Ex.11.1 Q4 Constructions Solution - NCERT Maths Class 10

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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.


Text Solution


What is known?

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

What is unknown?



  • 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 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

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


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


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

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