Ex.11.1 Q3 Constructions Solution - NCERT Maths Class 10

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Construct a triangle with sides \(5\, \rm{cm},\) \(6\, \rm{cm}\) and \(7 \,\rm{cm}\) and then another triangle whose sides are \(\begin{align}\frac{7}{5}\end{align}\) of the corresponding sides of the first triangle.


 Video Solution
Ex 11.1 | Question 3

Text Solution

What is known?

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

What is unknown?



  • Draw the line segment of largest length \(7 \,\rm{cm}\). Measure \(5 \,\rm{cm}\) and \(6 \,\rm{cm}\) 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 (\(7 \,\rm{cm}\)).
  • 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 of construction:

(i) Draw \({{BC = 7}}\,{\rm{cm}}\) with \(B\) and \(C\) as centres and radii \(5\, \rm{cm}\) and \(6 \,\rm{cm}\) respectively. Draw arcs to intersect at \(A.\) \(\,\Delta {{ABC}}\) is obtained.

(ii) Draw ray \(BX\) making \(\,\angle {{CBX}}\) acute.

(iii) Mark \(7\) points (greater of \(7\) and \(5\) in \(\begin{align}\frac{7}{5}\end{align}\) ) \({{{B}}_{{I}}}{{,}}\,{{{B}}_{{2}}}{{,}}\,{{ \ldots \ldots \ldots }}{{{B}}_{{7}}}\) on \(BX\) such that \(\,{{B}}{{{B}}_{{I}}}{{ = }}{{{B}}_{{I}}}{{{B}}_{{2}}}{{ = \ldots \ldots \ldots \ldots = }}{{{B}}_{{6}}}{{{B}}_{{7}}}\)

(iv) Join \({{{B}}_{{5}}}\) (smaller of \(7\) and \(5\) in \(\begin{align}\frac{7}{5}\end{align}\) and so the \(5^\rm{th}\) point) to \(C\) and draw \({{{B}}_{{7}}}{{C'}}\) parallel to \({{{B}}_{{5}}}{{C}}\) intersecting the extension of \(BC\) at  \(C'\).

(v) Through \({{C'}}\) draw \({{C'A'}}\) parallel to \(CA\) to meet the extension of \(BA\) at \(A’.\)

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

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{7}}}{{{5}}}\end{align}\]


In \(\Delta {B}{B}_{7}{C',}\;{B}_{3}{C}\) is parallel to \({B}_{7}{C'}\)

Hence by Basic proportionality theorem,

\[\begin{align}\frac{{{{{B}}_{{6}}}{{{B}}_{{7}}}}}{{{{B}}{{{B}}_{{5}}}}}&{{ = }}\frac{{{{CC'}}}}{{{{BC}}}}\\&{{ = }}\frac{{{2}}}{{{5}}}\end{align}\]

Adding \(1\),

\[\begin{align}\frac{{{{CC'}}}}{{{{BC}}}}{{ + 1}}& = \frac{{{2}}}{{{5}}}{{ + 1}}\\\frac{{{{BC + CC'}}}}{{{{BC}}}}{{ }}&=\frac{{{7}}}{{{5}}}\\\frac{{{{BC'}}}}{{{{BC}}}}{{ }}&= \frac{{{7}}}{{{5}}}\end{align}\]

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

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

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

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

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

\(∴ \)  Corresponding sides are proportional


\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{7}}}{{{5}}}\end{align}\]

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