\circ - 60^\circ = 120^\circ\).
4. Draw perpendicular lines (tangents) at points A and B to the radii OA and OB respectively.
5. These two tangents intersect at an external point P, forming an angle of \(60^\circ\) at P.
In simple words: Draw two radii at an angle supplementary to the given inclination angle (\(180^\circ - 60^\circ = 120^\circ\)), then draw perpendiculars at their endpoints to form the tangents.

Exam Tip: The angle between the two tangents and the angle between the two radii are supplementary (their sum is \(180^\circ\)).

Question 5. Draw a circle of radius 5cm from a point 8cm away from its centre construct the pair of tangents to the circle and measure their length.
Answer: Steps of construction:
1. Draw a circle with centre O and radius \(r = 5\) cm.
2. Take a point P at a distance of 8 cm from centre O and join OP.
3. Draw the perpendicular bisector of OP, letting M be the midpoint of OP.
4. With M as centre and radius equal to PM (or OM), draw a circle that intersects the given circle at two points, say T and T'.
5. Join PT and PT'. These are the required tangents.
6. Measuring the length PT with a ruler gives approximately \(\sqrt{8^2 - 5^2} = \sqrt{64 - 25} = \sqrt{39} \approx 6.24\) cm.
In simple words: Draw the outer point, find the midpoint of the distance to the center, draw a reference circle, and connect the outer point to the intersection points.

Calculation verification:
\(\text{Tangent length} = \sqrt{(\text{Distance})^2 - (\text{Radius})^2}\)
\(= \sqrt{8^2 - 5^2} = \sqrt{64 - 25} = \sqrt{39} \approx 6.24 \text{ cm}\)

Exam Tip: Always verify your geometric construction length analytically using Pythagoras theorem in triangle OPT (\(PT^2 = OP^2 - OT^2\)).

Question 6. Construct a triangle PQR in which QR=6cm \(\angle Q = 60^\circ\) and \(\angle R = 45^\circ\). Construct another triangle similar to \(\Delta PQR\) such that its sides are 5/6 of the corresponding sides of \(\Delta PQR\).
Answer: Steps of construction:
1. Draw a line segment \(QR = 6\) cm.
2. At point Q, construct an angle of \(60^\circ\).
3. At point R, construct an angle of \(45^\circ\).
4. Let the rays intersect at vertex P. Triangle PQR is formed.
5. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
6. Locate 6 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6\) on QX at equal intervals.
7. Join \(Q_6\) to R.
8. Through \(Q_5\), draw a line parallel to \(Q_6R\) meeting QR at R'.
9. From R', draw a line parallel to RP meeting QP at P'.
10. Triangle P'QR' is the required similar triangle.
In simple words: Construct the triangle using the base and two angles, then use a 6-division auxiliary ray to draw a smaller similar triangle inside it with ratio 5/6.

Exam Tip: Ensure the sum of angles in the triangle is verified: \(\angle P = 180^\circ - (60^\circ + 45^\circ) = 75^\circ\).