# Ex.11.1 Q5 Constructions Solution - NCERT Maths Class 10

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## Question

Draw a triangle $$ABC$$ with side $${{BC = 6 \,\rm{cm,}\;AB = 5\,\rm{cm}}}$$ and $$\angle {{ABC = 6}}{{{0}}^0}.$$  Then construct a triangle whose sides are \begin{align}\frac{3}{2}\end{align} of the corresponding sides of the triangle $$ABC.$$

Video Solution
Constructions
Ex 11.1 | Question 5

## Text Solution

#### What is known?

$$2$$ sides and the angle between them and the ratio of corresponding sides of $$2$$ triangles.

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 $${{BC = 6 \,\rm{cm}}}.$$ At $$B,$$ make $$\angle {{CBY = }}\,{{60^\circ }}$$ and cut an arc at $$A$$ so that $$BA = 5 \,\rm{cm}.$$ Join $$\,{{AC}}\,,\,\,{{\Delta ABC}}$$ is obtained.

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

(iii) Mark $$4$$ ($${4} > 3$$ in $$\frac{3}{4}$$) 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 $${{{B}}_{{4}}}$$ to $$C$$ and draw $${{{B}}_{{3}}}{{C'}}$$ parallel to $${{{B}}_{{4}}}{{C}}$$ to intersect $$BC$$ at $${{C'}}{{.}}$$

(v) Draw $${{C'A'}}$$ parallel to $$CA$$ to intersect $$BA$$ 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{{{3}}}{{{4}}}\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{{{{C'C}}}}{{{{BC'}}}}= \frac{{{1}}}{{{3}}}\\\frac{{{{C'C}}}}{{{{BC'}}}}{{ + 1}}& = \frac{{{1}}}{{{3}}}{{ + 1 \quad \text{(Adding 1)}}}\\\frac{{{{C'C + BC'}}}}{{{{BC'}}}}& = \frac{{{4}}}{{{3}}}\\\frac{{{{BC}}}}{{{{BC'}}}}& =\frac{{{4}}}{{{3}}}\,\, {\rm{(or)}} \,\frac{{{{BC'}}}}{{{{BC}}}} =\frac{{{3}}}{{{4}}}\end{align}

Consider $${{\Delta BA'C'}}\,\,{{and}}\,\,{{\Delta BAC}}$$

\begin{align} & \angle \text{A }\!\!'\!\!\text{ BC }\!\!'\!\!\text{ }=\angle \text{ABC}={{60}^{\circ }} \\ & \angle \text{BCA }\!\!'\!\!\text{ }=\angle \text{BCA}\,\,\\&\left( \text{Corresponding}\,\text{angles}\because \,\text{CA}||\text{CA} \right) \\ & \angle \text{BA }\!\!'\!\!\text{ C }\!\!'\!\!\text{ }=\angle \text{BAC}\\&\left( \text{Corresponding}\,\text{angles} \right) \\ \end{align}

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

Therefore corresponding sides are proportional,

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

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