RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2

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 M₁ and M₂ respectively.
4. Taking M₁ as centre and M₁P as radius, draw a circle which intersects the above circle at T₁ and T₂.
5. Join PT₁ and PT₂.
Then PT₁ and PT₂ are the required tangent lines.
Similarly the tangent lines QT₃ and QT₄ can be obtained.

Justification—
Joining OT₁ ∠PT₁O = 90°, since it is an angle formed in a semi-circle.
∴ PT₁ ⊥ OT₁
Since OT₁ is a radius of the given circle, therefore PT₁ will be a tangent line to the circle.
Similarly, PT₂, QT₃ and QT₄ 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 T₁ and T₂ and the circle with centre A at T₃ and T₄.
4. Join AT₁, AT₂, BT₃ and BT₄.
Then there will be the required tangents.
Justification—
Joining BT₁ ∠BT₁A = 90°, since ∠BT₁A is an angle situated in a semi-circle.

∴ AT₁ ⊥ BT₁
Since BT₁ is a radius of the given circle, therefore AT₁ will be a tangent line to the circle.
Similarly, AT₂, BT₃ and BT₄ 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.