Ex.11.1 Q7 Constructions Solution - NCERT Maths Class 10

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Question

Draw a right triangle in which the sides (other than hypotenuse) are of lengths \(4 \,\rm{cm}\) and \(3 \,\rm{cm}\). Then construct another triangle whose sides are \(\begin{align}\frac{5}{3}\end{align}\) times the corresponding sides of the given triangle.

 

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 \({{BC = 4\, \rm{cm}}}{{.}}\) At  \(B,\) make an angle \(\angle {{CBY}} = {90^ \circ }\) and mark \(A\) on \(BY\) such that \({{BA = 3 }}\,{\rm{cm}}{{.}}\) Join \(A\) to \(C. \) Thus \({{\Delta ABC}}\) is constructed.

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

(iii) Mark \(5\) (\(5  > 3\) in \(\frac{5}{3}\) ) points \({{{B}}_{{I}}}{{,}}\,{{{B}}_{{2}}}{{,}}\,{{{B}}_{{3}}}{{,}}\,{{{B}}_{{4}}}{{,}}{{{B}}_{{5}}}\) on \(BX\) so that \(\,{{B}}{{{B}}_{{1}}}{{ = }}{{{B}}_{{1}}}{{{B}}_{{2}}}{{ = }}{{{B}}_{{2}}}{{{B}}_{{3}}}{{ = }}{{{B}}_{{3}}}{{{B}}_{{4}}}{{ = }}{{{B}}_{{4}}}{{{B}}_{{5}}}\)

(iv) Join\({B}_{3}\) (\(3^\rm{rd}\) point on \(BX\) as \(3<5\)) to \(C\) and draw \({{{B}}_{{5}}}{{C'}}\) parallel to \({{{B}}_{{3}}}{{C}}\) so that \({{C'}}\) lies on the extension of \(BC.\)

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

Now \({{\Delta BA'C'}}\) is the required triangle similar to \({{\Delta BAC}}\) where

\(\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{5}}}{{{3}}}\end{align}\)

Proof:

In \({{\Delta B}}{{{B}}_{{3}}}{{C',}}\,\,{{{B}}_{{3}}}{{C}}\,\,{{||}}\,\,{{{B}}_{{3}}}{{{C}}^{{1}}}\)

Hence by Basic proportionality theorem,

\[\begin{align} \frac{{{{B}}_{{3}}}{{{B}}_{{5}}}}{{B}{{{B}}_{{3}}}}&{=}\frac{{CC }\!\!'\!\!{ }}{{BC}}{=}\frac{{2}}{{3}} \\ \frac{{CC }\!\!'\!\!{ }}{{BC}} {+1}&=\frac{{2}}{{3}}{+1}\qquad \text{(Adding}\,\,{1)} \\ \frac{{CC }\!\!'\!\!{ +BC}}{{BC}} &{=}\frac{{5}}{{3}} \\ \frac{{BC }\!\!'\!\!{ }}{{BC}}&{=}\frac{{5}}{{3}} \end{align}\]

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

\(\angle {{ABC}} = \angle {{A'BC'}} = 90^\circ \)

\(\angle {{BCA}} = \angle {{BC'A'}} \) (Corresponding angles as \({{CA}}\,\,\,{{||}}\,\,{{C'A'}} \) )

\(\angle {{BAC}} = \angle {{BA'C'}}\)

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

Therefore corresponding sides are proportional,

Hence,

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}&{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}\\&{{ = }}\frac{{{5}}}{{{3}}}\end{align}\]

  
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