**Download NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2 Constructions**

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

**1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.**

**Solution:**

**Steps of construction:**

**Step 1.**Draw a circle of 6 cm radius with centre O.**Step 2.**Locate a point P such that PO = 10 cm.**Step 3.**Join PO.**Step 4.**Bisect the line segment OP and mark the mid-point of OP as M.**Step 5.**With M as centre and MO as radius, draw a circle.**Step 6.**Mark the points of intersection of the two circles as Q and R.**Step 7.**Join PR and PQ.

PQ and PQ are the tangents to the circle with centre O.

The length of PQ and PR is 8 cm.

**Justification**

Join OQ and OR

∠PQO = ∠PRO = 90° (Angle of a semi-circle is right angle)

∴ OQ ⊥ PQ and OR ⊥ PR

Hence, PQ and PR are the tangent to the circle with centre O and radius 6 cm.

**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:**

**Step 1.**Draw two circles of 4cm and 6 cm radius with centre O.**Step 2.**Locate a point P on the larger circle.**Step 3.**Join PO.**Step 4.**Bisect the line segment OP and mark the mid-point of OP as M.**Step 5.**With M as centre and MO as radius, draw a circle.**Step 6.**Mark the points of intersection of the previous circle and the smaller circle as Q and R.**Step 7.**Join PR and PQ.

PQ and PQ are the tangents to the circle with centre O.

In △PQO

PQ^{2} + OQ^{2} = OP^{2} (Pythagoras theorem)

PQ^{2} + 4^{2} = 6^{2}PQ^{2} = 36 – 16

PQ^{2} = 20

PQ = √20 cm = 2√5 cm = 4.4721 cm

**Justification**

Join OQ and OR

∠PQO = ∠PRO = 90° (Angle of a semi-circle is right angle)

∴ OQ ⊥ PQ and OR ⊥ PR

Hence, PQ and PR are the tangent to the circle with centre O and radius 4 cm.

**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:**

**Step 1.**Draw a circle of radius 3 cm and centre O.**Step 2.**Extend a diameter and mark the points P and Q each at a distance of 7 cm from O.**Step 3.**Bisect OP and OQ**Step 4.**Mark the mid-point of OP as X and OQ as Y.**Step 5.**With X as centre and XO as radius, draw a circle intersecting the circle with centre O at A and B.**Step 6.**With Y as centre and YO as radius, draw a circle intersecting the circle with centre O at C and D.**Step 7.**Join PA, PB, QC and QD.

PA, PB, QC and QD are the required tangents to the circle.

**Justification**

Join OA, OB, OC and OD

∠PAO = ∠PBO = 90° (Angle of a semi-circle is right angle)

∴ OA ⊥ PA and OB ⊥ PB

Similarly, OC ⊥ CQ and OD ⊥ DQ

Hence, PA, PB, QC and QD are the tangent to the circle with centre O and radius 3 cm.

**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:**

**Step 1.**Draw a circle of radius 5 cm and centre O.**Step 2.**Draw two radius OA and OB inclined to each other at an angle of 120°, i.e. ∠BOA = 120°**Step 3.**Draw perpendiculars to OA and OB at A and B respectively.**Step 4.**Mark the point of intersection of the two perpendiculars as P.

PA and PB are the required tangents to the circle.

**Justification**

In the quadrilateral OAPB

∠P + ∠B + ∠O + ∠A = 360° (Angle sum of a quadrilateral)

∠P + 90° + 120° + 90° = 360°

∠P + 300° = 360°

∠P = 60°

Hence, PA and PB are the tangent to the circle with centre O and radius 5 cm inclined to each other at 60°.

**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:**

**Step 1.**Draw a line segment AB = 8 cm.**Step 2.**With A and B as centre and radius 4 cm and 3 cm, respectively, draw circles.**Step 3.**Bisect AB and Mark the mid-point of AB as C.**Step 4.**With C as centre and AC as radius, draw a circle.**Step 5.**Mark the points of intersection of circles with centres A and C as P and Q.**Step 6.**Mark the points of intersection of circles with centres B and C as R and S.**Step 7.**Join AR, AS, BP and BQ.

AR, AS, BP and BQ are the required tangents to the circles.

**Justification**

Join AP, AQ, BR and BS

∠APB = ∠AQB = 90° (Angle of a semi-circle is right angle)

∴ AP ⊥ PB and AQ ⊥ QB

Similarly, BR ⊥RA and BS ⊥ SA

Hence, AR, AS, BP and BQ are the required tangents.

**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:**

∵ ∠BDC = 90° and the angle of a semi-circle is also a right angle

∴ BC will be the diameter of the circle passing through B, C and D

Hence, the mid-point of BC will be the centre of the circle.

**Steps of construction:**

**Step 1.**Bisect BC and mark the mid-point of BC as O.**Step 2.**Bisect OA and mark the mid-point of OA as M.**Step 3.**With M as centre and MO as radius, draw a circle.**Step 4.**The two circles will intersect at B and P.**Step 5.**Join AP.

AP and AB are the required tangents to the circles.

**Justification**

Join OP

∠ABC = 90°

∠APO = 90° (Angle of a semi-circle is right angle)

∴ AP ⊥ PO and AB ⊥ OB

Hence, AP and AB are the required tangents.

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

**Solution:**

**Steps of construction:**

**Step 1.**Draw two non-parallel chords.**Step 2.**Draw the perpendicular bisectors of the two chords and mark their intersection as O.**Step 3.**Take a point P outside the circle and join OP.**Step 4.**Bisect OP and mark the point of intersection of OP as M.**Step 5.**With M as centre and MO as radius, draw a circle intersecting the previous circle at Q and R.**Step 6.**Join PQ an PR.

PQ and PR are the required tangents to the circles.

**Justification**

The perpendicular bisector of a chord of a circle passes through the centre of the circle. Therefore, the point of intersection of the perpendicular bisectors will be the centre of the circle.

∴ O is the centre of the circle.

Join OQ and OR

∠OQP = ∠ORP = 90° (Angle of a semi-circle is right angle)

∴ PQ ⊥ QO and PR ⊥ RO

Hence, PQ and PR are the required tangents.

**Download NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2 Constructions**

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