Ex.11.1 Q5 Constructions Solution - NCERT Maths Class 10

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Draw a triangle \(ABC\) with side \({{BC = 6 \,\rm{cm,}\;AB = 5\,\rm{cm}}}\) and \(\angle {{ABC = 6}}{{{0}}^0}.\)  Then construct a triangle whose sides are \(\begin{align}\frac{3}{2}\end{align}\) of the corresponding sides of the triangle \(ABC.\)


Text Solution

What is known?

\(2\) sides and the angle between them and the ratio of corresponding sides of \(2\) triangles.

What is unknown?



  • Draw the triangle with the given conditions.
  • Then draw another line which makes an acute angle with the base line. Divide the line into \(m + n\) parts where \(m\) and \(n\) are the ratio given.
  • Two triangles are said to be similar if their corresponding angles are equal. They are said to satisfy Angle-Angle-Angle (AAA) Axiom.
  • Basic proportionality theorem states that, “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally".


Steps of constructions:

(i) Draw \({{BC = 6 \,\rm{cm}}}.\) At \(B,\) make \(\angle {{CBY = }}\,{{60^\circ }}\) and cut an arc at \(A\) so that \(BA = 5 \,\rm{cm}.\) Join \(\,{{AC}}\,,\,\,{{\Delta ABC}}\) is obtained.

(ii) Draw the ray \(BX\) such that \(\angle {{CBX}}\) is acute.

(iii) Mark \(4 \) (\({4} > 3\) in \(\frac{3}{4}\)) points \({{{B}}_1},{{{B}}_{{2}}},\,{{{B}}_{{3}}},\,{{{B}}_{{4}}}\) on \(BX\) such that \({{B}}{{{B}}_{{1}}}{{ = }}{{{B}}_{{1}}}{{{B}}_{{2}}}{{ = }}{{{B}}_{{2}}}{{{B}}_{{3}}}{{ = }}{{{B}}_{{3}}}{{{B}}_{{4}}}\)

(iv) Join \({{{B}}_{{4}}}\) to \(C\) and draw \({{{B}}_{{3}}}{{C'}}\) parallel to \({{{B}}_{{4}}}{{C}}\) to intersect \(BC\) at \({{C'}}{{.}}\)

(v) Draw \({{C'A'}}\) parallel to \(CA\) to intersect \(BA\) at \(A’.\)

Now, \(\,{{\Delta A'BC'}}\) is the required triangle similar to \({{\Delta ABC}}\) where

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{3}}}{{{4}}}\end{align}\]


In \({{\Delta }}\,{{B}}{{{B}}_4}{{C'}}\,\,,\,\,{{{B}}_3}{{C' || }}{{{B}}_4}{{C}}\)

Hence by Basic proportionality theorem,

\[\begin{align}\frac{{{{{B}}_{{3}}}{{{B}}_{{4}}}}}{{{{B}}{{{B}}_{{3}}}}}&= \frac{{{{C'C}}}}{{{{BC'}}}}= \frac{{{1}}}{{{3}}}\\\frac{{{{C'C}}}}{{{{BC'}}}}{{ + 1}}& = \frac{{{1}}}{{{3}}}{{ + 1 \quad \text{(Adding 1)}}}\\\frac{{{{C'C + BC'}}}}{{{{BC'}}}}& = \frac{{{4}}}{{{3}}}\\\frac{{{{BC}}}}{{{{BC'}}}}& =\frac{{{4}}}{{{3}}}\,\, {\rm{(or)}} \,\frac{{{{BC'}}}}{{{{BC}}}} =\frac{{{3}}}{{{4}}}\end{align}\]

Consider \({{\Delta BA'C'}}\,\,{{and}}\,\,{{\Delta BAC}}\)

\[\begin{align} & \angle \text{A }\!\!'\!\!\text{ BC }\!\!'\!\!\text{ }=\angle \text{ABC}={{60}^{\circ }} \\ & \angle \text{BCA }\!\!'\!\!\text{ }=\angle \text{BCA}\,\,\\&\left( \text{Corresponding}\,\text{angles}\because \,\text{CA}||\text{CA} \right) \\ & \angle \text{BA }\!\!'\!\!\text{ C }\!\!'\!\!\text{ }=\angle \text{BAC}\\&\left( \text{Corresponding}\,\text{angles} \right) \\ \end{align}\]

By AAA axiom, \({{\Delta BA'C' \sim \Delta BAC}}\)

Therefore corresponding sides are proportional,

\[\begin{align}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{3}}}{{{4}}}\end{align}\]

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