In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Ex.11.1 Q7 Constructions Solution - NCERT Maths Class 10

Go back to  'Ex.11.1'

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.

 

 Video Solution
Constructions
Ex 11.1 | Question 7

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

\(\begin{align}&{{B}}{{{B}}_{{1}}}\\=&{{{B}}_{{1}}}{{{B}}_{{2}}}\\=&{{{B}}_{{2}}}{{{B}}_{{3}}}\\=&{{{B}}_{{3}}}{{{B}}_{{4}}}\\=&{{{B}}_{{4}}}{{{B}}_{{5}}}\end{align}\)

(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,

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

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school