Question

Construct a triangle $\mathrm{ABC}$ in which $\mathrm{BC}=8\mathrm{cm}$, $\angle \mathrm{B}=45°$ and $\mathrm{AB}-\mathrm{AC}=3.5\mathrm{cm}$.

Answer 1

Step 1: Draw a line segment of base $\mathrm{BC}=8\mathrm{cm}$.

Measure and draw $\angle \mathrm{B}=45°$ and extend the ray $\mathrm{BX}$.

Step 2: Open a compass and measure $\mathrm{AB}-\mathrm{AC}=3.5\mathrm{cm}$. Since $\mathrm{AB}-\mathrm{AC}=3.5\mathrm{cm}$ is positive,

$\therefore$$\mathrm{BD}$ will be above the line $\mathrm{BC}$.

Considering $\mathrm{B}$ as a centre, draw an arc on the ray $\mathrm{BX}$. Let the arc intersect $\mathrm{BX}$ at point $\mathrm{D}$ such that $BX=3.5$. Join $\mathrm{DC}$.

Step 3: Now draw the perpendicular bisector of the line $\mathrm{DC}$ and the intersection point at the ray $\mathrm{BX}$ is assumed as $\mathrm{A}$. Join $\mathrm{AC}$.

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