Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Constructions : Exercise 11.2 (Mathematics NCERT Class 10th)

Q.1      In each of the following, give the justification of the construction :

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.
Sol.         Step of Construction :

1. Take a point O and draw a circle of radius 6 cm.
2. Mark a point P at a distance of 10 cm from the centre O.
3. Join OP and bisect it. Let M be its mid- point.
4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
5. Join PQ and PR. Then PQ and PR are the required tangents.
On measuring, we find PQ = PR = 8 cm.
Justification :
On joining OQ, we find that $\angle PQO = {90^o}$, as $\angle PQO$ is the angle in the semi-circle.
Therefore, $PQ \bot OQ$
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.

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.
Sol.         Steps of Construction :-

1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
2. Mark a point P on the circumference of the bigger circle.
3. Join OP and bisect it. Let M be its mid- point.
4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
5. Join PQ and PR. Then, PQ and PR are the required tangents.
On measuring, we find that PQ = PR = 4.8 cm (approx).
Justification : -
On joining OQ, we find that $\angle PQO = {90^o}$, as $\angle PQO$ is the angle in the semi-circle.
Therefore, $PQ \bot OQ$
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.

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.
Sol.        Steps of Construction : 1. Take a point O, draw a circle of radius 3 cm with this point as centre.
2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
3. Bisect OP and OQ. Let their respective mid- points be  ${M_1}\,and\,{M_2}$.
4. With ${M_1}$ as centre and ${M_1}P$ as radius , draw a circle to intersect the circle at ${T_1}\,and\,{T_2}$
5. Join $P{T_1}\,and\,P{T_2}$. Then $P{T_1}\,and\,P{T_2}$ are the required tangents.
Similarly, the tangents $Q{T_3}\,and\,Q{T_4}$ can be obtained.
Justification :
On joining $O{T_1}$ we find $\angle P{T_1}O = {90^o}$, as an angle in the semi- circle.
Therefore, $P{T_1} \bot \,O{T_1}$
Since $O{T_1}$ is a radius of the given circle, so $P{T_1}$ has to be a tangent to the circle.
Similarly, $P{T_2},\,Q{T_3}\,and\,Q{T_4}$ are also tangents 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 60º
Sol.         Steps of Construction :
1. With O as centre of radius = 5 cm, draw a circle.
2. Draw any diameter AOC.
3. Draw a radius OL such that $\angle COL = {60^o}$ (i.e., the given angle). 4. At L draw $LM \bot OL$
5. At A, draw $AN \bot OA$
6. These two perpendiculars intersect each other at P. Then PA and PL are the required tangents.
Justification :
Since, OA is the radius , so PA has to be a tangent to the circle.
Similarly, PL is the tangent to the circle.
$\angle APL = {360^o} - \angle OAP - \angle OLP - \angle AOL$
= 360º – 90º – 90º – (180º – 60º)
= 360º – 360º + 60º = 60º
Thus tangents PA and PL are inclined to each other at an angle of 60º.

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.
Sol.         Steps of Construction :

1. Draw a line segment 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. 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 ${T_3}\,and\,{T_4}$.
4. Join $A{T_1}\,,\,A{T_2},\,B{T_3}\,and\,B{T_4}$. Then these are the required tangents.
Justification :
On joining $B{T_1}$, we find that, $B{T_1}A = 90^\circ$, as $\angle B{T_1}A$ is the angle in the semi- circle.
Therefore, $A{T_1} \bot B{T_1}$
Since, $B{T_1}$ is the radius of the given circle, so $A{T_1}$ has to be a tangent to the circle.
Similarly, $A{T_2},B{T_3}\,and\,B{T_4}$ are the tangents.

Q.6      Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and $\angle B = {90^o}$. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Sol.        Steps of a Construction : 1. With the given data, draw a $\Delta \,$ABC, in which AB = 6 cm, BC = 8 cm and $\angle B = {90^o}$.
2. Draw $BD \bot AC$.
3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C,D.
5. Join OA.
6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at points P and Q.
8. Join AP and AQ.
These are the required tangents from A.

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.
Sol.        Steps of Construction :

1. Draw a circle with the help of a bangle.
2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.
3. With CS as diameter, draw a semi-circle.
4. At the point A, draw $AB \bot AS$, cutting the semi- circle at B.
5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T'. Join AT and AT'. Then, AT and AT' are the required tangent lines.

• Anshu chaudhary

Very good nice its helpful THANK YOU

• Anshu chaudhary

Thank you it is helpful

• Soumy salvi

Bohot sahi he

• yogitha

Nice I want extra question on the these lesson

• Thanks for help me

• neha singh

• Mayank

Mast hai teacher

• It is just like a second teacher.......... Very useful

• Nikita shukla

Very helful i am proude of

• Anonymous

Gud becoz whenever I have any problem or confusion then I use to open it immediately

• Kunal Vashisht

Thanx sir very good

• Rupali

Useful and good

• Aashish Negi

very good site & very useful

• Mandeep singh

Very useful site and very precious to me .If I have any problem I always open it

• Vry help ful book

• really very usefull for me

• Naman puri

Not satisfied..... Question..no..7

• Vijay Washilkar

It is helpful in justification

• kharbanda

nice

• Anonymous

Message *

• Aishu a s it is very useful

Thank you

• Nisha mandal

It is very help ful

• Nisha mandal

It is very osum book

• Shivanshi Saxena

Very helpful for me

• Shivanshi Saxena

Very very helpful for me

• • Bhupesh

Very good book

• • 