Question

Construct each of the following and give justification:

A triangle if its perimeter is $10.4\mathrm{cm}$ and two angles are $45°$ and $120°$.

Answer 1

Step 1: Constructing triangle $ABC$:

Draw $\mathrm{XY}$ $=10.4\mathrm{cm}$.

Draw $\angle \mathrm{LXY}=45°$ at $\mathrm{X}$ and $\angle \mathrm{MYX}=120°$ at $\mathrm{Y}$.

Bisect $\angle \mathrm{LXY}$ and $\angle \mathrm{MYX}$ and intersect them at $\mathrm{A}$.

Draw perpendicular bisector of $\mathrm{AX}$ such that it meets $\mathrm{XY}$ at $\mathrm{B}$ and perpendicular bisector of $\mathrm{AY}$ such that it meets $\mathrm{XY}$ at $\mathrm{C}$. Join $\mathrm{AB}$ and $\mathrm{AC}$.

Thus, $\mathrm{ABC}$ is the required triangle.

Step 2: Justification for the triangle constructed:

We know that, $\mathrm{B}$ and $C$ lies on the perpendicular bisector of $\mathrm{AX}$ and $AY$ respectively.

$\mathrm{AB}+\mathrm{BC}+\mathrm{CA}=\mathrm{XB}+\mathrm{BC}+\mathrm{CY}=\mathrm{XY}$ (Distance from the perpendicular bisector is equal)

In triangle $\mathrm{AXB},\mathrm{AB}=\mathrm{XB}$

So, $\angle \mathrm{BAX}=\angle \mathrm{AXB}..................\left(1\right)$

As $\angle \mathrm{ABC}$ is the exterior angle of triangle $\mathrm{AXB}$.

$\angle \mathrm{ABC}$ $=\angle \mathrm{BAX}+\angle \mathrm{AXB}$

Using equation $\left(1\right)$

$\angle \mathrm{ABC}$ $=\angle \mathrm{AXB}+\angle \mathrm{AXB}$

$\angle \mathrm{ABC}=2\angle \mathrm{AXB}$

$\angle \mathrm{ABC}=\angle \mathrm{LXY}$ $(\mathrm{AX}$ is the bisector of $\angle \mathrm{LXB})$

In triangle $\mathrm{AYC}$, $\mathrm{AC}=\mathrm{CY}$

$\angle \mathrm{CAY}=\angle \mathrm{AYC}$

As $\angle \mathrm{ACB}$ is an exterior angle of triangle $\mathrm{AYC}$

$\angle \mathrm{ACB}$ $=\angle \mathrm{CAY}+\angle \mathrm{AYC}$

$\angle \mathrm{ACB}$$=\angle \mathrm{CAY}+\angle \mathrm{CAY}$

$\angle \mathrm{ACB}$ $=2\angle \mathrm{CAY}$

$\angle \mathrm{ACB}$ $=\angle \mathrm{MYX}$ $(\mathrm{AY}$ is the bisector of $\angle \mathrm{MYX})$

Thus, the construction is justified.