Question

Construct a triangle $\mathrm{ABC}$ in which $\mathrm{BC}=7\mathrm{cm}$, $\angle \mathrm{B}=75°$and $\mathrm{AB}+\mathrm{AC}=13\mathrm{cm}$.

Answer 1

To construct a $△$$\mathrm{ABC}$ where $\mathrm{BC}=7\mathrm{cm}$, $\angle \mathrm{B}=75°$ and $\mathrm{AB}+\mathrm{AC}=13\mathrm{cm}$,we will the following steps:

Steps to construct a triangle:

Draw a line segment of base $\mathrm{BC}=7\mathrm{cm}$ as a base.

Measure and draw $\angle \mathrm{B}=75°$ and extend the ray $\mathrm{BX}$.

Now, by taking radius $13cm$ draw an arc from $B$ on the line of angle just projected.

Now mark it as '$D$', & join $C$ to $D$.

Now draw the perpendicular bisector of the line $\mathrm{DC}$ and the intersection point at $\mathrm{BD}$ is taken as $\mathrm{A}$. Join $\mathrm{AC}$.

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

Hence, the required $∆ABC$ will $\mathrm{BC}=7cm$,$\angle B={45}^{°}$and & $\mathrm{AB}+\mathrm{AC}=13\mathrm{cm}$ is constructed.