NCERT Solutions Class 10 Mathematics Chapter 11 Construction

Exercise 11.1

Q.1) In each of the following, give the justification of the construction also:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two arts.
Sol.1) Steps of Construction:
Step I: 𝐴𝐵 = 7.6 𝑐𝑚 is drawn.
Step II: A ray 𝐴𝑋 making an acute angle with
𝐴𝐵 is drawn.
Step III: After that, a ray BY is drawn parallel to
𝐴𝑋 making equal acute angle as in the previous step.
Step IV: Point 𝐴1, 𝐴2, 𝐴3, 𝐴4 and 𝐴5 is marked on 𝐴𝑋
and point 𝐵1, 𝐵2. ... to 𝐵8 is marked on BY such that 𝐴𝐴1 = 𝐴1𝐴2 = 𝐴2𝐴3 =
. . . . 𝐵𝐵1 = 𝐵1𝐵2 = . . . . 𝐵7𝐵8
Step V: 𝐴5 and 𝐵8 is joined and it intersected AB at point C diving it in the ratio 5:8.
𝐴𝐶 ∶ 𝐶𝐵 = 5 ∶ 8
Justification:
𝛥𝐴𝐴5𝐶 ~ 𝛥𝐵𝐵8𝐶
∴ 𝐴𝐴5/𝐵𝐵8 = 𝐴𝐶/𝐵𝐶 = 5/8

Q.2) Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle
Sol.2) Steps of Construction:
Step I: 𝐴𝐵 = 6 𝑐𝑚 is drawn.
Step II: With A as a centre and radius equal to 4 cm, an arc is draw
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form 𝛥𝐴𝐵𝐶.
Step V: A ray 𝐴𝑋 is drawn making an acute angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on 𝐴𝑋 as A1 A2....A5
Step VII: 𝐴5𝐵 is joined. 𝐴2𝐵′ is drawn parallel to 𝐴5𝐵 and 𝐵′𝐶′ is drawn parallel to 𝐵𝐶.
𝛥𝐴𝐵′𝐶′ is the required triangle
Justification:
∠𝐴 (Common)
∠𝐶 = ∠𝐶′ and
∠𝐵 = ∠ 𝐵′ (corresponding angles)
Thus 𝛥𝐴𝐵′𝐶′ ~ 𝛥𝐴𝐵𝐶 by AAA similarity condition

Q.3) Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Sol.3) Steps of Construction:
Step I: A triangle 𝐴𝐵𝐶 with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray 𝐵𝑋 making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as 𝐵1, 𝐵2, 𝐵3, 𝐵4, 𝐵5, 𝐵6 and 𝐵7 are marked on 𝐵𝑋.
Step IV; Point 𝐵5 is joined with C to draw 𝐵5𝐶.
Step V: 𝐵7𝐶′ is drawn parallel to 𝐵5𝐶 and 𝐶′𝐴′ is parallel to 𝐶𝐴.
Thus 𝐴′𝐵𝐶′ is the required triangle.
Justification
𝛥𝐴𝐵′𝐶′ ~ 𝛥𝐴𝐵𝐶 by AAA similarity condition

Q.4) Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Sol.4) Steps of Construction:
Step I: 𝐵𝐶 = 5 𝑐𝑚 is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects 𝐵𝐶 at O.
Step III: At a distance of 4 cm, a point A is marked on perpendicular bisector of BC.
Step IV: AB and AC is joined to form 𝛥𝐴𝐵𝐶.
Step V: A ray 𝐵𝑋 is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points 𝐵1, 𝐵2 and B3 is marked 𝐵𝑋.
Step VII: 𝐵2 is joined with C to form 𝐵2𝐶.
Step VIII: 𝐵3𝐶′ is drawn parallel to 𝐵2𝐶 and 𝐶′𝐴′ is drawn parallel to 𝐶𝐴.
Thus, 𝐴′𝐵𝐶′ is the required triangle formed.
Justification:
𝛥𝐴𝐵′𝐶′ ~ 𝛥𝐴𝐵𝐶 by AA similarity condition.

Q.5) Draw a triangle ABC with side 𝐵𝐶 = 6 𝑐𝑚, 𝐴𝐵 = 5 𝑐𝑚 and ∠𝐴𝐵𝐶 = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Sol.5) Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an ∠𝐴𝐵𝐶 = 60° with BC.
Step III: AC is joined to form 𝛥𝐴𝐵𝐶.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points 𝐵1, 𝐵2, 𝐵3 and 𝐵4 at equal distance is marked on BX.
Step VII: B3 is joined with C' to form 𝐵3𝐶′.
Step VIII: 𝐶′𝐴′ is drawn parallel CA.
Thus, A'BC' is the required triangle.
Justification:
∠A = 60° (Common)
∠𝐶 = ∠𝐶′ 𝛥𝐴𝐵′𝐶′ ~ 𝛥𝐴𝐵𝐶 by AA similarity condition.

Q.6) Draw a triangle ABC with side 𝐵𝐶 = 7 𝑐𝑚, ∠𝐵 = 45°, ∠𝐴 = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC
Sol.6) Sum of all side of triangle = 180°
∴ ∠𝐴 + ∠𝐵 + ∠𝐶 = 180°
∠𝐶 = 180° − 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45° with BC.
Step III: At C, a ray making an angle of 30° with BC is drawn intersect ting the previous s ray at A.
Thus, ∠𝐴 = 105°.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1, B2, B3 and B4 at equal distance is marked on 𝐵𝑋.
Step VI: 𝐵3𝐶 is joined and 𝐵4𝐶′ is made parallel to 𝐵3𝐶.
Step VII: 𝐶′𝐴′ is made parallel 𝐶𝐴.
Thus, 𝐴′𝐵𝐶′ is the required triangle.
Justification:
∠𝐵 = 45° (Common)
∠𝐶 = ∠𝐶′ 𝛥𝐴𝐵′𝐶′ ~ 𝛥𝐴𝐵𝐶 by AA similarity condition.

Q.7) Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Sol.7) Steps of Construction:
Step I: 𝐵𝐶 = 3 𝑐𝑚 is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is made on previous ray intersecting it at point A.

Step IV: AC is joined to form 𝛥𝐴𝐵𝐶.
Step V: A ray 𝐵𝑋 is drawn making an acute angle with BC opposite to vertex A.
Step VI: 5 points B1, B2, B3, B4 and B5 at equal distance is marked on BX.
Step VII: 𝐵3𝐶 is joined 𝐵5𝐶′ is made parallel to 𝐵3𝐶.
Step VIII: 𝐴′𝐶′ is joined together.
Thus, 𝛥𝐴′𝐵𝐶′ is the required triangle.
Justification:
As in the previous question 6.

Exercise 11.2

Q.1) In each of the following, give also the justification of the construction: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Sol.1) Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is drawn.
Step II: A point P at a distance of 10 𝑐𝑚 from the centre O is taken. OP is joined.
Step III: Perpendicular bisector 𝑂𝑃 is drawn and let it intersected at M.
Step IV: With M as a centre and 𝑂𝑀 as a radius, a circle is drawn intersecting previous circle at Q and R.
Step V: 𝑃𝑄 and 𝑃𝑅 are joined.
Thus, 𝑃𝑄 and 𝑃𝑅 are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
𝑃𝑄 = 𝑃𝑅 = 8𝑐𝑚.
Justification:
𝑂𝑄 is joined.
∠𝑃𝑄𝑂 = 90°                                          (Angle in the semi circle)
∴ 𝑂𝑄 ⊥ 𝑃𝑄
Therefore, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.

Q.2) Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Sol.2) Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and 𝑂𝑃 is joined
Step IV: Perpendicular bisector of 𝑂𝑃 is drawn which cuts it at M
Step V: With M as a centre and 𝑂𝑀 as a radius, a circle is drawn which intersect the circle of radius 4 cm at Q and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement: 𝑂𝑄 = 4 𝑐𝑚           (Radius of the circle)
𝑃𝑄 = 6 𝑐𝑚                                    (Radius of the circle)
∠𝑃𝑄𝑂 = 90°                                 (Angle in the semi circle)
Applying Pythagoras theorem in 𝛥𝑃𝑄𝑂,

𝑃𝑄² + 𝑄𝑂² = 𝑃𝑂²
⇒ 𝑃𝑄² + 42 = 62
⇒ 𝑃𝑄² + 16 = 36
⇒ 𝑃𝑄² = 36 − 16
⇒ 𝑃𝑄² = 20
⇒ 𝑃𝑄 = 2√5
Justification: ∠𝑃𝑄𝑂 = 90° (Angle in the semi circle)
∴ 𝑂𝑄 ⊥ 𝑃𝑄
Therefore, 𝑂𝑄 is the radius of the circle then 𝑃𝑄 has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.

Q.3) Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Sol.3) Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 𝑐𝑚.
Step III: Perpendicular bisector of 𝑂𝑃 and 𝑂𝑄 is drawn and 𝑥 and 𝑦 be its mid-point.
Step IV: With O as a centre and 𝑂𝑥 be its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as centre and 𝑂𝑦 as radius and it intersected the circle at R and T.
Step VI: 𝑃𝑀 and 𝑃𝑁 are joined also 𝑄𝑅 and 𝑄𝑇 are joined.
Thus, 𝑃𝑀 and 𝑃𝑁 are tangents to the circle from P and 𝑄𝑅 and 𝑄𝑇 are tangents to the circle from point Q.
Justification:
∠PMO = 90° (Angle in the semi circle)
∴ 𝑂𝑀 ⊥ 𝑃𝑀
Therefore, 𝑂𝑀 is the radius of the circle then PM has to be a tangent of the circle.
Similarly, 𝑃𝑁, 𝑄𝑅 and 𝑄𝑇 are tangents of the circle.

Q.4) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Sol.4) We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠𝑂) = 360° − (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and 𝑂𝑄 is joined. 𝑂𝑄 is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular
OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.
Justification:
Sum of all angles in the quadrilateral
𝑃𝑄𝑂𝑅 = 360°
∠𝑄𝑂𝑅 + ∠𝑂𝑅𝑃 + ∠𝑂𝑄𝑅 + ∠𝑅𝑃𝑄 = 360°
⇒ 120° + 90° + 90° + ∠𝑅𝑃𝑄 = 360°
⇒ ∠𝑅𝑃𝑄 = 360° − 300°
⇒ ∠𝑅𝑃𝑄 = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°

Q.5) Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Sol.5) Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4 cm,
a circle is drawn which cut the line at point O.
Step III: With B as a centre and radius equal to 3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a radius, a circle is drawn which intersect the previous two circles at P, Q and R, S.
Step V: 𝐴𝑃, 𝐴𝑄, 𝐵𝑅 and BS are joined. Thus, AP, AQ, BR and BS are the required tangents.
Justification:
∠𝐵𝑃𝐴 = 90° (Angle in the semi circle)
∴ 𝐴𝑃 ⊥ 𝑃𝐵 Therefore, BP is the radius of the circle then AP has to be a tangent of the circle. Similarly, AQ, BR and BS are tangents of the circle

Q.6) Let 𝐴𝐵𝐶 be a right triangle in which 𝐴𝐵 = 6 𝑐𝑚, 𝐵𝐶 = 8 𝑐𝑚 and ∠𝐵 = 90𝑜. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Sol.6) Steps of Construction:
Step I: A 𝛥𝐴𝐵𝐶 is drawn.
Step II: Perpendicular to 𝐴𝐶 is drawn to point B which intersected it at D.
Step III: With O as a centre and 𝑂𝐶 as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: 𝑂𝐴 is joined and a circle is drawn with diameter 𝑂𝐴 which intersected the previous circle at B and E.
Step V: 𝐴𝐸 is joined.
Thus, 𝐴𝐵 and 𝐴𝐸 are the required tangents to the circle from A.
Justification:
∠𝑂𝐸𝐴 = 90° (Angle in the semi circle)
∴ 𝑂𝐸 ⊥ 𝐴𝐸
Therefore, 𝑂𝐸 is the radius of the circle then 𝐴𝐸 has to be a tangent of the circle.
Similarly, 𝐴𝐵 is tangent of the circle.