In **NCERT Solution For Class 10, Maths, Chapter 11, Constructions, Exercise 11.2,** we shall discuss questions related construction of Tangents to a Circle. Solutions of chapter 11, Exercise 11.2 is well explained which are given below.

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Toggle## Class 10, Maths, Chapter 11, Exercise 11.2, Solutions

**Q.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.**

**Ans:**

**Step of Construction:**

- Take a point O and draw a circle of radius 6 cm.
- Locate a point P at a distance of 10 cm from the centre O. (OP = 10 cm)
- Join OP and bisect it. Let M be its mid-point.
- With M as centre and OM as radius, draw a circle to intersect the circle at Q and R.
- Join PQ and PR, Then, PQ and PR are the required tangents.

On measurement, we find length of tangent 8 cm (PQ = PR = 8 cm).

**Justification: To prove: **PQ and PR are tangents.

On joining OQ, we find that

∠PQO = 90° [Angle in the semi-circle]

∴ PQ ⊥ OQ.

Since OQ is the radius of the given circle, so PQ must be a tangent to the circle. Similarly, PR is also a tangent to 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.**

**Ans:** **Steps of Construction**:

- Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
- Locate a point P on the circumference of the bigger circle.
- Join OP and bisect it. then M is the mid-point.
- With M as centre and OM as radius, draw a circle to intersect the circle at Q and R.
- Join PQ and PR where circle intersect.
- PQ and PR are the required tangents.

On measuring, we find that PQ = PR = 4.8 cm (approx.)

**Justification: To prove: **PQ and PR are tangents.

On joining OQ, we find that

∠PQO = 90° [Angle in the 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.

**Measurement:**

In ΔPQO, we have, OP = 6 cm and OQ = 4 cm

By using Pythagoras theorem,

⟹ (6)^{2} = (4)^{2} + PQ^{2}

⟹ 36 = 16 + PQ^{2}

⟹ PQ^{2} = 36 – 16 = 20

⟹ PQ = $\sqrt{20}$ = 4.47 cm

Hence, length of tangent is 4.47 cm.

**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.**

**Ans:**

**Steps of Construction:**

- Take a point O, draw a circle of radium 3 cm with this point as centre.
- Take two points P and P’ on one of its extended diameter such that OP = OP’ = 7 cm.
- Bisect OP and OP’. Let their respective mid-points be M and N.
- With M as centre and OM as radius draw a circle to intersect the circle at Q and R and with N as centre and ON as radius draw a circle to intersect the circle at T and U.

- Join PQ and PR.
- Join P’T and P’U. now we have our required tangents.

**Justification:**

** To prove: **PQ, PR, P’T and P’U are tangents.

** **∠PQO = 90° [Angle in the semi-circle]

∴ PQ ⊥ OQ.

Since OQ is the radius of the given circle, so PQ must be a tangent to the circle. Similarly, PR, P’T and P’U are also a tangent to 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°.**

**Ans: ****Steps of Construction:**

(1.) With O as centre, draw a circle of radius 5 cm.

(2.) by taking point B on circumference of circle and join OB.

(3.) Draw a radius OB and make angle 120^{0} (180^{0} – 60^{0}).

(4.) Draw any diameter AOC.

(5.) At A, draw AP ⊥ OA.

(6.) At B, draw BP ⊥ OB.

(7.) These two perpendiculars intersect each other at P.

Then, AP and BP are the required tangents.

**Justification:**

**To prove: ****∠****APB = 60 ^{0}**

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

Similarly, PB is the tangent to the circle

∵ sum of angles of quadrilateral = 360^{0}

⟹ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360^{0}

⟹ 90° + ∠APB + 90° + 180° = 360^{0}

⟹ ∠APB = 360^{0} – 90° – 90° – 180°

⟹ ∠APB = 360^{0} – 300°

⟹ ∠APB = 60^{0} (Hence proved)

Thus, tangents PA and PB are inclined to each other at an angle of 60^{0}.

**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.**

**Ans: ****Step of Construction:**

(1.) Draw a line AB of 8 cm (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.) Bisect the line AB at point O.

(4.) Clearly, O is the centre of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at T_{1} and T_{2}, circle with centre A at T_{3} and T_{4}.

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

**Justification:**

**To prove: **AT_{1, }AT_{2} , BT_{3}, BT_{4} are tangents.

On joining BT_{1}, we find that,

∠BT_{1}A = 90°, [Angle in the semi-circle]

∴ AT_{1} ⊥ BT_{1}

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

Similarly, AT_{2}, BT_{3} and BT_{4} are the tangents.

**Q.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.**

**Ans:**

**Step of Construction:**

(1.) With the given data, take a line BC of 8 cm (BC = 8cm).

(2.) taking O as centre OB as radius, draw a circle.

- Draw an angle of 90
^{0}at point B. draw an arc of 6 cm from point B. - Join AB and AC.
- Join B to D where AC and circle intersect.
- Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
- Taking M as centre and MA as radius, draw a circle which intersects the circle at P and B.
- Join AP and AB. These are the required tangents from A.

**Justification:**

**To prove: **AB and AP are tangents.

On joining PO, we find that,

∠APO = 90°, [Angle in the semi-circle]

∴ PO ⊥ AP

Since OP is the radius of the given circle, so AP must be a tangent to the circle.

Similarly, AB is the tangent.

**Q.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.**

**Ans:**

**Steps of Construction:**

- Draw a circle with the help of a bangle.
- Draw two secant or chord AB and BC on the circle.
- Draw a bisector of AB and BC, which meets at point O consider as centre of circle.
- Take point P outside the circle. Join OP.
- Bisect OP line by making arc and this way we get out bisector at point M.
- Take MP as radius and M as a centre, draw a circle which intersects the circle at Q and R. Then, PQ and PR are the required tangent lines.

**Justification:**

**To prove: **PQ and PR are tangents.

On joining QO, we find that,

∠OQP = 90°, [Angle in the semi-circle]

∴ PQ ⊥ OQ

Since OQ is the radius of the given circle, so PQ must be a tangent to the circle.Similarly, PR is the tangent.