Ex.11.1 Q2 Constructions Solution - NCERT Maths Class 10

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Question

Construct a triangle of sides \(4 \;\rm{cm}\), \(5 \;\rm{cm}\) and \(6\; \rm{cm}\) and then a triangle similar to it whose sides are \(\begin{align}\frac{2}{3}\end{align}\) of the corresponding sides of the first triangle.

 

Text Solution

 

What is known?

Sides of the triangle and the ratio of corresponding sides of \(2\) triangles.

What is unknown?

Construction.

Reasoning:

  • Draw the line segment of largest length \(6\, \rm{cm}\). Measure \(5 \;\rm{cm}\) and \(4 \;\rm{cm}\) separately and cut arcs from \(2\) ends of the line segment such that they cross each other at one point. Connect this point from both the ends.
  • Then draw another line which makes an acute angle with the given line (\(6\; \rm{cm}\)).
  • 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 \(BC = 6 \text{cm.}\) With \(B\) and \(C\) as centres and radii \(5 \;\rm{cm}\) and \(4\,\rm{cm}\) respectively draw arcs to intersect at \(A.\)  \({\rm{\Delta ABC}}\) is obtained.

(ii) Draw ray \(BX\) making an acute angle with \(BC.\)

(iii) Mark \(3\) ( {\(3> 2\) in the ratio \(\begin{align}\frac{2}{3}\end{align}\)points \({{{B}}_{{1}}}{{,}}\,\,{{{B}}_{{2}}}{{,}}\,\,{{{B}}_{{3}}}\) on \(BX\) such that \(\,{{B}}{{{B}}_{{1}}}{{ = }}{{{B}}_{{1}}}{{{B}}_{{2}}}{{ = }}{{{B}}_{{2}}}{{{B}}_{{3}}}\) .

(iv) Join \(\,{{{B}}_{\rm{3}}}{{C}}\) and draw the line through \({B_2}\)  (2nd point where \(2 < 3\) in the ratio \(\begin{align}\frac{2}{3}\end{align}\)) parallel to \(B_3C\) meeting \(BC\) at \({{C'.}}\)

(v) Draw a line thorough \(\,{{C'}}\) parallel to \(CA\) to meet \(BA\) at \(A’.\) Now \(\,{{\Delta A'BC'}}\) is the required triangle similar to \(\,{{\Delta ABC}}\) where

\[\begin{align}\frac{{{{BC'}}}}{{{{BC}}}}&{{ = }}\frac{{{{BA'}}}}{{{\rm{BA}}}}{\rm{ = }}\frac{{{{C'A'}}}}{{{\rm{CA}}}}\\&{{ = }}\frac{{{2}}}{{{3}}}\end{align}\]

Proof:

In \(\Delta B B_{3} C, B_{2} C\) is parallel to \(\,{{{B}}_{{3}}}{{C}} .\)

Hence by Basic proportionality theorem,

\[\begin{align}\frac{{{{{B}}_{{2}}}{{{B}}_{{2}}}}}{{{{B}}{{{B}}_{{2}}}}}&{{ = }}\frac{{{{C'C}}}}{{{{BC'}}}}\\&{{ = }}\frac{{{1}}}{{{2}}}\end{align}\]

Adding \(1\),

\[\begin{align}\frac{{{{C'C}}}}{{{{BC'}}}}{{ + 1}}& = \frac{{{1}}}{{{2}}}{{ + 1}}\\\frac{{{{C'C + BC'}}}}{{{{BC'}}}}{{ }}&=\frac{{{3}}}{{{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{{BC}}}}{{{{BC'}}}}{{}}&= \frac{{{3}}}{{{2}}}\\\,\,\,\,\,\text{(or)}\,\,\,\,\,\,\frac{{{{BC'}}}}{{{{BC}}}}{{ }}&=\frac{{{2}}}{{{3}}} & & {{ \ldots (1)}}\end{align}\]

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

\(\angle \text{A }\!\!'\!\!\text{ BC }\!\!'\!\!\text{ }=\angle \text{ABC}\) (Common)

\(\angle \text{BA }\!\!'\!\!\text{ C }\!\!'\!\!\text{ }=\angle \text{BAC}\) (Corresponding angles \(∵\)  \(\,\,\text{C }\!\!'\!\!\text{ }A'||\,\text{CA}\) )

\(\angle \text{BA }\!\!'\!\!\text{ C }\!\!'\!\!\text{ }=\angle \text{BCA}\) (Corresponding angles \(∵\) \( \,\,\text{C }\!\!'\!\!\text{ }A'||CA  \) )

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

Corresponding sides are proportional 

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{2}}}{{{3}}}\,\,\left( {{\text{from}}\,{{(1)}}} \right)\end{align}\]

  
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