Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Questions and Answers.

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

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. Take a point 0 as centre and draw a circle of radius 6 cm.

2. Mark a point P at a distance of 10 cm from the centre O.

3. Join OP and bisect it. Let M be its mid-point of OP.

4. With M as centre and MP as radius, draw a circle to intersect the previous circle at Q and R.

5. Join PQ and PR. Then, PQ and PR are the required tangents.

On measurement, we find that PQ = PR = 8 cm.

Justification :

On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.

∴ PQ ⊥ OQ.

Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the 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. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.

2. Mark a point P on the bigger circle.

3. Join OP and bisect it. Let M be its mid-point.

4. With M as centre and MP as radius, draw a circle to intersect the smaller circle at Q and R.

5. Join PQ and PR. Then, PQ and PR are the required tangents.

On measuring, we find that PQ = PR = 4.5 cm (approx.).

Calculation : From A OQP,

OP² = OQ² + PQ²

or 6² = 4² + PQ²

or PQ² = 36 – 16 = 20

So, PQ = \(\sqrt{20}\)cm = 4-47 cm (approx.)

Similarly, PR = 4.47 cm (approx.).

Justification :

On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.

∴ PQ ⊥ OQ.

Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the 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. Take a point O, draw a circle of radius 3 cm with this point as centre.

2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.

3. Bisect OP and OQ. Let their respective mid-points be M_{1} and M_{2}.

4. With M_{1} as centre and M_{1}P as radius, draw a circle to intersect the circle at T_{1} and T_{2}.

5. Join PT_{1} and PT_{2}. Then, PT_{1} and PT_{2} are the required tangents.

Similarly, the tangents QTS and QT4 can be obtained.

Justification :

On joining OT4, we find ∠PT10 = 90°, as an angle in a semi-circle.

∴ PT_{1} ⊥ OT_{1}

Since OT_{1} is a radius of the given circle, so PT_{1} has to be a tangent to the circle.

Similarly, PT_{2}, QT_{3} and QT_{4} are also tangents to the 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. With O as centre and radius = 5 cm, draw a circle.

2. Draw any diameter AOC.

3. Draw a radius OL such that ∠COL = 60° (i.e., the given angle).

4. At L, draw LM ⊥ OL.

5. At A, draw AN ⊥ OA.

6. These two perpendiculars intersect each other at P. Then, PA and PL are the required tangents.

Justification :

Since OA is the radius, so PA has to be a tangent to the circle.

Similarly, PL is the tangent to the circle.

∠APL = 360° – ∠OAP – ∠OLP – ∠AOL

= 360° – 90° – 90° – (180° – 60°)

= 360° – 360° + 60° = 60°

Thus, tangents PA and PL are inclined to each other at an angle of 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. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.

3. Draw perpendicular bisector of the line segment AB. It intersects the line segment AB at O. Clearly, O is the mid-point of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at T_{4} and T_{4}, circle with centre A at T_{3} and T_{4}.

4. Join AT_{1}( AT_{2}, BT_{3 }and BT_{4)}. Then, these are the required tangents.

Justification :

On joining BT_{1} we find that ∠BT_{1}A = 90°, as ∠BT_{1}A is the angle in a semi-circle.

∴ AT_{1} ⊥ BT_{1}

Since BT_{1} is the radius of the given circle, so AT_{1} has to be a tangent to the circle.

Similarly, AT_{2}, BT_{3} and BT_{4} are the 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. With the given data, draw a A ABC, in which AB = 6 cm, BC = 8 cm and ∠B = 90°.

2. Draw BD ⊥ AC.

3. Draw perpendicular bisectors of BC and BD. They meet at a point O.

4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C and D.

5. Join OA.

6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.

7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at point B and P.

8. Join AB and AP.

These are the required tangents from A.

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 :

1. Draw a circle with the help of a bangle.

2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.

3. With CS as diameter, draw a semi-circle.

4. At the point A, draw AB ⊥ AS, cutting the semi-circle at B.

5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T’. Join AT and AT’. Then, AT and AT’ are the required pair of tangents.

Aliter:

1. Mark any three points P, Q and R on the circle.

2. Draw the perpendicular bisectors of PQ and QR to meet at O, the centre of the circle.

3. Join OA and locate its mid-point M.

4. With M as centre and OM as radius, draw a circle to intersect the previous circle at T_{1} and T_{2}.

5. Join AT_{1} and AT_{2}, which are the required tangents.