Question

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

Answer 1

$1.$Draw base $BC$ of length $8$cm

$2.$Now, let's draw $\angle B={45}^{\circ }$.Let the ray be $BX$

$3.$Open the compass to length $AB-AC=3.5$cm.

Since $AB-AC=3.5$cm is positive.

So, $BD$ will be above line $BC$

From point $B$ as center, cut an arc on ray $BX$.

Let the arc intersect $BX$ at $D$

$4.$Join $CD$

$5.$Now, we will draw perpendicular bisector of $CD$

$6.$Mark point $A$ where perpendicular bisector intersects $BD$

$7.$Join $AC$

Result:$\therefore △ABC$ is the required triangle.