# Constructions : Exercise 11.1 (Mathematics NCERT Class 10th)

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

Q.1      Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Sol.      Steps of Construction : 1. Draw a line segment AB = 7.6 cm
2. Draw a ray AC making any acute angle with AB, as shown in the figure.
3. On ray AC, starting from A, mark 5 + 8 = 13 equal line segments :
$A{A_1},\,{A_1}{A_2},{A_2}{A_3},{A_3}{A_4},{A_4}{A_5},{A_5}{A_6},{A_6}{A_7},{A_7}{A_8},{A_8}{A_9},{A_9}{A_{10}},{A_{10}}{A_{11}},{A_{11}}{A_{12}}\,and\,{A_{12}}{A_{13}}$
4. Join ${A_{13}}B$
5. From ${A_5},\,draw\,{A_5}P||\,{A_{13}}B$, meeting AB at P.
6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts, we find AP = 2.9 cm and PB = 4.7 cm (approx).
Justification :
In  $\Delta\,AB{A_{13}},P{A_5}||\,B{A_{13}}$
Therefore, $\Delta {\mkern 1mu} AP{A_5}$  ~ $\Delta AB{A_3}$
$\Rightarrow$ ${{AP} \over {PB}} = {{A{A_5}} \over {{A_5}{A_{13}}}} = {5 \over 8}$
$\Rightarrow$ AP : PB = 5 : 8

Q.2      Construct a triangle of sides 4 cm, 5 cm and 6cm and then a triangle similar to it whose sides are ${2 \over 3}$ of the corresponding sides of it.
Sol.          Steps of Construction :
1. Draw a line segment BC = 6 cm
2. With B as centre and radius equal to 5 cm, draw an arc.
3. With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.
4. Join AB and AC, then $\Delta$ ABC is the required triangle. 5. Below BC, make an acute angle CBX.
6. Along BX, mark off three points : ${B_1},{B_2}\,and\,{B_3}$ such that $B{B_1} = {B_1}{B_2} = {B_2}{B_3}$.
7. Join ${B_3}C$.
8. From ${B_2},draw\,{B_2}D||{B_3}C$, meeting BC at D.
9. From D, draw ED || AC, meeting BA at E. Then,
EBD is the required triangle whose sides are ${2 \over 3}rd$ of the corresponding sides of $\Delta$ ABC.
Justification : Since DE || CA
Therefore,  $\Delta ABC$  ~  $\Delta {\mkern 1mu} EBD$
and ${{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {2 \over 3}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${2 \over 3}$ rd of the corresponding sides of $\Delta$ ABC.

Q.3      Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are ${7 \over 5}$ of the corresponding sides of the first triangle.
Sol.         Steps of construction : 1. With the given data, construct $\Delta$ ABC in which BC = 7 cm , CA = 6 cm and AB = 5 cm
2. Below BC, make an acute $\angle CBX$.
3. Along BX, mark off seven points : ${B_1},{B_2},{B_3},{B_4},{B_5},{B_6}\,and\,{B_7}$ such that     $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4} = {B_4}{B_5} = {B_5}{B_6} = {B_6}{B_7}$
4. Join ${B_5}C$
5. From ${B_7}draw\,{B_7}D||{B_5}C$, meeting BC produced at D.
6. From D, draw DE || CA, meeting BA produced at E.
Then EBD is the required triangle whose sides are ${7 \over 5}$ th of the corresponding sides of $\Delta \,ABC$.
Justification :
Since DE || CA
Therefore,  $\Delta ABC \sim \Delta EBD\,\,and\,\,{{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {7 \over 5}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${7 \over 5}$ the of the corresponding sides of $\Delta \,ABC$.

Q.4      Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1{1 \over 2}$ times the corresponding sides of the isosceles triangle.
Sol.        Steps of Construction :
1. Draw BC = 8 cm
2. Construct PQ the perpendicular bisector of line segment BC meeting BC at M.
3. Along MP cut off MA = 4 cm
4. Join BA and CA. Then $\Delta \,ABC$ so obtained is the required $\Delta \,ABC$
5. Extend BC to D, such that BD = 12 cm.
6. Draw DE || CA, meeting BA produced at E.
Then $\Delta \,EBD$ is the required triangle. Justification :
Since DE || CA
Therefore, $\Delta {\mkern 1mu} ABC$  ~ $\Delta EBD$
and ${{EB} \over {AB}} = {{DE} \over {CA}} = {{BD} \over {BC}} = {{12} \over 8} = {3 \over 2}$
Hence, we get the new triangle similar to the given triangle whose sides are ${3 \over 2},$, i.e., $1{1 \over 2}$ times of the corresponding sides of the isosceles $\Delta \,ABC$.

Q.5      Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and $\angle ABC = {60^0}$. Then construct a triangle whose sides are ${3 \over 4}$ of the corresponding sides of the triangle ABC.
Sol.       Steps of Construction :

1. With the given data, construct $\Delta ABC$ in which BC = 6 cm, $\angle ABC = {60^0}$ and AB = 5 cm.
2. Below BC, make an acute $\angle CBX$.
3. Along BX, mark off 4 points : ${B_1},{B_2},{B_3}\,and\,\,{B_4}$ such that $B{B_1} = {B_1}{B_2}\, = {B_2}{B_3}\,\,\, = {B_3}{B_4}$.
4. Join ${B_4}C$.
5. From D, draw ED || AC, meeting BA at E.Then, EBD is the required triangle whose sides are ${3 \over 4}$ th of the corresponding sides of $\Delta ABC$.
Justification :
Since, DE || CA Therefore  $\Delta ABC$${\mkern 1mu} \Delta EBD$
and, ${{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {3 \over 4}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${3 \over 4}$th of the corresponding sides of $\Delta ABC.$

Q.6      Draw a triangle ABC with side BC = 7 cm, $\angle B = {45^o},\angle A = {105^o}$. Then construct a triangle whose sides are ${4 \over 3}$ times the corresponding sides of $\Delta \,ABC$.

1. With the given data, construct $\Delta \,ABC$ in which BC = 7 cm ,
$\angle B = {45^o},\,\angle C = {180^o} - \left( {\angle A + \angle B} \right)$

$\Rightarrow$ $\angle C = {180^o} - \left( {{{105}^o} + {{45}^o}} \right)$
$= {180^o} - {150^o} = {30^o}$
2. Below BC, make an acute $\angle CBX$
3. Along BX, mark off four points : ${B_1},{B_2},{B_3}\,and\,{B_4}$ such that  $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4}$
4. Join ${B_3}C$
5. From ${B_4}$ draw ${B_4}D||{B_3}C$ meeting BC produced at D.
6. From D, draw ED || AC, meeting BA produced at E.
Then EBD is the required triangle whose sides are ${4 \over 3}$ times of the corresponding sides of  $\Delta \,ABC$
Justification :
Since, DE || CA Therefore, $\Delta {\mkern 1mu} ABC$$\Delta {\mkern 1mu} EBD$
and ${{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {4 \over 3}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${4 \over 3}$ times of the corresponding sides of $\Delta \,ABC$

Q.7      Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are ${5 \over 3}$ times the corresponding sides of the given triangle.
Sol.        Steps of Construction :
1. With given data, construct $\Delta \,ABC$ in which BC = 4 cm $\angle B = {90^o}$ and BA = 3 cm. 2. Below BC, make an acute $\angle CBX$
3. Along BX, mark off five points :
${B_1},{B_2},{B_3},{B_4}\,and\,{B_5}$ such that  $B{B_1} = {B_1}{B_2} = {B_2}{B_3} = {B_3}{B_4} = {B_4}{B_5}$
4. Join ${B_3}C$
5. From ${B_5}$ draw ${B_5}D||{B_3}C$ meeting BC produced at D.
6. From D, draw ED ||AC, meeting BA produced at E.
Then EBD is the required triangle whose sides are ${5 \over 3}$ times of the corresponding sides of $\Delta \,ABC$
Justification :
Since, DE || CA Therefore, $\Delta {\mkern 1mu} ABC$$\Delta {\mkern 1mu} EBD$
and ${{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {5 \over 3}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${5 \over 3}$ times of the corresponding sides of $\Delta \,ABC$.

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