These NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

Question 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:

Given: A circle of radius 6 cm and a point P is located outside the circle at a distance of 10 cm from the centre O of the circle, i.e.. OP = 10 cm.

Required : To draw two tangents to the circle from P and to measure their lengths also.

Steps of Construction :

(i) Draw a circle of radius 6 cm arid let O be the centre of it

(ii) Produce O to X and from OX cut OP = 10 cm.

(iii) Draw a perpendicular bisector ST of OP to intersect OP at M.

(iv) Taking M as centre and MO and MP as radius draw another circle so as to intersect the previously drawn circle at Q and R.

(v) Join PQ and PR.

Hence. PQ and PR and the required tangents.

(vi) Then measure the tangents FQ and PR. You will find that length of tangents PQ and PR will be 8 cm each.

Calculation:

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:

(i) Draw a circle of radius OP = 6 cm.

(ii) Draw a consentric circle of radius ON = 4 cm

(iii) Draw the ⊥ bisector of OP which and it at M.

(iv) Taking M as the centre and radius equal to NQ.

Draw a circle which intersect the inner circle at P and Q.

Justification:

In dotted circle PO is the diameter and ∠PRQ = 90° (Angle in semi-circle)

∴ OQ ⊥ PQ

Thus, PQ will be the tangent because we know that angle between radius (OQ) and tangent (PQ) is 90°.

Question 3.

Draw a circle with 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 tangent to the circle from these two points P and Q.

Solution:

Steps of Construction:

(i) Draw a circle with radius 3 cm and with centre O.

(ii) Taking two points P and Q ouside the circle such that OP = 7 cm and OQ = 7 cm and bisect OP and OQuestion Let the mid-points of OP be M and OQ be N.

(iii) Taking M and N as centres draw two circles as MO and NO are the radius respectively. The two circles cut the main circles at the points R, S, T and U.

(iv) Join P to R and S and Join Q to T and U, Then PR, PS and NT, NU are the required tangents of the circle,

Justification:

In left circle of is the diameter

∴ ∠PRO = 90° (angle in a semi-circle)

∴ OR x PR

Thus, PR is tangent (∵ Angle between radius and tangent is 90°)

Similarly PS, QT and QU are tangents.

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 in Construction:

(i) Draw a circle of radius 5 cm.

(ii) Draw a diameter QM.

(iii) At Q draw a perpendicular QA.

(iv) At O make the angle of 60° so that ∠QOR = 120°

(v) At R draw a perpendicular RB.

(vi) Both the perpendicular intersect at P.

(vii) Thus, PQ and PR are the required tangents which are inclined at 60°.

Justification:

In PQOR

∠P + ∠Q + ∠O + ∠R = 360°

or ∠P + 90° + 120° + 90° = 360°

or ∠P + 300° = 360°

or ∠P = 360° – 300° = 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:

(i) Draw a line segment AB = 8 cm and bisect it. Let M be the mid-point of AB.

(ii) Taking A as centre draw a circle with j radius 4 cm and taking B as centre draw another circle with radius 3 cm and taking M as centre draw’ circle with MA as radius it cuts both circles at the points P, Q and R. S.

(iii) Join A to R and S. Join B to P and Q.

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

Justification : In middle class circle AB is the diameter

∠APB = 90° and ∠APB = 90° (Angie in a semi-circle)

∴ AP ⊥ BP and BR ⊥ AR (Angle between radius and tangent is 90°)

Thus, BP and AR are tangents similarly AS and BQ are 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:

(i) Construct a AABC in which BC = 8 cm, AB = 6 cm and ∠B = 90°

(ii) From the point B draw a perpendicular BP which cut the AC at D.

(iii) Now draw the per perpendicular bisector of CD and BC which intersect at M.

∴ BM ⊥ AB (Angie between radius and tangent is 90°)

Thus, AB is tangent.

AQ is also tangent which is equal to AB because tangents drawn from the outer point are equal.

Question 7.

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

Solution:

Steps of Construction;

(i) Draw’ a circle of any radius.

(ii) Take a point P outside the circle.

(iii) Take PM = PA.

(iv) Now take BA as diameter and draw its perpendicular bisector which cut B at M.

(v) Taking M as centre draw the semi-circle.

(vi) From P draw a perpendicular which met the semi-circle at D.

(vii) Now taking PD as radius draw two arc which cut the circle at Q and R.

(viii) join P to Q and P to R.

(ix) Thus we get PQ and PR as the required tangents.

(x) Taking M as the centre draw a circle which passes through the B, C and D.

(xi) Taking AB = 6 cm in compass draw an arc which cut the circle at Q.

(xii) AB and AQ are the required tangents.

justification : Since ∠BDC = 90° (Angle in semi-circle)

∴ BM is-the radius.

and ∠B = 90°