Haryana State Board HBSE 10th Class Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of construction :

Question 1.

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.

Solution:

Steps of Construction :

1. Draw a circle with centre O of radius 6 cm.

2. Mark a point A, 10 cm away from the centre.

3. Join AO and bisect it at M.

4. Draw a circle with M as a centre and radius equal to AM intersects the given circle at points P and Q.

5. Join AP and AQ.

Then AP and AQ are the required tangents. Lengths of AP and AQ are 8-2 cm. (approx.)

Justification : Join OP

∠APO = 90° [angle in a semi-circle]

AP ⊥ OP

Therefore, AP is a tangent to the given circle.

Similarly AQ is a also tangent to the given circle.

Question 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.

Solution:

Steps of Construction :

1. O as the centre and draw two circles having radii 4 cm and 6 cm.

2. Taking point P on the larger circle.

3. Join OP and bisect it at M.

4. Taking M as the centre and radius equal to PM draw the circle which intersects the smaller circle at A and B.

5. Join PA and PB.

Then PA and PB are the required tangents.

On measuring the lengths of tangents

PA = PB = 4.47 (Approx.)

By calculation we have the length of tangent.

Joining AO

AO ⊥ PA [By theorem 10.1]

In right ∆PAO, we have

PO^{2} = PA^{2} + AO^{2}

⇒ (6)^{2} = PA^{2} + (4)^{2}

PA^{2} = (6)^{2} – (4)^{2}

PA^{2} = 36 – 16 = 20

⇒ PA = √20 = 4.472 cm

Justification : PA ⊥ OA

AO is the radius of the circle.

PA is tangent to a given circle.

Similarly PB is the tangent to the given circle.

Question 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.

Solution:

Steps of Construction :

1. Draw a circle with centre O and radius equal to 3 cm.

2. Take points P and Q on extended diameter of the circle such that OP = OQ = 7 cm.

3. Mark mid point M_{1} of OP and mid point of OQ as M_{2}.

4. Draw the circles as M_{1} and M_{2} centres and radii equal to PM_{1} and OM_{2} respectively, which intersects A and B (M_{1} as centre) and C, D (M_{2} as centre).

5. Join PA, PB and QC, QD.

Then PA, PB, QC and QD are tangents from the points P and Q respectively.

Justification : Join OA, OB, OC and OD ∠PAO = 90°

[Angle in a semi circle is 90°]

OA is the radius of circle.

PA is tangent to the given circle.

Similarly PB is the tangent to the given circle.

Again ∠OCQ = 90°

[Angle in a semi circle is 90°]

OC is the radius.

QC is the tangent to the given circle.

Similarly, QD is the tangent to the given circle.

Question 4.

Draw a pair of tangents to a circle of radius 5 cm. which are inclined to each other at an angle of 60°.

Solution:

Steps of Construction :

1. Draw a circle with centre O and radius 5 cm.

2. Construct radii OA and OB such that ∠AOB = 360° – (90° + 90° + 60°) = 120°.

3. Draw AL ⊥ OA at A and BM ⊥ OB at B. They intersect at P.

Then PA and PB Eire the required tangents inclined to each other at 60°.

Justification : PA ⊥ OA and PB ⊥ OB and ∠AOB = 120°.

⇒ ∠APB = 360° – 300°

⇒ ∠APB = 60°.

Therefore, tangents PA and PB are inclined at 60°.

Question 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.

Solution:

Steps of Construction :

1. Draw a line segment AB = 8 cm.

2. A as centre and radius equal to 4 cm draw a circle.

3. B as centre and radius equal to 3 cm draw another circle.

4. Bisect line segment AB at M.

5. M is the centre and radius equal to AM draw the circle which intersects the previous circles at P, Q (A as centre) and R, S (B as centre).

6. Join BP, BQ, AR and AS.

7. Then, BP, BQ, AR and AS are the required tangents.

Justification : Join AP, RB

∠APB = 90° [Angle in a semi circle is 90°]

PB ⊥ AP.

Therefore, AP is radius and BP is tangent.

Similarly, AR, AS and BQ are tangents.

Question 6.

Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Solution:

Steps of Construction:

1. Construct ∆ABC such that BC = 8 cm, AB = 6 cm and ∠ABC = 90°.

2. Draw BD ⊥ AC.

3. ∠CDB = 90° So, BC is the diameter of circle passing through B, C, D.

4. Bisect line BC at O.

5. O as the centre, CO = OB as radius draw a circle which passes through B, C, D.

6. Join AO which intersects the circle at M.

7. M as centre and MA = OM as radius draw another circle which intersect the previous circle at P and B.

8. Join PA.

Then, AP and AB are the required tangents.

Justification:

∠ABC = 90° [By Construction]

AB ⊥ OB.

OB is radius and AB is tangent.

Similarly, AP is tangent.

Question 7.

Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Solution:

Steps of Construction :

Take four points A, B, C and D on the given circle and join AB and CD.

1. Draw the perpendicular bisectors of non parallel chords AB and CD which intersect at O.

2. O is the centre of circle.

3. Take a point P outside the circle.

4. Join OP and bisected it at M.

5. M as the centre and radius PM = MO, draw another circle which intersect previous circle at R.

6. Join PR and PQ.

Then PR and PQ are the required tangents.

Justification : Join OR,

∠PRO = 90° [Angle in a semi circle is 90°]

PR ⊥ OR

OR is a radius and PR is a tangent.

Similarly, PQ is a tangent.