Ex.11.1 Q5 Constructions Solution - NCERT Maths Class 10

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Question

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 \(ABC.\)

 

 Video Solution
Constructions
Ex 11.1 | Question 5

Text Solution

What is known?

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

What is unknown?

Construction.

Reasoning:

  • 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:

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

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

Proof:

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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