NCERT Solutions Class 10 Mathematics Chapter 11 Construction

NCERT Solutions Class 10 Mathematics Chapter 11 Construction have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 10 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 10 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 10 Mathematics are an important part of exams for Class 10 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 10 Mathematics and also download more latest study material for all subjects. Chapter 11 Construction is an important topic in Class 10, please refer to answers provided below to help you score better in exams

Chapter 11 Construction Class 10 Mathematics NCERT Solutions

Class 10 Mathematics students should refer to the following NCERT questions with answers for Chapter 11 Construction in Class 10. These NCERT Solutions with answers for Class 10 Mathematics will come in exams and help you to score good marks

Chapter 11 Construction NCERT Solutions Class 10 Mathematics

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: π΄π΅ = 7.6 ππ is drawn.
Step II: A ray π΄π making an acute angle with
π΄π΅ is drawn.
Step III: After that, a ray BY is drawn parallel to
π΄π making equal acute angle as in the previous step.
Step IV: Point π΄1, π΄2, π΄3, π΄4 and π΄5 is marked on π΄π
and point π΅1, π΅2. ... to π΅8 is marked on BY such that π΄π΄1 = π΄1π΄2 = π΄2π΄3 =
. . . . π΅π΅1 = π΅1π΅2 = . . . . π΅7π΅8
Step V: π΄5 and π΅8 is joined and it intersected AB at point C diving it in the ratio 5:8.
π΄πΆ βΆ πΆπ΅ = 5 βΆ 8
Justification:
π₯π΄π΄5πΆ ~ π₯π΅π΅8πΆ
β΄ π΄π΄5/π΅π΅8 = π΄πΆ/π΅πΆ = 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: π΄π΅ = 6 ππ 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 π₯π΄π΅πΆ.
Step V: A ray π΄π 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 π΄π as A1 A2....A5
Step VII: π΄5π΅ is joined. π΄2π΅β² is drawn parallel to π΄5π΅ and π΅β²πΆβ² is drawn parallel to π΅πΆ.
π₯π΄π΅β²πΆβ² is the required triangle
Justification:
β π΄ (Common)
β πΆ = β πΆβ² and
β π΅ = β  π΅β² (corresponding angles)
Thus π₯π΄π΅β²πΆβ² ~ π₯π΄π΅πΆ 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 π΄π΅πΆ with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray π΅π making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as π΅1, π΅2, π΅3, π΅4, π΅5, π΅6 and π΅7 are marked on π΅π.
Step IV; Point π΅5 is joined with C to draw π΅5πΆ.
Step V: π΅7πΆβ² is drawn parallel to π΅5πΆ and πΆβ²π΄β² is parallel to πΆπ΄.
Thus π΄β²π΅πΆβ² is the required triangle.
Justification
π₯π΄π΅β²πΆβ² ~ π₯π΄π΅πΆ 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: π΅πΆ = 5 ππ is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects π΅πΆ 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 π₯π΄π΅πΆ.
Step V: A ray π΅π is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points π΅1, π΅2 and B3 is marked π΅π.
Step VII: π΅2 is joined with C to form π΅2πΆ.
Step VIII: π΅3πΆβ² is drawn parallel to π΅2πΆ and πΆβ²π΄β² is drawn parallel to πΆπ΄.
Thus, π΄β²π΅πΆβ² is the required triangle formed.
Justification:
π₯π΄π΅β²πΆβ² ~ π₯π΄π΅πΆ by AA similarity condition.

Q.5) Draw a triangle ABC with side π΅πΆ = 6 ππ, π΄π΅ = 5 ππ and β π΄π΅πΆ = 60Β°. 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 β π΄π΅πΆ = 60Β° with BC.
Step III: AC is joined to form π₯π΄π΅πΆ.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points π΅1, π΅2, π΅3 and π΅4 at equal distance is marked on BX.
Step VII: B3 is joined with C' to form π΅3πΆβ².
Step VIII: πΆβ²π΄β² is drawn parallel CA.
Thus, A'BC' is the required triangle.
Justification:
β A = 60Β° (Common)
β πΆ = β πΆβ² π₯π΄π΅β²πΆβ² ~ π₯π΄π΅πΆ by AA similarity condition.

Q.6) Draw a triangle ABC with side π΅πΆ = 7 ππ, β π΅ = 45Β°, β π΄ = 105Β°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Ξ ABC
Sol.6) Sum of all side of triangle = 180Β°
β΄ β π΄ + β π΅ + β πΆ = 180Β°
β πΆ = 180Β° β 150Β° = 30Β°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45Β° with BC.
Step III: At C, a ray making an angle of 30Β° with BC is drawn intersect ting the previous s ray at A.
Thus, β π΄ = 105Β°.
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 π΅π.
Step VI: π΅3πΆ is joined and π΅4πΆβ² is made parallel to π΅3πΆ.
Step VII: πΆβ²π΄β² is made parallel πΆπ΄.
Thus, π΄β²π΅πΆβ² is the required triangle.
Justification:
β π΅ = 45Β° (Common)
β πΆ = β πΆβ² π₯π΄π΅β²πΆβ² ~ π₯π΄π΅πΆ 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: π΅πΆ = 3 ππ is drawn.
Step II: At B, A ray making an angle of 90Β° 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 π₯π΄π΅πΆ.
Step V: A ray π΅π 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: π΅3πΆ is joined π΅5πΆβ² is made parallel to π΅3πΆ.
Step VIII: π΄β²πΆβ² is joined together.
Thus, π₯π΄β²π΅πΆβ² 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 ππ from the centre O is taken. OP is joined.
Step III: Perpendicular bisector ππ is drawn and let it intersected at M.
Step IV: With M as a centre and ππ as a radius, a circle is drawn intersecting previous circle at Q and R.
Step V: ππ and ππ are joined.
Thus, ππ and ππ are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
ππ = ππ = 8ππ.
Justification:
ππ is joined.
β πππ = 90Β°                                          (Angle in the semi circle)
β΄ ππ β₯ ππ
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 ππ is joined
Step IV: Perpendicular bisector of ππ is drawn which cuts it at M
Step V: With M as a centre and ππ 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: ππ = 4 ππ           (Radius of the circle)
ππ = 6 ππ                                    (Radius of the circle)
β πππ = 90Β°                                 (Angle in the semi circle)
Applying Pythagoras theorem in π₯πππ,

ππ2 + ππ2 = ππ2
β ππ2 + 42 = 62
β ππ2 + 16 = 36
β ππ2 = 36 β 16
β ππ2 = 20
β ππ = 2β5
Justification: β πππ = 90Β° (Angle in the semi circle)
β΄ ππ β₯ ππ
Therefore, ππ is the radius of the circle then ππ 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 ππ.
Step III: Perpendicular bisector of ππ and ππ is drawn and π₯ and π¦ be its mid-point.
Step IV: With O as a centre and ππ₯ 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 ππ¦ as radius and it intersected the circle at R and T.
Step VI: ππ and ππ are joined also ππ and ππ are joined.
Thus, ππ and ππ are tangents to the circle from P and ππ and ππ are tangents to the circle from point Q.
Justification:
β PMO = 90Β° (Angle in the semi circle)
β΄ ππ β₯ ππ
Therefore, ππ is the radius of the circle then PM has to be a tangent of the circle.
Similarly, ππ, ππ and ππ 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 60Β°.
Sol.4) We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360Β°
β΄ Angle between the radius (β π) = 360Β° β (90Β° + 90Β° + 60Β°) = 120Β°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and ππ is joined. ππ is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120Β° 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 60Β°.
Justification:
Sum of all angles in the quadrilateral
ππππ = 360Β°
β πππ + β πππ + β πππ + β πππ = 360Β°
β 120Β° + 90Β° + 90Β° + β πππ = 360Β°
β β πππ = 360Β° β 300Β°
β β πππ = 60Β°
Hence, QP and PR are tangents inclined to each other at an angle of 60Β°

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: π΄π, π΄π, π΅π and BS are joined. Thus, AP, AQ, BR and BS are the required tangents.
Justification:
β π΅ππ΄ = 90Β° (Angle in the semi circle)
β΄ π΄π β₯ ππ΅ 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 π΄π΅πΆ be a right triangle in which π΄π΅ = 6 ππ, π΅πΆ = 8 ππ and β π΅ = 90π. 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 π₯π΄π΅πΆ is drawn.
Step II: Perpendicular to π΄πΆ is drawn to point B which intersected it at D.
Step III: With O as a centre and ππΆ as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: ππ΄ is joined and a circle is drawn with diameter ππ΄ which intersected the previous circle at B and E.
Step V: π΄πΈ is joined.
Thus, π΄π΅ and π΄πΈ are the required tangents to the circle from A.
Justification:
β ππΈπ΄ = 90Β° (Angle in the semi circle)
β΄ ππΈ β₯ π΄πΈ
Therefore, ππΈ is the radius of the circle then π΄πΈ has to be a tangent of the circle.
Similarly, π΄π΅ is tangent of the circle.

 NCERT Solutions Class 10 Mathematics Chapter 1 Real Numbers
 NCERT Solutions Class 10 Mathematics Chapter 2 Polynomials
 NCERT Solutions Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables
 NCERT Solutions Class 10 Mathematics Chapter 4 Quadratic Equations
 NCERT Solutions Class 10 Mathematics Chapter 5 Arithmetic Progressions
 NCERT Solutions Class 10 Mathematics Chapter 6 Triangles
 NCERT Solutions Class 10 Mathematics Chapter 7 Coordinate Geometry
 NCERT Solutions Class 10 Mathematics Chapter 8 Introduction to Trigonometry
 NCERT Solutions Class 10 Mathematics Chapter 9 Some Application of Trigonometry
 NCERT Solutions Class 10 Mathematics Chapter 10 Circles
 NCERT Solutions Class 10 Mathematics Chapter 11 Construction
 NCERT Solutions Class 10 Mathematics Chapter 12 Areas Related to Circles
 NCERT Solutions Class 10 Mathematics Chapter 13 Surface Area and Volume
 NCERT Solutions Class 10 Mathematics Chapter 14 Statistics
 NCERT Solutions Class 10 Mathematics Chapter 15 Probability

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NCERT Solutions Class 10 Mathematics Chapter 11 Construction

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