Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.2

In each of the following, give also the justification of the 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. First of all taking O as centre and 6 cm as radius draw a circle.

2. Locate a point P at a distance of 10 cm from 0.

3. Now join OP and bisect it. Denote its mid-point by M.

4. Taking M as centre and MP as radius, draw a circle which intersects the above circle at Q and R.

5. Join PQ and PR

Then, PQ and PR are the required tangent lines on measurement.

PQ = PR = 8 cm.

Justification—

Join OQ. ∠PQO = 90°, since angle ∠PQO is an angle formed in a semi-circle.

∴ PQ ⊥ OQ

Since OQ is a radius of the given circle, therefore PQ will be the tangent line to the circle.

Similarly, PR will also be a tangent line 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. First of all taking O as centre, draw two concentric circles of radii 4 cm and 6 cm.

2. Locate a point P on the circum-ference of the larger circle.

3. Now join OP and bisect it. Take its mid-point as M.

4. Taking M as centre and MP as radius, draw a circle which intersects the above circle at Q and R.

5. Join PQ and PR.

Then PQ and PR are the required tangent lines.

On measurement

PQ = PR = 4.5 cm. (approx.)

Justification—

Join OQ, ∠PQO = 90°, since ∠PQO is an angle formed in a semi-circle.

∴ PQ ⊥ OQ

Since OQ is a radius of the given circle, therefore PQ will be a tangent line to the circle.

Similarly, PR will also be a tangent line to the circle.

On measuring the length of the tangent line

PQ = PR = 4.5 cm.

Calculation—

Tangent line PQ = \(\sqrt{(O P)^{2}-(O Q)^{2}}\)

= \(\sqrt{(6)^{2}-(4)^{2}}\)

= \(\sqrt{36-16}\)

= \(\sqrt{20}\) = 4.47 cm.

≃ 4.5 cm.

Question 3.

Draw a circle of radius of 3 cm. Take two points P and Q on one of its extended diameter each as 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. First of all take a point O and taking it as centre draw a circle of radius 3 cm.

2. Take two points P and Q on its diameter produced. Thus OP = OQ = 7 cm.

3. Now bisect OP and OQ. Take their mid-points as M_{1} and M_{2} respectively.

4. Taking M_{1} as centre and M_{1}P as radius, draw a circle which intersects the above circle at T_{1} and T_{2}.

5. Join PT_{1} and PT_{2}.

Then PT_{1} and PT_{2} are the required tangent lines.

Similarly the tangent lines QT_{3} and QT_{4} can be obtained.

Justification—

Joining OT_{1} ∠PT_{1}O = 90°, since it is an angle formed in a semi-circle.

∴ PT_{1} ⊥ OT_{1}

Since OT_{1} is a radius of the given circle, therefore PT_{1} will be a tangent line to the circle.

Similarly, PT_{2}, QT_{3} and QT_{4} will also be the tangent lines 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. First of all draw a circle with centre O and radius 5 cm.

2. Draw any diameter AOC.

3. Now draw radius OL such that DCOL = 60° (i.e., given angle)

4. At L drawU LM ^ OL.

5. At A drawU AN ^ OA.

6. Both of these perpendiculars intersect at P.

Then PA and PB will be the required tangent lines.

Justification—

Since OA is a radius, therefore PA will be a tangent line to the circle. Similarly, PL will also be a tangent line to the circle.

∠APL = 360° – ∠OAP – ∠OLP – ∠AOL

= 360° – 90° – 90° – (180° – 60°)

= 360° – 360° + 60° = 60°

Thus, the tangent lines PA and PL will be inclined at an angle of 60° with each other.

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. First of all draw a line segment AB = 8 cm.

2. Now taking A as centre and 4 cm as radius draw a circle and taking B as centre and 3 cm as radius draw a circle. Bisect AB and find its mid-point O.

3. Clearly O is the mid-point of AB. With centre O draw a circle of radius OA or OB which intersects the circle with centre B at T_{1} and T_{2} and the circle with centre A at T_{3} and T_{4}.

4. Join AT_{1}, AT_{2}, BT_{3} and BT_{4}.

Then there will be the required tangents.

Justification—

Joining BT_{1} ∠BT_{1}A = 90°, since ∠BT_{1}A is an angle situated in a semi-circle.

∴ AT_{1} ⊥ BT_{1}

Since BT_{1} is a radius of the given circle, therefore AT_{1} will be a tangent line to the circle.

Similarly, AT_{2}, BT_{3} and BT_{4} will also be the tangent lines.

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 drawn. Construct the tangents from A to this circle.

Solution:

Steps of Construction :

1. First of all construct a DABC with given values in which AB = 6 cm., BC = 8 cm. and DB = 90°.

2. Draw BD ^ AC.

3. Now draw a circle with D as centre and DA, DB or DC as radius.

4. Bisect BC and BD. Let their bisectors meet at O.

5. With O as centre draw a circle with centre OB. This circle passes through the points B, C and D.

6. Join OA.

7. Draw perpendicular bisector of OA. This perpendicular bisector is obtained at point M.

8. Taking M as centre draw a circle with radius MA which intersects the above circle at points P and Q.

9. Join AP and AQ which are the required tangent lines to this circle 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. First of all draw a circle with the help of a bangle.

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

3. Now taking CS as diameter, draw a semi-circle.

4. At point A draw AB ⊥ AS which intersects the semi-circle at B.

5. Draw an arc with centre A and radius as AB which cuts the given circle at T and T’. Join A and AT’.

Then, AT and AT’ will be the required tangent lines.

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