NCERT Solutions for Class 10 Maths Chapter 11 Constructions
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 ππ 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΄πΆ ∶ πΆπ΅ = 5 ∶ 8
Justification:
π₯π΄π΄5πΆ ~ π₯π΅π΅8πΆ
∴ π΄π΄5/π΅π΅8 = π΄πΆ/π΅πΆ = 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 ππ 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΅πΆ.
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΅′ is drawn parallel to π΄5π΅ and π΅′πΆ′ is drawn parallel to π΅πΆ.
π₯π΄π΅′πΆ′ is the required triangle
Justification:
∠π΄ (Common)
∠πΆ = ∠πΆ′ and
∠π΅ = ∠ π΅′ (corresponding angles)
Thus π₯π΄π΅′πΆ′ ~ π₯π΄π΅πΆ 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)
To construct a triangle with sides 5 cm, 6 cm, and 7 cm, follow these steps:
1. Draw the Base:
- Start by drawing a horizontal line segment AB, representing the base of the triangle. Let's say AB is 7 cm.
2. Construct the Other Sides:
- From point A, use a protractor to draw an angle of approximately 60 degrees. This angle will be between side AB and a new line segment AC.
- Measure 5 cm along the line AC from point A, and mark the endpoint as C.
- Draw a line segment BC to complete the triangle.
3. Measure and Confirm:
- Measure the lengths of sides AB, AC, and BC to ensure they are approximately 7 cm, 5 cm, and 6 cm, respectively.
Now, to construct another triangle whose sides are 7/5 of the corresponding sides of the first triangle:\
1. Scale the Triangle:
- Measure the length of each side (AB, AC, and BC) in the first triangle.
- Multiply each length by 7/5 to get the corresponding lengths for the second triangle.
2.Construct the Second Triangle:
- Start with a new base line segment, representing the base of the second triangle.
- Use the scaled lengths to construct the other sides of the triangle.
By following these steps, you will have constructed two triangles: one with sides 5 cm, 6 cm, and 7 cm, and another with sides that are 7/5 of the corresponding sides of the first triangle.
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΅πΆ = 5 ππ is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects π΅πΆ 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΅πΆ.
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΅2πΆ.
Step VIII: π΅3πΆ′ is drawn parallel to π΅2πΆ and πΆ′π΄′ is drawn parallel to πΆπ΄.
Thus, π΄′π΅πΆ′ is the required triangle formed.
Justification:
π₯π΄π΅′πΆ′ ~ π₯π΄π΅πΆ by AA similarity condition.
Q.5) Draw a triangle ABC with side π΅πΆ = 6 ππ, π΄π΅ = 5 ππ and ∠π΄π΅πΆ = 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 ∠π΄π΅πΆ = 60° with BC.
Step III: AC is joined to form π₯π΄π΅πΆ.
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΅3πΆ′.
Step VIII: πΆ′π΄′ is drawn parallel CA.
Thus, A'BC' is the required triangle.
Justification:
∠A = 60° (Common)
∠πΆ = ∠πΆ′ π₯π΄π΅′πΆ′ ~ π₯π΄π΅πΆ by AA similarity condition.
Q.6) Draw a triangle ABC with side π΅πΆ = 7 ππ, ∠π΅ = 45°, ∠π΄ = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Ξ ABC
Sol.6) Sum of all side of triangle = 180°
∴ ∠π΄ + ∠π΅ + ∠πΆ = 180°
∠πΆ = 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, ∠π΄ = 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 π΅π.
Step VI: π΅3πΆ is joined and π΅4πΆ′ is made parallel to π΅3πΆ.
Step VII: πΆ′π΄′ is made parallel πΆπ΄.
Thus, π΄′π΅πΆ′ is the required triangle.
Justification:
∠π΅ = 45° (Common)
∠πΆ = ∠πΆ′ π₯π΄π΅′πΆ′ ~ π₯π΄π΅πΆ 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΅πΆ = 3 ππ 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 π₯π΄π΅πΆ.
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΅3πΆ is joined π΅5πΆ′ is made parallel to π΅3πΆ.
Step VIII: π΄′πΆ′ is joined together.
Thus, π₯π΄′π΅πΆ′ 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 ππ from the centre O is taken. OP is joined.
Step III: Perpendicular bisector ππ is drawn and let it intersected at M.
Step IV: With M as a centre and ππ as a radius, a circle is drawn intersecting previous circle at Q and R.
Step V: ππ and ππ
are joined.
Thus, ππ and ππ
are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
ππ = ππ
= 8ππ.
Justification:
ππ is joined.
∠πππ = 90° (Angle in the semi circle)
∴ ππ ⊥ ππ
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 ππ is joined
Step IV: Perpendicular bisector of ππ is drawn which cuts it at M
Step V: With M as a centre and ππ 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: ππ = 4 ππ (Radius of the circle)
ππ = 6 ππ (Radius of the circle)
∠πππ = 90° (Angle in the semi circle)
Applying Pythagoras theorem in π₯πππ,
ππ2 + ππ2 = ππ2
⇒ ππ2 + 42 = 62
⇒ ππ2 + 16 = 36
⇒ ππ2 = 36 − 16
⇒ ππ2 = 20
⇒ ππ = 2√5
Justification: ∠πππ = 90° (Angle in the semi circle)
∴ ππ ⊥ ππ
Therefore, ππ is the radius of the circle then ππ 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 ππ.
Step III: Perpendicular bisector of ππ and ππ is drawn and π₯ and π¦ be its mid-point.
Step IV: With O as a centre and ππ₯ 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 ππ¦ as radius and it intersected the circle at R and T.
Step VI: ππ and ππ are joined also ππ
and ππ are joined.
Thus, ππ and ππ are tangents to the circle from P and ππ
and ππ are tangents to the circle from point Q.
Justification:
∠PMO = 90° (Angle in the semi circle)
∴ ππ ⊥ ππ
Therefore, ππ is the radius of the circle then PM has to be a tangent of the circle.
Similarly, ππ, ππ
and ππ 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 (∠π) = 360° − (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and ππ is joined. ππ 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
ππππ
= 360°
∠πππ
+ ∠ππ
π + ∠πππ
+ ∠π
ππ = 360°
⇒ 120° + 90° + 90° + ∠π
ππ = 360°
⇒ ∠π
ππ = 360° − 300°
⇒ ∠π
ππ = 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: π΄π, π΄π, π΅π
and BS are joined. Thus, AP, AQ, BR and BS are the required tangents.
Justification:
∠π΅ππ΄ = 90° (Angle in the semi circle)
∴ π΄π ⊥ ππ΅ 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΅πΆ be a right triangle in which π΄π΅ = 6 ππ, π΅πΆ = 8 ππ and ∠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΅πΆ is drawn.
Step II: Perpendicular to π΄πΆ is drawn to point B which intersected it at D.
Step III: With O as a centre and ππΆ as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: ππ΄ is joined and a circle is drawn with diameter ππ΄ which intersected the previous circle at B and E.
Step V: π΄πΈ is joined.
Thus, π΄π΅ and π΄πΈ are the required tangents to the circle from A.
Justification:
∠ππΈπ΄ = 90° (Angle in the semi circle)
∴ ππΈ ⊥ π΄πΈ
Therefore, ππΈ is the radius of the circle then π΄πΈ has to be a tangent of the circle.
Similarly, π΄π΅ is tangent of the circle.
Either way the teacher or student will get the solution to the problem within 24 hours.