Ex.11.1 Q6 Constructions Solution - NCERT Maths Class 10

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Question

Draw a triangle ABC with side \(\begin{align}{{BC }} = {{ }}7\,{\rm{ cm,}}\,\,\angle {{B }} = {{ }}45^\circ ,\,\,\angle {{A }} = {{ }}105^\circ .\end{align}\) Then, construct a triangle whose sides are \(\begin{align}\frac{4}{3}\end{align}\) times the corresponding sides of \(\begin{align}\Delta{ ABC}\end{align}\).

 

Text Solution

 

What is known?

One side and \(2\) angles of a triangle 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 construction:

(i) Draw \({{BC = 7}}\,{\rm{cm}}{{.}}\) At \(B,\) make an angle \(\angle {{CBY}} = {45^ \circ }\) and at \(C,\) make \(\angle {{BCZ}} = 30^\circ \,\,\left[ {{{180}^ \circ } - \left( {{{45}^ \circ } + {{105}^ \circ }} \right)} \right]\) . Both \(BY\) and \(CZ\) intersect at \(A\) and thus \({{\Delta ABC}}\) is constructed.

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

(iii) Mark \(4\) \(\begin{align}\left( {4{{ }} > {{ }}3\,{\rm{ in}}\,\,\frac{4}{3}} \right)\end{align}\) points \({{{B}}_{{1}}}{{,}}{{{B}}_{{2}}}{{,}}{{{B}}_{{3}}}{{,}}{{{B}}_{{4}}}\) on BX such that \({{B}}{{{B}}_{{1}}}{{ = }}{{{B}}_{{1}}}{{{B}}_{{2}}}{{ = }}{{{B}}_{{2}}}{{{B}}_{{3}}}{{ = }}{{{B}}_{{3}}}{{{B}}_{{4}}}\)

(iv) Join \(B_3\) (third point on \(BX,\;3\, < 4\) in \(\begin{align}\frac{4}{3}\end{align}\)) to \(C\) and draw \({{{B}}_{{4}}}{{C'}}\) parallel to \(BC\) such that \(C’\) lies on the extension of \(BC.\)

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

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

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{4}}}{{{3}}}\end{align}\]

Proof:

In \({{\Delta B}}{{{B}}_{{4}}}{{C',}}\,\,{{{B}}_{{3}}}{{C}}\,\,{{||}}\,\,{{{B}}_{{4}}}{{C'}}\)

Hence by Basic proportionality theorem,

\[\begin{align}\frac{{{{{B}}_{{3}}}{{{B}}_{{4}}}}}{{{{B}}{{{B}}_{{3}}}}}&= \frac{{{{CC'}}}}{{{{BC}}}}{{ = }}\frac{{{1}}}{{{3}}}\\\frac{{{{CC'}}}}{{{{BC}}}}{{ + 1 }}&=\frac{{{1}}}{{{3}}}{{ + 1}}\,\qquad{\rm{(Adding}}\,{{1)}}\\\frac{{{{BC + CC'}}}}{{{{BC}}}}& = \frac{{{4}}}{{{3}}}\\\frac{{{{BC'}}}}{{{{BC}}}}& = \frac{{{4}}}{{{3}}}\end{align}\]

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

\(\angle {{A'BC'}}\, = \angle {{ABC}} = 45^\circ \)

\(\angle {{BC'A'}} = \angle {{BCA}} = 30^\circ \) (Correspomding angles as \( \,\,{{CA}}\,{{||}}\,\,{{C'A'}}\) )

\(\angle {{BA'C'}} = \angle {{BAC}} = 105^\circ \) (Corresponding angles )

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

Hence corresponding sides are proportional

\[\begin{align}\frac{{{{BA'}}}}{{{{BA}}}}{{ = }}\frac{{{{BC'}}}}{{{{BC}}}}{{ = }}\frac{{{{C'A'}}}}{{{{CA}}}}{{ = }}\frac{{{4}}}{{{3}}}\end{align}\]

  
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