NCERT Solutions for Class 10 Maths Chapter 11: Constructions - Exercise 11.2

Exercise No: 11.2

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. Give the justification of the construction.

Ans: A pair of tangents to the given circle can be constructed as follows.

Step 1. Draw a circle with the centre as $O$ and its radius is $6$ cm.

Step 2. Mark a point $10$ cm away from $O$ and name it as $P$. Now, join the points $OP$.

Step 3. Draw the perpendicular bisector on $OP$ and name the mid-point as $M$. 

Step 4. Taking $M$ as a centre and $MO$ as a radius, draw a circle.

Step 5. And new circle intersects the previous circle at points \[Q\] and \[R\].

Step 6. Now join $PQ$ and $PR$. Then the required tangents are $PQ$ and $PR$.

The lengths of tangents $PQ$ and $PR$ are $8$ cm each.

Justification

The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle, whose centre is $O$ and its radius is $6$ cm. 

Now, join $OQ$ and $OR$.

Now in triangle $PQO$,

$\angle PQO=90{}^\circ $

Then, by using Pythagoras’ theorem.

$PQ=\sqrt{O{{P}^{2}}-O{{Q}^{2}}}$

$=\sqrt{{{\left( 10 \right)}^{2}}-{{6}^{2}}}$

$=\sqrt{100-36}$

$=\sqrt{64}$

$=8$ cm. 

Similarly, $PR=8$cm.

$\therefore OQ\bot PQ$.

Since $OQ$ is the radius of the circle, $PQ$ has to be a tangent of the circle.

Similarly, $PR$ is a tangent of the circle.

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. Give the justification of the construction.

Ans: Tangent on the given circle can be drawn as follows.

Step 1. Draw a circle with the centre as $O$ and its radius is $4$ cm.

Step 2. Again, draw a circle with the centre as $O$ and its radius is $6$ cm. Locate a point $P$ on this circle and join $OP$.

Step 3. Draw a bisector on $OP$. And $M$be the midpoint of $OP$.

Step 4. Taking $M$ as the centre and \[MO\] as its radius, draw a circle. And this circle intersects at the previous circle at the points $Q$ and $R$.

Step 5. Now, join $PQ$ and $PR$. These are the required tangents.

Let us find the length of $PQ$ and $PR$. Then,

In $\Delta PQO$,

Since $PQ$ is a tangent,

$\angle PQO=90{}^\circ $

$PO=6$ cm and $QO=4$ cm

By using Pythagoras theorem, we get

$P{{Q}^{2}}=P{{O}^{2}}-Q{{O}^{2}}$

$={{6}^{2}}-{{4}^{2}}$

$=36-16=20$

$\Rightarrow PQ=\sqrt{20}=4.47$

$\therefore PQ=4.47$ cm

Similarly, $PR=4.47$ cm

Therefore, the length of each tangent is $4.47$ cm.

Justification:

The construction can be justified by proving that $PQ$ and $PR$ are the tangents of the circle and whose centre is $O$ and radius is $4$ cm. Now, let us join $OQ$ and $OR$.

$\angle PQO$ is an angle in the semi-circle. And the angle in a semicircle is a right angle.

$\therefore \angle PQO=90{}^\circ $ 

We know that $OQ\bot PQ$.

Since $OQ$ is the radius of the circle, then $PQ$ has to be a tangent of the circle. Similarly, $PR$ is also a tangent of the circle.

Hence proved.

3. Draw a circle of radius $3$ cm. Take two points $P$ and $Q$ on one of its extended diameters each at a distance of $7$ cm from its centre. Draw tangents to the circle from these two points $P$ and $Q$. Give the justification of the construction.

Ans: The tangent can be constructed on the given circle as follows.

Step 1. Draw a circle with the centre as $O$ and its radius is $3$ cm.

Step 2. Now, take one of its diameters, $PQ$ and extend it on both sides. Then, locate two points on this diameter such that $OR=OS=7$ cm.

Step 3. Draw a bisector on the $OR$ and similarly on the $OS$. Let $T$ and $U$be the mid-points of $OR$ and $OS$.

Step 4. Taking $T$ and $U$ as its centre, then with $TO$ and $UO$ as radius, draw two circles. These two circles will intersect at the previous circle at points $V,W,X,$ and $Y$ respectively.

Step 5. Now, join $RV,RW,SX,$ and $SY$. Then, these are the required tangents.

Justification 

The construction can be justified by proving that $RV,RW,SX,$ and $SY$ are the tangents to the circle and whose centre is $O$, then its radius is $3$ cm. Now, let us join $OV,OW,OX,$ and $OY$.

$\angle RVO$ is an angle in the semi-circle. And the angle in a semicircle is a right angle.

$\therefore \angle RVO=90{}^\circ $.

We know that $OV\bot RV$.

Since $OV$ is the radius of the circle, $RV$ has to be a tangent of the circle. Similarly, $OW,OX,$ and $OY$ are the tangents of the circle.

Hence proved.

4. Draw a pair of tangents to a circle of radius $5$ cm which are inclined to each other at an angle of $60{}^\circ $. Give the justification of the construction.

Ans: The tangents can be constructed as follows.

Step 1. Draw a circle with the centre as $O$ and its radius is $5$ cm.

Step 2. Take a point $A$ on the circumference of the circle and join $OA$. Draw a perpendicular to $OA$ at point $A$.

Step 3. Draw a radius of $OB$, making an angle of $120{}^\circ $ with $OA$.

Step 4. Draw a perpendicular line to $OB$ at point \[B\]. Let both the perpendiculars intersect at a point \[P\].

Step 5. Then, \[PA\] and \[PB\] are the required tangents at an angle of  $60{}^\circ $.

Justification

The construction can be justified by proving that $\angle APB=60{}^\circ $

By our construction

$\angle OAP=90{}^\circ $, $\angle OBP=90{}^\circ $ and $\angle AOB=120{}^\circ $.

We know that the sum of all interior angles of a quadrilateral is $360{}^\circ $

Then, $\angle OAP+\angle AOB+\angle OBP+\angle APB=360{}^\circ $

$90{}^\circ +120{}^\circ +90{}^\circ +\angle APB=360{}^\circ $

$300{}^\circ +\angle APB=360{}^\circ $

$\angle APB=60{}^\circ $

Hence proved.

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. Give the justification of the construction.

Ans: The tangents can be constructed on the given circles as follows.

Step 1. Draw a line segment $AB$ of $8$cm. Taking $A$ and $B$ as centre, draw two circles of radii are $4$cm and $3$ cm.

Step 2. Draw the bisector on the line $AB$. Let $C$ be the midpoint of $AB$.

Step 3. Taking $C$ as centre, draw a circle of radius as $AC$. Then, it intersects the circles at points of $P,Q,R,$ and $S$.

Step 4. Now, join $BP,BQ,AS,$ and $AR$. Then, these are the required tangents.

Justification

The construction can be justified by proving that \[AS\] and \[AR\] are the tangents of the circle and whose centre is $B$ and radius is $3$ cm and \[BP\] and \[BQ\] are the tangents of the circle and whose centre is $A$ and radius is $4$ cm. For this, join \[AP,\text{ }AQ,\text{ }BS,\]and \[BR\].

\[\angle ASB\]is an angle in the semi-circle. We know that an angle in a semicircle is a right angle.

\[\therefore \angle ASB=90{}^\circ \]

\[\Rightarrow BS\bot AS\]

Since \[BS\] is the radius of the circle, \[AS\] has to be a tangent of the circle. Similarly, \[AR,\text{ }BP,\]and \[BQ\] are the tangents.

Hence proved.

6. Let \[\mathbf{ABC}\] be a right triangle in which \[\mathbf{AB}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{8}\text{ }\mathbf{cm}\] and \[\angle \mathbf{B}\text{ }=\text{ }\mathbf{90}{}^\circ .\text{ }\mathbf{BD}\] is the perpendicular from \[\mathbf{B}\] on \[\mathbf{AC}\]. The circle through \[\mathbf{B},\text{ }\mathbf{C},\] and \[\mathbf{D}\] is drawn. Construct the tangents from \[\mathbf{A}\] to this circle. Give the justification of the construction. 

Ans: Consider the case below. If you draw a circle across \[B,\text{ }D,\] and \[C\], the diameter will be \[BC\] because \[BDC\] is \[90{}^\circ \]. The midpoint of \[BC\] will be the centre \[E\] of this circle.

On the provided circle, the needed tangents can be created as follows.

Step 1. Join \[AE\] and cut it in half. Let \[F\] be the midpoint of \[AE\].

Step 2. Draw a circle with \[F\] as the centre and \[FE\] as the radius, intersecting the circle at points \[B\] and \[G\]. Participate in \[AG\].

The required tangents are \[AB\] and \[AG\].

Justification

The structure can be justified by demonstrating that \[AG\] and \[AB\] are the circle's tangents. Join \[EG\] for this.

\[\angle AGE\] is an angle in the semi-circle. We know that an angle in a semicircle is a right angle. 

\[\therefore \angle AGE=90{}^\circ \]

\[\Rightarrow EG\bot AG\]

Since \[EG\] is the radius of the circle, \[AG\] has to be a tangent of the circle. 

Already, ∠B = 90° 

\[\Rightarrow AB\bot BE\]

Since \[BE\] is the radius of the circle, \[AB\] has to be a tangent of the circle. 

Hence proved.

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 circles. Give the justification of the construction. 

Ans: On the provided circle, the needed tangents can be created as follows. Step 1. With the help of a bangle, draw a circle.

Step 2. Draw two chords \[QR\] and \[\text{ }ST\] from a point \[P\] outside the circle.

Step 3. Draw perpendicular bisectors of these chords and allow them to connect at point \[O\].

Step 4. Join \[PO\] and cut it in half. Let \[U\] be the midpoint of \[PO\]. Draw a circle   

with a radius of \[OU\] around \[U\], intersecting the circle at \[V\] and \[W\]. \[PV\]and    

\[PW\] should team together.

The required tangents are \[PV\] and \[PW\].

Justification 

The structure can be justified by demonstrating that \[PV\] and \[PW\] are the circle's tangents. To do so, it must first be established that \[O\] is the circle's centre. Let's join forces with \[OV\] and \[OW\].

The perpendicular bisector of a chord is known to travel through its centre.

As a result, the centre is bisected by the perpendicular bisector of chords \[QR\] and \[ST\].

The junction point of these perpendicular bisectors is clearly the circle's centre. \[\angle PVO\] denotes a semi-circular angle. A right angle is defined as an angle in a semicircle.

\[\therefore \angle PVO=90{}^\circ \]

\[\Rightarrow OV\bot PV\]

\[PV\] must be a tangent of the circle since \[OV\] is the radius of the circle. \[PW\] is also a tangent to the circle.

Hence proved.

NCERT Solutions for Class 10 Maths PDFs (Chapter-wise)

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 10 of CBSE Class 10 Maths Solutions –

• Chapter 11: Important Questions

• Chapter 11: Revision Notes

• Chapter 11: NCERT Exemplar Solutions

• Chapter 11: RD Sharma Solutions

Points Discussed in Exercise 11.2

Exercise 11.2 is based on the construction of a tangent on a circle from a point that lies outside the circle. Earlier, we have learnt that we cannot draw the tangent of a circle from a point that lies inside the circle. Also, we can draw only one tangent from the point that lies on the circle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.2 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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