Ex.11.1 Q4 Constructions Solution - NCERT Maths Class 10

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Question

Construct an isosceles triangle whose base is \(8 \;\rm{cm}\) and altitude \(4 \;\rm{cm}\) and then another triangle whose sides are \(\begin{align}1\frac{1}{2} \end{align}\) times the corresponding sides of the isosceles triangle.

 

 Video Solution
Constructions
Ex 11.1 | Question 4

Text Solution

What is known?

Base and altitude of an isosceles triangle and the ratio of corresponding sides of \(2\) triangles.

What is unknown?

Construction.

Reasoning:

  • Draw the line segment of base \(8 \;\rm{cm}\). Draw perpendicular bisector of the line. Mark a point on the bisector which measures \(4 \;\rm{cm}\) from the base. Connect this point from both the ends.
  • Then draw another line which makes an acute angle with the given 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 construction:

(i) Draw \(\,{{BC = 8}}\,{\rm{cm}}{{.}}\) Through \(D,\) the mid-point of \(BC,\) draw the perpendicular to \(BC\) and cut an arc from \(D\) on it such that \({{DA = 4}}\,{{cm}}{{.}}\) Join \(BA\) and \(CA.\) \({{\Delta ABC}}\) is obtained.

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

(iii) Mark \(3\) \(\begin{align}\left( {3{{ }} > {{ }}\;2\text{ in}\,\,1\frac{1}{2} = \,\frac{3}{2}} \right)\end{align}\)  points \({{{B}}_{{I}}}{{,}}\,{{{B}}_{{2}}}{{,}}\,{{{B}}_{{3}}}\) on BX such that \({{B}}{{{B}}_{{1}}}{{ = }}{{{B}}_{{1}}}{{{B}}_{{2}}}{{ = }}{{{B}}_{{2}}}{{{B}}_{{3}}}\)

(iv) Join \({B}_{2}\) (\( \;2^\rm{nd}\)  point \(∵ \;2<3\)) to C and draw \({B}_{3} {C}^{\prime}\) parallel to \({B}_{2} {C}\) intersect \(BC\) extended at \(C’\).

(v)  Through \(C’\) draw \({{C'A'}}\) parallel to \(CA\) to intersect \(BA\) extended to \(A’.\)

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

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

Proof:

In \({{\Delta B}}{{{B}}_{{3}}}{{C',}}\,\,{{{B}}_{{2}}}{{C}}\,\,{{||}}\,\,{{{B}}_{{3}}}{{C'}}\) ,

Hence by Basic proportionality theorem,

\[\begin{align}\frac{{{{{B}}_{{2}}}{{{B}}_{{3}}}}}{{{{B}}{{{B}}_{{2}}}}}{{ = }}\frac{{{{CC'}}}}{{{{BC}}}}{{ = }}\frac{{{1}}}{{{2}}}\end{align}\]

Adding \(1\),

\[\begin{align}\frac{{{{CC'}}}}{{{{BC}}}}{{ + 1}}&= \frac{{{1}}}{{{2}}}{{ + 1}}\\\frac{{{{BC + CC'}}}}{{{{BC}}}}{{ }}&=\frac{{{3}}}{{{2}}}\\\frac{{{{BC'}}}}{{{{BC}}}}{{}}&= \frac{{{3}}}{{{2}}}\end{align}\]

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

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

\(\angle {BCA}=\angle {BC' } { A '}\) (Corresponding angles \(\therefore \mathrm{CA} \| \mathrm{C}^{\prime} \mathrm{A}^{\prime}\)

\(\angle {{BAC}}\, = \,\angle {{BA'C'}} \) (Corresponding angles)

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

\(∴\) Corresponding sides are proportional

Hence,

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

  
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