Important Questions for CBSE Class 10 Maths Chapter 11 - Constructions

Practice Questions of Chapter 11 - Constructions

Some of the questions that can help students with their preparations for upcoming board examinations are mentioned below.

Question 1: Construct a triangle ABC in which BC = 10 cm, AB = 5.7 cm, and ∠ABC = 60°. Then construct another triangle whose sides are 2 times the corresponding sides of ΔABC.

Question 2: Draw a 7.6-cm line segment and divide it in the ratio of 5:8. Take measurements for the two components.

Question 3: Draw a pair of tangents to a circle with a radius of 5 cm and a 60-degree angle between them.

Question 4: Make a triangle with 5 cm, 6 cm, and 7 cm sides. Create a new triangle with sides that are 4/5 of the first triangle's equivalent sides.

Question 5: Make an isosceles ABC with BC = 5.5 cm and AL = 3 cm altitude. Then make another triangle with sides that are 3/4 the length of the corresponding sides of ABC.

Related Study Resources

Apart from the resources provided on the page, the following resources are also helpful for students preparing for the upcoming board examinations.

• NCERT Solutions for Chapter 11 Constructions

• Revision Notes for Constructions of Chapter 11

1 Marks Questions

1. The length of the tangent from a point A at a distance of $5\;{\text{cm}}$ from the centre of the circle is $4\;{\text{cm}}.$ What will be the radius of the circle?

(a) $1\;{\text{cm}}$

(b) $2\;{\text{cm}}$

(c) $3\;{\text{cm}}$

(d) none of these

Ans. c) $3\;{\text{cm}}$

Let's draw a tangent from point $A$ to the circle as shown below.

A tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, $\angle {\text{OTA}} = {90^\circ }$ and $\Delta {\text{OTA}}$ is a right-angled triangle.

By Pythagoras theorem,

${\text{O}}{{\text{A}}^2} = {\text{O}}{{\text{T}}^2} + {\text{A}}{{\text{T}}^2}$

${5^2} = {\text{O}}{{\text{T}}^2} + {4^2}$

${\text{O}}{{\text{T}}^2} = 25 - 16$

${\text{O}}{{\text{T}}^2} = 9$

${\text{OT}} =  \pm 3$

Radius OT cannot be negative.

Hence, the radius is $3\;{\text{cm}}$.

2. In the figure given below, PA and $PB$are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at $Q$. If $PB = 12\;{\text{cm}}$ and $CQ = 3$ ${\text{cm}}$, what is the length of ${\text{PC}}$ ?

(a) $9\;{\text{cm}}$

(b) $10\;{\text{cm}}$

(c) $1\;{\text{cm}}$

(d) $13\;{\text{cm}}$

Ans. (a) $9\;{\text{cm}}$

${\text{PA}} = {\text{PB}}$$({\text{Tangents}}{\kern 1pt} \,{\kern 1pt} {\text{from}}{\kern 1pt} \,{\kern 1pt} {\text{external}}{\kern 1pt} {\kern 1pt} {\text{point }}{\kern 1pt} P)$ 

$ \Rightarrow {\text{PC}} + {\text{CA}} = 12$

$ \Rightarrow {\text{PC}} + {\text{CQ}} = 12$

$(\because {\text{CA}} = {\text{CQ (Tangents from external point C}}))$

$ \Rightarrow {\text{PC}} + 3 = 12$

$ \Rightarrow {\text{PC}} = 9{\text{cm}}$

3. The tangent of a circle makes angle with radius at point of contact

(a) ${60^\circ }$

(b) ${30^\circ }$

(c) ${90^\circ }$

(d) none of these

Ans. (c) ${90^\circ }$

The angle between a tangent and the radius drawn to the point of contact is ${90^\circ }$.

4. If tangent PA and PB from a point P to a circle with centre 0 are inclined to each other at an angle of ${80^\circ }$, then what is the value of $\angle POA$ ?

(a) ${30^\circ }$

(b) ${50^\circ }$

(c) ${70^\circ }$

(d) ${90^\circ }$

Ans. (b) ${50^\circ }$

The lengths of tangents drawn from an external point to a circle are equal.

A tangent at any point of a circle is perpendicular to the radius at the point of contact.

In $\vartriangle {\text{OAP}}$ and in $\vartriangle {\text{OBP}}$

${\text{OA}} = {\text{OB}}$ (radii of the circle are always equal)

${\text{AP}} = {\text{BP}}$ (length of the tangents)

${\text{OP}} = {\text{OP}}$ (common)

Therefore, by SSS congruency $\vartriangle {\text{OAP}} \cong \vartriangle {\text{OBP}}$

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent then their corresponding parts are equal.

Hence,

$\angle {\text{POA}} = \angle {\text{POB}}$

$\angle {\text{OPA}} = \angle {\text{OPB}}$

Therefore, OP is the angle bisector of $\angle {\text{APB}}$and $\angle AOB$

Hence, $\angle OPA = \angle OPB = {\dfrac{1}{2}}(\angle APB)$

$ = {\dfrac{1}{2}} \times {80^\circ }$

$ = {40^\circ }$

By angle sum property of a triangle,

$\ln \vartriangle {\text{OAP}}$

$\angle A + \angle POA + \angle OPA = {180^\circ }$

$OA \bot AP$ 

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

${90^\circ } + \angle {\text{POA}} + {40^\circ } = {180^\circ }$

${130^\circ } + \angle {\text{POA}} = {180^\circ }$

$\angle {\text{POA}} = {180^\circ } - {130^\circ }$

$\angle {\text{POA}} = {50^\circ }$

Thus, option (A) ${50^\circ }$ is the correct answer.

2 Marks Questions

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

Draw a circle of radius $6\;{\text{cm}}$. From a point $10\;{\text{cm}}$ away from its centre, construct the pair of tangents to the circle and measure their lengths.

Ans. Given: A circle whose centre is ${\text{O}}$ and radius is $6\;{\text{cm}}$ and a point ${\text{P}}$ is $10\;{\text{cm}}$ away from its centre.

To construct: To construct the pair of tangents to the circle and measure their lengths.

Steps of Construction:

(a) Join PO and bisect it. Let ${\text{M}}$ be the mid-point of ${\text{PO}}$.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points $Q$ and $R$.

(c) Join ${\text{PQ}}$ and PR.

Then PQ and PR are the required two tangents.

By measurement, ${\text{PQ}} = {\text{PR}} = 8\;{\text{cm}}$

Justification: Join OQ and OR.

Since $\angle {\text{OQP}}$ and $\angle {\text{ORP}}$ are the angles in semicircles.

$\therefore \angle {\text{OQP}} = {90^\circ } = \angle {\text{ORP}}$

Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at $Q$ and ${\text{R}}$ respectively.

$\therefore $ We may see that the circle with OP as diameter increases the given circle in two points. Therefore, only two tangents can be draw.

2. In figure, PA and PB are tangents from $P$ to the circle with centre $O.R$is a point on the circle, prove that ${\mathbf{PC}} + {\mathbf{CR}} = {\mathbf{PD}} + {\mathbf{DR}}$.

Ans. Since length of tangents from an external point to a circle are equal in length

$\therefore PA = PB$

$CA = CR$

$And{\kern 1pt} $${\kern 1pt} {\kern 1pt} DB = DR$

$Now{\kern 1pt} \,{\kern 1pt} PA = PB$

$ \Rightarrow {\text{PC}} + {\text{CA}} = {\text{PD}} + {\text{DB}}$

$ \Rightarrow {\text{PC}} + {\text{CR}} = {\text{PD}} + {\text{DR}}({\text{By}}({\text{i}}))$

3. The length of tangents from a point $A$ at distance of $26\;{\text{cm}}$ from the centre of the circle is $10\;{\text{cm}}$, what will be the radius of the circle?

Ans. Since tangents to a circle is perpendicular to radius through the point of contact

$\therefore \angle OTA = {90^\circ }$

In right $\Delta OTA = {90^\circ }$, we have

$O{A^2} = O{T^2} + A{T^2}$

$ \Rightarrow {(26)^2} = O{T^2} + {(10)^2}$

$ \Rightarrow O{T^2} = 676 - 100$

$ \Rightarrow O{T^2} = 576$

$ \Rightarrow OT = 24$

Hence, radius of circle is $24\;{\text{cm}}$

4. In the given figure, if TP and TQ are the two tangents to a circle with centre 0 so that $\angle POQ = {110^\circ }$, then find $\angle PTO$.

Ans. Since $\angle POQ + \angle PTO = {180^\circ }$

$\left( {\because \angle OPT = {{90}^\circ },\angle OQT = {{90}^\circ }} \right)$

$ \Rightarrow {110^\circ } + \angle PTQ = {180^\circ }$

$ \Rightarrow \angle PTQ = {180^\circ } - {110^\circ } = {70^\circ }$

5. In the figure, given below PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at $Q.$ If $PB = 10\;{\text{cm}}$ and $CQ = 2$ ${\text{cm}}$, what is the length of PC?

Ans. ${\text{PA}} = {\text{PB}} = 10\;{\text{cm}}$

${\text{CQ}} = {\text{CA}} = 2\;{\text{cm}}$

${\text{PC}} = {\text{PA}} - {\text{CA}}$

$ = 10 - 2 = 8\;{\text{cm}}$

3 Marks Questions

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

Draw a line segment of length $7.6\;{\text{cm}}$ and divide it in the ratio $5:8.$ Measure the two parts.

Ans. Given: A line segment of length $7.6\;{\text{cm}}$.

To construct: To divide it in the ration $5:8.$and to measure the two parts.

Steps of construction:

(a) From any ray AX, making an acute angle with ${\text{AB}}$.

(b) Locate $13( = 5 + 8)$ points ${A_1},{A_2},{A_3},{A_4},{A_5},{A_6},{A_7},{A_8},{A_9},{A_{10}},{A_{11}},{A_{12}}$ and ${A_{13}}$ on AX such that ${\text{A}}{{\text{A}}_1} = {{\text{A}}_1}\;{{\text{A}}_2} = {{\text{A}}_2}\;{{\text{A}}_3} = {{\text{A}}_3}\;{{\text{A}}_4} = {{\text{A}}_4}\;{{\text{A}}_5} = {{\text{A}}_5}\;{{\text{A}}_6} = {{\text{A}}_6}\;{{\text{A}}_7} = {{\text{A}}_7}\;{{\text{A}}_8} = {{\text{A}}_8}\;{{\text{A}}_9} = {{\text{A}}_9}\;{{\text{A}}_{10}} = {{\text{A}}_{10}}\;{{\text{A}}_{11}} = {{\text{A}}_{11}}\;{{\text{A}}_{12}}$ $ = {{\text{A}}_{12}}\;{{\text{A}}_{13}}$

(c) Join ${\text{B}}{{\text{A}}_{13}}$

(d) Through the point ${{\text{A}}_5}$, draw a line parallel to ${{\text{A}}_{13}}\;{\text{B}}$ intersecting ${\text{AB}}$ at the point ${\text{C}}$.

Then, $AC:CB = 5:8$

On measurement, ${\text{AC}} = 3.1\;{\text{cm}},{\text{CB}} = 4.5\;{\text{cm}}$

 (By construction)

$\therefore \dfrac{{{\text{A}}{{\text{A}}_5}}}{{\;{{\text{A}}_5}\;{{\text{A}}_{13}}}} = \dfrac{{{\text{AC}}}}{{{\text{CB}}}}$ (By Basic Proportionality Theorem)

But $\dfrac{{{\text{A}}{{\text{A}}_5}}}{{\;{{\text{A}}_5}\;{{\text{A}}_{13}}}} = \dfrac{5}{8}$ (By construction)

Therefore, $\dfrac{{{\text{AC}}}}{{{\text{CB}}}} = \dfrac{5}{8}$

$ \Rightarrow {\text{AC}}:{\text{CB}} = 5:8$

2. Construct a triangle of sides $4\;{\text{cm}},5\;{\text{cm}}$ and $6\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle.

Ans. To construct: To construct a triangle of sides $4\;{\text{cm}},5\;{\text{cm}}$ and $6\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle.

Steps of construction:

(a) Draw a triangle ${\text{ABC}}$ of sides $4\;{\text{cm}},5\;{\text{cm}}$ and $6\;{\text{cm}}$.

(b) From any ray BX, making an acute angle with ${\text{BC}}$ on the side opposite to the vertex ${\text{A}}$.

(c) Locate 3 points ${{\text{B}}_1},\;{{\text{B}}_2}$ and ${{\text{B}}_3}$ on ${\text{BX}}$ such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3}$.

(d) Join ${{\text{B}}_3}{\text{C}}$ and draw a line through the point ${{\text{B}}_2}$, draw a line parallel to ${{\text{B}}_3}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${\text{C}}$ '.

(e) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ is the required triangle.

 (By construction)

$\therefore \dfrac{{{\text{B}}{{\text{B}}_2}}}{{\;{{\text{B}}_2}\;{{\text{B}}_3}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{{\text{C}}^\prime }{\text{C}}}}$ (By Basic Proportionality Theorem)

But $\dfrac{{{\text{B}}{{\text{B}}_2}}}{{\;{{\text{B}}_2}\;{{\text{B}}_3}}} = \dfrac{2}{1}$ (By construction)

Therefore, $\dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{CC}}}} = \dfrac{2}{1}$

$ \Rightarrow \dfrac{{{\text{CC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{1}{2}$ 

$ \Rightarrow \dfrac{{{{\text{C}}^\prime }{\text{C}}}}{{{\text{B}}{{\text{C}}^\prime }}} + 1 = \dfrac{1}{2} + 1$ 

$ \Rightarrow \dfrac{{{{\text{C}}^\prime }{\text{C}} + {\text{B}}{{\text{C}}^\prime }}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{{1 + 2}}{2}$

 $ \Rightarrow \dfrac{{{\text{BC}}}}{{{\text{B}}{{\text{C}}^\prime }}} = \dfrac{3}{2}$

 $ \Rightarrow \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{2}{3} \cdots  \ldots ...$ (i)

  (By construction)

$\therefore \Delta {\text{B}}{{\text{C}}^\prime }{{\text{A}}^\prime }~\Delta {\text{BCA}}({\text{AA}}$ similarity)

$\therefore \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{2}{3}$

3. Construct a triangle with sides $6\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$ and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle.

Ans. To construct: To construct a triangle of sides $5\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle.

Steps of construction:

(a) Draw a triangle ${\text{ABC}}$ of sides $5\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$.

(b) From any ray BX, making an acute angle with ${\text{BC}}$ on the side opposite to the vertex ${\text{A}}$.

(c) Locate 7 points ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3},\;{{\text{B}}_4},\;{{\text{B}}_5},\;{{\text{B}}_6}$ and ${{\text{B}}_7}$ on BX such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3} = {{\text{B}}_3}\;{{\text{B}}_4} = $ ${{\text{B}}_4}\;{{\text{B}}_5} = {{\text{B}}_5}\;{{\text{B}}_6} = {{\text{B}}_6}\;{{\text{B}}_7}$

(d) Join ${{\text{B}}_5}{\text{C}}$ and draw a line through the point ${{\text{B}}_7}$, draw a line parallel to ${{\text{B}}_5}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${\text{C}}$ '.

(e) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${A^\prime }B{C^\prime }$ is the required triangle.

 (By construction)

$\therefore \Delta {\text{ABC}}~\Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ (AA similarity)

$\therefore \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}}$ (By Basic Proportionality Theorem)

(By construction)

$\therefore \Delta {\text{B}}{{\text{B}}_7}{{\text{C}}^\prime }~\Delta {\text{B}}{{\text{B}}_5}{\text{C}}$ (AA similarity)

But $\dfrac{{{\text{B}}{{\text{B}}_5}}}{{{\text{B}}{{\text{B}}_7}}} = \dfrac{5}{7}($ By construction)

Therefore, $\dfrac{{BC}}{{B{C^\prime }}} = \dfrac{5}{7}$

$ \Rightarrow \dfrac{{B{C^\prime }}}{{BC}} = \dfrac{7}{5}$

$\therefore \dfrac{{AB}}{{AB}} = \dfrac{{{A^\prime }{C^\prime }}}{{AC}} = \dfrac{{B{C^\prime }}}{{BC}} = \dfrac{7}{5}$

4. Construct a tangent to a circle of radius $4\;{\text{cm}}$ from a point on the concentric circle of radius $6\;{\text{cm}}$ and measure its length. Also verify the measurement by actual calculation.

Ans. To construct: To construct a tangent to a circle of radius $4\;{\text{cm}}$ from a point on the concentric circle of radius $6\;{\text{cm}}$ and measure its lengths. Also to verify the measurements by actual calculation.

Steps of Construction:

(a) Join PO and bisect it. Let ${\text{M}}$ be the mid-point of ${\text{PO}}$.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the point $Q$ and $R$.

(c) Join PQ.

Then PQ is the required tangent.

By measurement, ${\text{PQ}} = 4.5\;{\text{cm}}$

By actual calculation,

${\text{PQ}} = \sqrt {{{({\text{OP}})}^2} + {{({\text{OQ}})}^2}} $

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

$ = \sqrt {36 - 16} $

$ = \sqrt {20}  = 4.47\;{\text{cm}}$

Justification: Join OQ. Then $\angle {\text{PQO}}$ is an angle in the semicircle and therefore,

$\angle {\text{PQO}} = {90^\circ } \Rightarrow {\text{PQ}} \bot {\text{OQ}}$

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

5. Draw a pair of tangents to a circle of radius $5\;{\text{cm}}$ which are inclined to each other at an angle of ${60^\circ }$.

Ans. To construct: A pair of tangents to a circle of radius $5\;{\text{cm}}$ which are inclined to each other at an angle of ${60^\circ }$

Steps of Construction:

(a) Draw a circle of radius $5\;{\text{cm}}$ with centre $O$.

(b) Draw an angle AOB of ${120^\circ }$

(c) At A and B, draw ${90^\circ }$ angles which meet at $C$.

Then ${\text{AC}}$ and ${\text{BC}}$ are the required tangents which are inclined to each other at an angle of ${60^\circ }$

Justification:

$\because \angle {\text{OAC}} = {90^\circ }$ and OA is a radius. (By construction)

$\therefore {\text{AC}}$ is a tangent to the circle.

$\because \angle {\text{OBC}} = {90^\circ }$ and ${\text{OB}}$ is a radius. (By construction)

$\therefore {\text{BC}}$ is a tangent to the circle.

Now, in quadrilateral OACB,

$\angle {\text{AOB}} + \angle {\text{OAC}} + \angle {\text{OBC}} + \angle {\text{ACB}} = {360^\circ }$

(Angle sum property of a quadrilateral)

$ \Rightarrow {120^\circ } + {90^\circ } + {90^\circ } + \angle {\text{ACB}} = {360^\circ }$

$ \Rightarrow {300^\circ } + \angle {\text{ACB}} = {360^\circ }$

$ \Rightarrow \angle {\text{ACB}} = {60^\circ }$

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

Ans. To construct: A right triangle $A B C$ with $AB = 6\;{\text{cm}},BC = 8\;{\text{cm}}$ and $\angle B = {90^\circ }. BD$ is the perpendicular from ${\text{B}}$ on ${\text{AC}}$ and the tangents from A to this circle.

Steps of Construction:

(a) Draw a right triangle ${\text{ABC}}$ with ${\text{AB}} = 6\;{\text{cm}},{\text{BC}} = 8\;{\text{cm}}$ and $\angle {\text{B}} = {90^\circ }$. Also, draw perpendicular BD on AC.

(b) Join AO and bisect it at M (here $O$ is the centre of circle through B, C, D).

(c) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points ${\text{B}}$ and ${\text{E}}$.

(d) Join ${\text{AB}}$ and ${\text{AE}}$.

Then ${\text{AB}}$ and ${\text{AE}}$ are the required two tangents.

Justification: Join OE.

Then, $\angle {\text{AEO}}$ is an angle in the semicircle.

$ \Rightarrow \angle {\text{AEO}} = {90^\circ }$

$ \Rightarrow {\text{AE}} \bot {\text{OE}}$

Since ${\text{OE}}$ is a radius of the given circle, AE has to be a tangent to the circle. Similarly ${\text{AB}}$ is also a tangent to the circle.

7. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord.

Ans. Let NM be chord of circle with centre $C$.

Let tangents at M.N meet at the point $O$.

Since OM is a tangent

$\therefore OM \bot CM\quad $

$\angle {\text{OMC}} = {90^\circ }$

$\because ON$ is a tangent

$\therefore ON \bot CN\quad $

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

Again in $\Delta CMN,CM = CN = r$

$\therefore \angle CMN = \angle CNM$

$\therefore \angle OMC - \angle CMN = \angle ONC - \angle CNM$

$ \Rightarrow \angle OML = \angle ONL$

Thus, tangents make equal angle with the chord

8. In the given figure, if ${\text{AB}} = {\text{AC}}$, prove that ${\text{BE}} = {\text{EC}}$.

Ans. Since tangents from an exterior point A to a circle are equal in length

$\therefore AD = AF \ldots ..(i)$

Similarly, tangents from an exterior point ${\text{B}}$ to a circle are equal in length

$\therefore BD = BE \ldots ..(ii)$

Similarly, for ${\text{C}}$

${\text{CE}} = {\text{CF}}.....(iii)$

Now ${\text{AB}} = {\text{AC}}$

$\therefore AB - AD = AC - AD$

$ \Rightarrow AB - AD = AC - AF \ldots  \ldots ..(By(i))$

$ \Rightarrow BD = CF$

$ \Rightarrow BE = CF \ldots  \ldots ..(By(ii))$

$ \Rightarrow BE = CE$

9. Find the locus of centre of circle with two intersecting lines.

Ans.

Let ${l_1},{l_2}$ be two intersection lines.

Let a circle with centre ${\text{P}}$ touch the two lines ${l_1}$ and ${l_2}$ at ${\text{M}}$ and ${\text{N}}$ respectively.

${\text{PM}} = {\text{PN}}($ Radii of same circle)

$\therefore {\text{P}}$ is equidistance from the lines ${l_1}$ and ${l_2}$

Similarly, centre of any other circle which touch the two intersecting lines ${l_1},{l_2}$ will be equidistant from ${l_1}$ and ${l_2}$.

$\therefore $ P lies on $l$ a bisector of the angle between ${l_1}$ and ${l_2}$

$(\because $ The locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines)

Hence, locus of centre of circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines.

10. In the given figure, a circle is inscribed in a quadrilateral ABCD in which $\angle B = {90^\circ }$. If $AD = 23\;{\text{cm}},AB = 29\;{\text{cm}}$ and $DS = 5\;{\text{cm}}$, find the radius of the circle.

Ans. In the given figure, $OP \bot BC$ and ${\text{O}}{{\text{Q}}^ \wedge }{\text{BA}}$

Also, ${\text{OP}} = {\text{OQ}} = {\text{r}}$

$\therefore OPBQ$ is a square

$\therefore {\text{BP}} = {\text{BQ}} = {\text{r}}$

But ${\text{DR}} = {\text{DS}} = 5\;{\text{cm}}$...(i)

$\therefore AR = AD - DR$

$\quad  = 23 - 5 = 18\;{\text{cm}}$

$AQ = AR = 18\;{\text{cm}}$

$BQ = AB - AQ$

$ = 20 - 18 = 11\;{\text{cm}}$

$r = 11\;{\text{cm}}$

4 Marks Questions

1. Construct a $\vartriangle ABC$ in which $AB = 6.5\;{\text{cm}},\angle B = {60^\circ }$ and $BC = 5.5\;{\text{cm}}$. Also construct a triangle ABC similar to $\vartriangle ABC$ whose each side is $\dfrac{3}{2}$ times the corresponding side of the $\vartriangle ABC$.

Ans. Steps of construction:

1. Draw a line segment ${\text{AB}} = 6.5\;{\text{cm}}$.

2. At B construct $\angle ABX = {60^\circ }$

3. With B as centre and radius ${\text{BC}} = 5.5\;{\text{cm}}$ draw an arc intersecting ${\text{BX}}$ at ${\text{C}}$.

4. Join AC.

Triangle so obtained is the required triangle.

5. Construct an acute angle ĐBAY at A on opposite side of vertex ${\text{C}}$ of $\Delta ABC$.

6. Locate 3 points ${A_1},{A_2},{A_3}$ on AY such that $A{A_1} = {A_1}{A_2} = {A_2}{A_3}$

7. Join ${A_2}$ to B and draw the line through ${A_9}$ parallel to ${A_2}B$ intersecting the extended line segment ${\text{AB}}$ at ${\text{B}}$ '.

8. Draw a line through B' parallel to BC intersecting the extended line seg.AC at C'.

9. $\Delta {\text{A}}{{\text{B}}^\prime }{{\text{C}}^\prime }$ so obtained is the required triangle

2. Draw a circle of radius $4\;{\text{cm}}$ from a point ${\text{P}},7\;{\text{cm}}$ from the centre of the circle, draw a pair of tangents to the circle measure the length of each tangent segment.

Ans. Steps of construction:

1. Take a point ${\text{O}}$ in the plane of a paper and draw a circle of the radius $4\;{\text{cm}}$.

2. Make a point ${\text{P}}$ at a distance of $7\;{\text{cm}}$ from the centre ${\text{O}}$ and Join OP.

3. Bisect the line segment OP. Let ${\text{M}}$ be the mid-point of OP.

4. Taking M as a centre and OM as radius draw a circle to intersect the given circle at the points ${\text{T}}$ and T'.

5. Join PT and PT', then PT and PT' are required tangents.

${\text{PT}} = {\text{P}}{{\text{T}}^\prime } = 5.75\;{\text{cm}}$

3. Draw a right triangle in which the sides containing the right angle are $5{\mathbf{cm}}$ and $4\;{\text{cm}}.$ Construct a similar triangle whose sides are $\dfrac{5}{3}$ times the sides of the above triangle.

Ans. Steps of construction:

1. Draw a line segment ${\text{BC}} = 5\;{\text{cm}}$.

2. At B construct $\angle CBX = {90^\circ }$.

3. With ${\text{B}}$ as centre and radius $ = 4\;{\text{cm}}$ draw an arc intersecting the ray ${\text{BX}}$ at ${\text{A}}$.

4. Join AC to obtain the required $\Delta ABC$.

5. Draw any ray BY making an acute angle with BC on the opposite side to the vertex A.

6. Locate 5 points ${B_1},{B_2},{B_3},{B_4}$ and ${B_5}$ on BY so that $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4} = {B_4}{B_5}$

7. Join ${B_3}$ to ${\text{C}}$ and draw a line through ${B_5}$ parallel to ${B_3}C$ intersecting the extended line segment ${\text{BC}}$ at ${{\text{C}}^\prime }$

8. Draw a line through ${\text{C}}$ ' parallel to CA intersecting the extended line segment ${\text{BA}}$ at A'.

Thus, DA'BC' is the required right triangle.

4. Construct a circle whose radius is equal to $4\;{\text{cm}}$. Let ${\text{P}}$ be a point whose distance from its centre is $6\;{\text{cm}}.$ Construct two tangents to it from ${\text{P}}$.

Ans. Steps of construction:

1. Take a point ${\text{O}}$ in the plane of the paper and draw a circle of radius $4\;{\text{cm}}$.

2. Make a point ${\text{P}}$ at a distance of $6\;{\text{cm}}$ from the centre ${\text{O}}$ and join OP.

3. Bisect the line segment OP. Let the point of bisection be M.

4. Taking M as centre and OM as radius, draw a circle to intersect the given circle at the point ${\text{T}}$ and ${{\text{T}}^\prime }$

5. Join PT and PT' to get the required tangents.

5. Draw a triangle ABC with sides ${\text{BC}} = 6.3\;{\text{cm}},{\text{AB}} = 5.2\;{\text{cm}}$ and $\angle ABC = {60^\circ }.$ Then construct a triangle whose sides are $\dfrac{4}{3}$ times the corresponding sides of $\Delta ABC$

Ans. Steps of construction:

1. Draw a line segment ${\text{BC}} = 6.3\;{\text{cm}}$.

2. At B make $\angle CBX = {60^\circ }$

3. With B as centre and radius equal to $5.2\;{\text{cm}}$, draw an arc intersecting ${\text{BX}}$ at ${\text{A}}$.

4. Join AC, then DABC is the required triangle.

5. Draw any ray by making an acute angle with ${\text{BC}}$ on the opposite side to the vertex ${\text{A}}$.

6. Locate the points ${B_1},{B_2},{B_3}$ and ${B_4}$ on BY so that $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4}$

7. Join ${B_3}$ to ${\text{C}}$ and draw a line through ${B_4}$ parallel to ${B_3}C$ intersecting the extended line segment ${\text{BC}}$ at ${{\text{C}}^\prime }$

8. Draw a line through ${\text{C}}$ ' parallel to CA intersecting the extended line segment BA at A'.

Thus, DA'BC' is the required triangle.

6. Draw a pair of tangents to a circle of radius $5\;{\text{cm}}$ which are inclined to each other at ${60^\circ }$

Ans. Steps of construction:

1. Draw a circle with center ${\text{O}}$ and radius $5\;{\text{cm}}$.

2. Draw any diameter ${\text{AOC}}$.

3. Construct $\angle BOC = {60^\circ }$ meeting the circle at ${\text{B}}$.

4. At A and B draw perpendiculars to OA and OB intersecting at P.

5. PA and PB are required tangents and $\angle APB = {60^\circ }$

7. Draw the tangents at the extremities of a diameter ${\text{AB}}$ of a circle of radius $2\;{\text{cm}}$. Are these tangents parallel? Given reasons.

Ans. Steps of construction:

1. Draw a circle of radius $2\;{\text{cm}}$.

2. Draw any diameter AOB.

3. Draw $AT \bot AB$ and $BM \bot AB$.

4. AT and BM are tangents extremities of the diameter $A B$.

5. $\because \angle 1 = {90^\circ },\angle 2 = {90^\circ },\therefore \angle 1 = \angle 2$, they are alternate angles.

8. Construct a $\Delta ABC$ in which $AB = 4\;{\text{cm}},BC = 5\;{\text{cm}}$ and $AC = 6\;{\text{cm}}.$ Now construct a triangle similar to $\vartriangle ABC$ such that each of its sides is two-third of the corresponding sides of $\Delta ABC$. Also prove your assertion.

Ans. Steps of construction:

1. Draw $\Delta ABC$ with sides ${\text{BC}} = 5\;{\text{cm}},{\text{AB}} = 4\;{\text{cm}}$ and ${\text{AC}} = 6\;{\text{cm}}$.

2. Below BC make acute $\angle CBX$.

3. Along BX mark off three points ${B_1},{B_2}$ and ${B_3}$ such that $B{B_1} = {B_1}{B_2} = {B_2}{B_3}$.

4. Join ${B_3}C$.

5. Draw  also  

Thus, D $\Delta B{C^\prime }{A^\prime }$ is the required triangle

9. Construct a $\vartriangle ABC$ in which ${\text{AB}} = 6.5\;{\text{cm}}\angle B = {60^\circ }$ and ${\text{BC}} = 5.5\;{\text{cm}}$. Also, construct a $\Delta A{B^\prime }{C^\prime }$ similar to $\Delta ABC$ whose each side is $\dfrac{3}{2}$ times the corresponding sides of the $\Delta ABC$

Ans. Steps of construction:

1. Construct a $\vartriangle ABC$ is which ${\text{AB}} = 6.5\;{\text{cm}},\angle B = {60^\circ }$ and ${\text{BC}} = 5.5\;{\text{cm}}$.

2. Draw a ray AX making any acute angle with ${\text{AB}}$ on the opposite side of the vertex ${\text{C}}$.

3. Cut three equal parts from AX say $A{X_1} = {X_1}{X_2} = {X_2}{X_3}$

4. Join ${X_2}$ to B.

5. From ${X_3}$ draw  at ${B^\prime }$.

6. At B' draw  intersecting ${\text{AY}}$ at ${{\text{C}}^\prime }$

7. $\Delta A{B^\prime }{C^\prime }$ is required triangle similar to $\Delta ABC$.

10. Draw a pair of tangents to a circle of radius $5\;{\text{cm}}$ which are inclined to each other at ${60^\circ }$

Ans. Steps of construction:

1. Draw a circle with centre ${\text{O}}$ and radius $5\;{\text{cm}}$.

2. Draw any radius OT.

3. Construct $\angle TO{T^\prime } = {180^\circ } - {60^\circ } = {120^\circ }$.

4. Draw $TP \bot OT$ and $TP \bot OT$. Then $P{T^\prime }$ and ${\text{PT}}$ are the two required tangents such that $\angle TP{T^\prime } = {60^\circ }.$ Here, $PT = P{T^\prime }$

11. Draw a circle of radius $4\;{\text{cm}}$ with centre $0.$Draw a diameter POQ. Through P or Q

draw tangent to the circle.

Ans. Steps of construction:

1. Draw a circle of radius $4\;{\text{cm}}$.

2. Draw diameter POQ.

3. Construct $\angle PQT = {90^\circ }$.

4. Produce ${\text{PQ}}$ to ${T^\prime }$, then $TQ{T^\prime }$ is the required tangent at the point $Q$.

12. Construct a triangle with sides $5\;{\text{cm}},6\;{\text{cm}}$ and $7\;{\text{cm}}$ and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of first triangle.

Ans. Steps of construction:

1. Draw a line segment ${\text{AB}} = 5\;{\text{cm}}$.

2. With A as centre and radius $6\;{\text{cm}}$ draw an arc.

3. Again B as centre and radius $7\;{\text{cm}}$ draw another arc cutting the previous arc at ${\text{C}}$. Join AC and BC, then DABC is required triangle.

4. Draw any ray AX making acute angle.

5. Locate 7 points ${A_1},{A_2},{A_3},{A_4},{A_j},{A_6}$ and ${A_7}$ on AX so that

$A{A_1} = {A_1}{A_2} = {A_2}{A_3} = {A_3}{A_4} = {A_4}{A_5} = {A_5}{A_6} = {A_6}{A_7}$

6. Join ${A_5}$ to B and draw a line through ${A_7}$ parallel to ${A_5}B$ intersecting the extended line segment ${\text{AB}}$ at ${B^\prime }$.

7. Draw , then $\Delta A{B^\prime }{C^\prime }$ is the required triangle

13. Let ABC be a right triangle in which ${\text{AB}} = 6\;{\text{cm}},{\text{BC}} = 8\;{\text{cm}}$ and $\angle B = {90^\circ } \cdot {\text{BD}}$ is the perpendicular from ${\text{B}}$ on ${\text{AC}}$. The circle through ${\text{B}},{\text{C}}$ and ${\text{D}}$ is drawn construct the tangents from A to this circle.

Ans. Steps of construction:

1. Draw $\Delta ABC$ with ${\text{BC}} = 8\;{\text{cm}},{\text{AB}} = 6\;{\text{cm}}$ and $\angle B = {90^\circ }$

2. Draw perpendicular BD from B to AC.

3. Let ${\text{O}}$ be the mid-point of ${\text{BC}}$. Draw a circle with centre ${\text{O}}$ and radius ${\text{OB}} = {\text{OC}}$. This circle will pass through the point D.

4. Join AO and bisect AO.

5. Draw a circle with centre $O$ ' and ${O^\prime }A$ as radius cuts the previous circle at ${\text{B}}$ and ${\text{P}}$.

6. Join AP, AP and AB are required tangents drawn from A to the circle passing through B,C and D.

14. Construct a triangle of sides $4\;{\text{cm}},5\;{\text{cm}}$ and $6\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle

Ans. Steps of construction:

1. Draw $\Delta ABC$ with ${\text{AB}} = 4\;{\text{cm}},{\text{BC}} = 6\;{\text{cm}}$ and ${\text{AC}} = 5\;{\text{cm}}$.

2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex $A$.

3. Locate 3 points ${B_1},{B_2}$ and ${B_3}$ on BX so that $B{B_1} = {B_1}{B_2} = {B_2}{B_3}$.

4. Join ${B_3}C$ and draw 

5. Draw a line through ${C^\prime }$ such that  Then, $\Delta {A^\prime }B{C^\prime }$ is the required triangle.

15. Construct an isosceles triangle whose base is $8\;{\text{cm}}$ and altitude $4\;{\text{cm}}$ and then another triangle whose sides are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle.

Ans. Steps of construction:

1. Draw line segment ${\text{AB}} = 8\;{\text{cm}}$.

2. Draw perpendicular bisector of ${\text{AB}}$ which intersects ${\text{AB}}$ at ${\text{D}}$.

3. Cut ${\text{DC}} = 4\;{\text{cm}}$. Join ${\text{AC}}$ and ${\text{BC}}$. Thus, $\Delta ABC$ is the required triangle.

4. Draw acute angle $\angle BAX$

5. Locate 3 points ${A_1},{A_2}$ and ${A_3}$ on AX so that $A{A_1} = {A_1}{A_2} = {A_2}{A_3}$

6. Join ${A_2}B$ and draw .

7. Draw .

8. $\Delta A{B^\prime }{C^\prime }$ is the required triangle

16. Draw a $\vartriangle ABC$ with side ${\text{BC}} = 7\;{\text{cm}},\angle B = {45^\circ }$ and $\angle A = {105^\circ }$, then construct a triangle whose sides are $\dfrac{4}{3}$ times the corresponding sides of $\Delta ABC$.

Ans. Steps of construction:

1. Draw $\vartriangle ABC$ in which ${\text{BC}} = 7\;{\text{cm}},\angle B = {45^\circ }$ and $\angle C = {30^\circ }$

2. Locate 4 points ${B_1},{B_2},{B_3}$ and ${B_4}$ on BZ such that $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4}$.

3. Join ${B_3}C$ and draw .

4. Now Draw 

5. $\Delta B{C^\prime }{A^\prime }$ is the required triangle.

17. Construct a circle whose radius is equal to $4\;{\text{cm}}$. Let P be a point whose distance from its centre is $6\;{\text{cm}}$. Construct two tangents to it from ${\text{P}}$.

Ans. Steps of construction:

1. Draw a circle with centre ${\text{O}}$ and radius $4\;{\text{cm}}$.

2. Mark point ${\text{P}}$ at a distance of $6\;{\text{cm}}$ from the centre ${\text{O}}$ and join OP.

3. Bisect the line segment OP. Let the point of bisection be M.

4. Take M as centre and OM as radius draw a circle to intersect the given circle at the points ${\text{T}}$ and ${\text{T}}$ '.

5. Join PT and PT' to get the required tangents.

18. 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 $3\;{\text{cm}}$. Construct tangents to each circle from the centre to the other circle.

Ans. Steps of construction:

1. Draw a line segment ${\text{AB}} = 8\;{\text{cm}}$.

2. Taking A as centre and radius = $4\;{\text{cm}}$ draw a circle.

3. Taking B as centre and radius = $3\;{\text{cm}}$ draw another circle.

4. Bisect the line segment AB, let the point of bisection be M.

5. Taking M as centre and MA as radius, draw a circle intersecting the given circles at the point ${T_1},{T_2},{T_3}$ and ${T_4}$ 

19. Construct a triangle of sides $4\;{\text{cm}},5\;{\text{cm}}$ and $6\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle.

Ans. Steps of construction:

1. Draw $\Delta ABC$ in which ${\text{AB}} = 4\;{\text{cm}},{\text{BC}} = 6\;{\text{cm}}$ and ${\text{AC}} = 5\;{\text{cm}}$.

2. Draw acute $\angle CBX$.

3. Locate 3 points ${B_1},{B_2}$ and ${B_3}$ on BX so that $B{B_1} = {B_1}{B_2} = {B_2}{B_3}$.

4. Join ${B_3}C$ and draw .

5. Now Draw 

Thus, $\Delta {A^\prime }B{C^\prime }$ is the required triangle.

20. Draw a pair of tangents to a circle of radius $5\;{\text{cm}}$ which are inclined to each other at an angle of ${60^\circ }$.

Ans. Steps of construction:

1. Draw a circle with centre $O$ and radius $OA = 5\;{\text{cm}}$.

2. Extend ${\text{OA}}$ to ${\text{B}}$ such that ${\text{OA}} = {\text{AB}} = 5\;{\text{cm}}$

3. With A as centre draw a circle of radius ${\text{OA}} = {\text{AB}} = 5\;{\text{cm}}$. 

Suppose it intersect the circle drawn in step 1 at the points ${\text{P}}$ and ${\text{Q}}$.

4. Join BP and BQ. Then BP and BQ are the required tangents.

Justification: In $\Delta OAP$

$OA = OP = 5\;{\text{cm}}(r)$

Also, ${\text{AP}} = 5\;{\text{cm}}$

$\Delta OAP$ is an equilateral triangle

$ \Rightarrow \angle PAO = {60^\circ }$

$ \Rightarrow \angle BAP = {120^\circ }$

In $\Delta BAP$,

${\text{AB}} = {\text{AP}}$ and $\angle BAP = {120^\circ }$

$\therefore \angle ABP = \angle APB = {30^\circ }$

Similarly, $\angle ABQ = \angle AQB = {30^\circ }$

$ \Rightarrow \angle PBQ = {60^\circ }$

21. From a point $P$ two tangents are drawn to a circle with centre $O$. If $OP = $ diameter of the circle, show that $\Delta APB$ is equilateral

Ans. Join OP.

Suppose OP meets the circle at Q. Join AQ.

We have i.e., ${\text{OP}} = $ diameter

$\therefore {\text{OQ}} + {\text{PQ}} = $ diameter

${\text{PQ}} = $ Diameter $ - $ radius $(\because OQ = r)$

$\therefore {\text{PQ}} = $ radius

Thus, ${\text{OQ}} = {\text{PQ}} = $ radius

Thus, OP is the hypotenuse of right triangle

OAP and $Q$ is the mid-point of OP

$\therefore O\;{\text{A}} = {\text{AQ}} = {\text{OQ}}$

$(\because $ mid-point of hypotenuse of a right triangle is equidistant from the vertices) $ \Rightarrow \Delta OAQ$ is equilateral

$ \Rightarrow \angle AOQ = {60^\circ }$

So, $\angle APO = {30^\circ }$

$\therefore \angle APB = 2\angle APO = {60^\circ }$

Also $PA = PB \Rightarrow \angle PAB = \angle PBA$

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

$\angle APB = {60^\circ }\therefore \angle PAB = \angle PBA = {60^\circ }$

Hence, $\vartriangle APB$ is equilateral.

22. Construct an isosceles triangle whose base is $8\;{\text{cm}}$ and altitude $4\;{\text{cm}}$ and then another triangle whose sides are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle.

Ans. To construct: To construct an isosceles triangle whose base is $8\;{\text{cm}}$ and altitude $4\;{\text{cm}}$ and then a triangle similar to it whose sides are $1\dfrac{1}{2}\left( {} \right.$ or $\left. {\dfrac{3}{2}} \right)$ of the corresponding sides of the first triangle.

Steps of construction:

(a) Draw ${\text{BC}} = 8\;{\text{cm}}$

(b) Draw perpendicular bisector of ${\text{BC}}$. Let it meets ${\text{BC}}$ at ${\text{D}}$.

(c) Mark a point A on the perpendicular bisector such that ${\text{AD}} = 4\;{\text{cm}}$.

(d) Join ${\text{AB}}$ and ${\text{AC}}$. Thus $\Delta {\text{ABC}}$ is the required isosceles triangle.

(e) From any ray BX, making an acute angle with BC on the side opposite to the vertex ${\text{A}}$.

(f) Locate 3 points ${{\text{B}}_1},\;{{\text{B}}_2}$ and ${{\text{B}}_3}$ on ${\text{BX}}$ such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3}$.

(g) Join ${{\text{B}}_2}{\text{C}}$ and draw a line through the point ${{\text{B}}_3}$, draw a line parallel to ${{\text{B}}_2}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${\text{C}}$ '.

(h) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ is the required triangle.

justification:

 (By construction)

$\therefore \Delta {\text{ABC}}~\Delta {{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ (AA similarity)

$\therefore \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}}$ (By Basic Proportionality Theorem)

 construction)

$\therefore \Delta {\text{B}}{{\text{B}}_3}{{\text{C}}^\prime }~\Delta {\text{B}}{{\text{B}}_2}{\text{C}}$ (AA similarity)

But $\dfrac{{{\text{B}}{{\text{B}}_3}}}{{{\text{B}}{{\text{B}}_2}}} = \dfrac{3}{2}$ (By construction)

Therefore,

$ \Rightarrow \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{3}{2}$

$\therefore \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{3}{2}$

23. Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = {60^\circ }.$ Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of triangle ${\text{ABC}}$.

Ans. To construct: To construct a triangle ${\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = $ ${60^\circ }$ and then a triangle similar to it whose sides are $\dfrac{3}{4}$ of the corresponding sides of the first triangle ABC.

Steps of construction:

(a) Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 6\;{\text{cm}},{\text{AB}} = 5\;{\text{cm}}$ and $\angle {\text{ABC}} = {60^\circ }$.

(b) From any ray BX, making an acute angle with ${\text{BC}}$ on the side opposite to the vertex ${\text{A}}$.

(c) Locate 4 points ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3}$ and ${{\text{B}}_4}$ on ${\text{BX}}$ such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3} = {{\text{B}}_3}\;{{\text{B}}_4}$.

(d) Join ${{\text{B}}_4}{\text{C}}$ and draw a line through the point ${{\text{B}}_3}$, draw a line parallel to ${{\text{B}}_4}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${{\text{C}}^\prime }$.

(e) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${{\text{A}}^\prime }{\text{BC}}$ ' is the required triangle.

Justification:

 By construction)

$\therefore \dfrac{{{\text{B}}{{\text{B}}_3}}}{{{\text{B}}{{\text{B}}_4}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}}$ (By Basic Proportionality Theorem)

But $\dfrac{{{\text{B}}{{\text{B}}_3}}}{{{\text{B}}{{\text{B}}_4}}} = \dfrac{3}{4}$ (By construction)

Therefore, $\dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{3}{4}$.........(i)

 construction $)$

$\therefore \Delta {\text{B}}{{\text{C}}^\prime }{{\text{A}}^\prime }~\Delta {\text{BCA}}$ (AA similarity)

$\therefore \dfrac{{{\text{A}}{{\text{B}}^\prime }}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{3}{4}$ (From eq. (i))

24. Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ },\angle {\text{A}} = {105^\circ }.$ Then construct a

triangle whose sides are $\dfrac{4}{3}$ times the corresponding sides of $\Delta {\text{ABC}}$

Ans. To construct: To construct a triangle ${\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ }$ and $\angle {\text{C}} = $ ${105^\circ }$ and then a triangle similar to it whose sides are $\dfrac{4}{3}$ of the corresponding sides of the first triangle ${\text{ABC}}$.

Steps of construction:

(a) Draw a triangle ${\text{ABC}}$ with side ${\text{BC}} = 7\;{\text{cm}},\angle {\text{B}} = {45^\circ }$ and $\angle {\text{C}} = {105^\circ }$.

(b) From any ray BX, making an acute angle with ${\text{BC}}$ on the side opposite to the vertex ${\text{A}}$.

(c) Locate 4 points ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3}$ and ${{\text{B}}_4}$ on ${\text{BX}}$ such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3} = {{\text{B}}_3}\;{{\text{B}}_4}$.

(d) Join ${{\text{B}}_3}{\text{C}}$ and draw a line through the point ${{\text{B}}_4}$, draw a line parallel to ${{\text{B}}_3}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${\text{C}}$ '.

(e) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${{\text{A}}^\prime }{\text{BC}}$ ' is the required triangle.

Justification:

 (By construction)

$\therefore \Delta {\text{B}}{{\text{B}}_4}{{\text{C}}^\prime }~\Delta {\text{B}}{{\text{B}}_3}{\text{C}}$ (AA similarity)

$\therefore \dfrac{{{\text{B}}{{\text{B}}_4}}}{{{\text{B}}{{\text{B}}_3}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}}$ (By Basic Proportionality Theorem)

But $\dfrac{{{\text{B}}{{\text{B}}_4}}}{{{\text{B}}{{\text{B}}_3}}} = \dfrac{4}{3}$ (By construction)

Therefore, $\dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{4}{3} \ldots  \ldots ..$ (i)

 construction)

$\therefore \Delta {\text{B}}{{\text{C}}^\prime }{{\text{A}}^\prime }~\Delta {\text{BCA}}$ (AA similarity)

$\therefore \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{4}{3}$ (From eq. (i))

25. Draw a right triangle in which the sides (other than hypotenuse) are of lengths $4\;{\text{cm}}$ and $3\;{\text{cm}}.$ Then construct another triangle whose sides are $\dfrac{5}{3}$ times the corresponding sides of the given triangle

Ans. To construct: To construct a right triangle in which sides (other than hypotenuse) are of lengths $4\;{\text{cm}}$ and $3\;{\text{cm}}$ and then a triangle similar to it whose sides are $\dfrac{5}{3}$ of the corresponding sides of the first triangle ${\text{ABC}}$.

Steps of construction:

(a) Draw a right triangle in which sides (other than hypotenuse) are of lengths $4\;{\text{cm}}$ and 3 ${\text{cm}}$.

(b) From any ray BX, making an acute angle with ${\text{BC}}$ on the side opposite to the vertex ${\text{A}}$.

(c) Locate 5 points ${{\text{B}}_1},\;{{\text{B}}_2},\;{{\text{B}}_3},\;{{\text{B}}_4}$ and ${{\text{B}}_5}$ on BX such that ${\text{B}}{{\text{B}}_1} = {{\text{B}}_1}\;{{\text{B}}_2} = {{\text{B}}_2}\;{{\text{B}}_3} = {{\text{B}}_3}\;{{\text{B}}_4} = {{\text{B}}_4}\;{{\text{B}}_5}$.

(d) Join ${{\text{B}}_3}{\text{C}}$ and draw a line through the point ${{\text{B}}_5}$, draw a line parallel to ${{\text{B}}_3}{\text{C}}$ intersecting ${\text{BC}}$ at the point ${\text{C}}$ '.

(e) Draw a line through ${\text{C}}$ ' parallel to the line CA to intersect BA at A'.

Then, ${{\text{A}}^\prime }{\text{B}}{{\text{C}}^\prime }$ is the required triangle.

Justification:

 (By construction)

$\therefore \Delta {\text{B}}{{\text{B}}_5}{{\text{C}}^\prime }~\Delta {\text{B}}{{\text{B}}_3}{\text{C}}$ (AA similarity)

$\therefore \dfrac{{{\text{B}}{{\text{B}}_5}}}{{{\text{B}}{{\text{B}}_3}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}}$ (By Basic Proportionality Theorem)

But $\dfrac{{{\text{B}}{{\text{B}}_5}}}{{{\text{B}}{{\text{B}}_3}}} = \dfrac{5}{3}$ (By construction)

Therefore, $\dfrac{{B{C^\prime }}}{{BC}} = \dfrac{5}{3}$..........(i)

 ( By construction)

$\therefore \Delta {\text{B}}{{\text{C}}^\prime }{{\text{A}}^\prime }~\Delta {\text{BCA}}$ (AA similarity)

$\therefore \dfrac{{{\text{AB}}}}{{{\text{AB}}}} = \dfrac{{{{\text{A}}^\prime }{{\text{C}}^\prime }}}{{{\text{AC}}}} = \dfrac{{{\text{B}}{{\text{C}}^\prime }}}{{{\text{BC}}}} = \dfrac{5}{3}$ (From eq. (i))

26. Draw a circle of radius $3\;{\text{cm}}$. Take two points ${\text{P}}$ and ${\text{Q}}$ on one of its extended diameter each at a distance of $7\;{\text{cm}}$ from its centre. Draw tangents to the circle from these two points ${\mathbf{P}}$ and $Q$.

Ans. To construct: A circle of radius $3\;{\text{cm}}$ and take two points ${\text{P}}$ and ${\text{Q}}$ on one of its extended diameter each at a distance of $7\;{\text{cm}}$ from its centre and then draw tangents to the circle from these two points ${\text{P}}$ and ${\text{Q}}$.

Steps of Construction:

(a) Bisect PO. Let ${\text{M}}$ be the mid-point of ${\text{PO}}$.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points ${\text{A}}$ and ${\text{B}}$.

(c) Join ${\text{PA}}$ and ${\text{PB}}$.

Then ${\text{PA}}$ and ${\text{PB}}$ are the required two tangents.

(d) Bisect QO. Let ${\text{N}}$ be the mid-point of ${\text{QO}}$.

(e) Taking ${\text{N}}$ as centre and NO as radius, draw a circle. Let it intersects the given circle at the points $C$ and $D$.

(f) Join ${\text{QC}}$ and ${\text{QD}}$.

Then QC and QD are the required two tangents.

Justification: Join OA and OB.

Then $\angle {\text{PAO}}$ is an angle in the semicircle and therefore $\angle {\text{PAO}} = {90^\circ }$.

$ \Rightarrow {\text{PA}} \bot {\text{OA}}$

Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, ${\text{PB}}$ is also a tangent to the circle.

Again, join OC and OD.

Then $\angle {\text{QCO}}$ is an angle in the semicircle and therefore $\angle {\text{QCO}} = {90^\circ }$.

Since ${\text{OC}}$ is a radius of the given circle, ${\text{QC}}$ has to be a tangent to the circle. Similarly, ${\text{QD}}$ is also a tangent to the circle.

27. Draw a line segment AB of length $8{\mathbf{cm}}$. Taking A as centre, draw a circle of radius 4 cm and taking $B$ as centre, draw another circle of radius $3\;{\text{cm}}.$ Construct tangents to each circle from the centre of the other circle.

Ans. To construct: A line segment of length $8\;{\text{cm}}$ and taking A as centre, to draw a circle of radius $4\;{\text{cm}}$ and taking ${\text{B}}$ as centre, draw another circle of radius $3\;{\text{cm}}$. Also, to construct tangents to each circle from the centre to the other circle.

Steps of Construction:

(a) Bisect BA. Let M be the mid-point of BA.

(b) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points ${\text{P}}$ and ${\text{Q}}$.

(c) Join BP and BQ.

Then, BP and BQ are the required two tangents from B to the circle with centre $A$.

(d) Again, Let ${\text{M}}$ be the mid-point of ${\text{AB}}$.

(e) Taking M as centre and MB as radius, draw a circle. Let it intersects the given circle at the points ${\text{R}}$ and ${\text{S}}$.

(f) Join AR and AS.

Then, AR and AS are the required two tangents from A to the circle with centre $B$.

Justification: Join BP and BQ.

Then $\angle $ APB being an angle in the semicircle is ${90^\circ }$

$$ \Rightarrow {\text{BP}} \bot {\text{AP}}$$

Since AP is a radius of the circle with centre A, BP has to be a tangent to a circle with centre A. Similarly, ${\text{BQ}}$ is also a tangent to the circle with centre ${\text{A}}$.

Again join AR and AS.

Then $\angle {\text{ARB}}$ being an angle in the semicircle is ${90^\circ }$

$ \Rightarrow {\text{AR}} \bot {\text{BR}}$

Since BR is a radius of the circle with centre B, AR has to be a tangent to a circle with centre B. Similarly, AS is also a tangent to the circle with centre B.

28. 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. To construct: 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.

Steps of Construction:

(a) Draw a circle with the help of a bangle.

(b) Take two non-parallel chords ${\text{AB}}$ and ${\text{CD}}$ of this circle.

(c) Draw the perpendicular bisectors of ${\text{AB}}$ and ${\text{CD}}$. Let these intersect at ${\text{O}}$. Then ${\text{O}}$ is the centre of the circle draw.

(d) Take a point P outside the circle.

(e) Join PO and bisect it. Let ${\text{M}}$ be the mid-point of ${\text{PO}}$.

(f) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points $Q$ and $R$.

(g) Join PQ and PR.

Then PQ and PR are the required two tangents.

Justification: Join OQ and OR.

Then, $\angle {\text{PQO}}$ is an angle in the semicircle.

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

$ \Rightarrow {\text{PQ}} \bot {\text{OQ}}$

Since ${\text{OQ}}$ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.

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• Focus on key topics for efficient studying.

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