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 :

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                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.
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               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 :

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                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.

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               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 :

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                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.
Sol.        Steps of Construction :
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              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.

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              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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