Question

Construct triangle $ABC$ having given $BC=7cm,AB-AC=1cm$ and $\angle ABC={45}^o$.

Answer 1

In $△ABC$, given $BC=7$ cm, $\angle B={45}^{\circ }$. and difference of two sides $AB$ and $AC$ is equal to $1$ cm.

Case $I:AB>AC$

Step $1:$ Draw the base $BC=7$ cm.

Step $2:$ Make $\angle XBC={45}^{\circ }$.

Step $3:$ Mark a point $D$ on ray $BX$ such that $BD=1$ cm.

Step $4:$ Join $DC$.

Step $5:$ Draw perpendicular bisector of $DC$ such that, it intersects ray $BX$ at a point $A$.

Step $6:$ Join $AC$.

Thus, $ABC$ is the required triangle.

Case $II:AB<AC$

Step $1:$ Draw the base $BC=7$ cm.

Step $2:$ Make $\angle XBC={45}^{\circ }$. and extend ray $BX$ in the opposite direction.

Step $3:$ Mark a point $D$ on the extended ray $BX$ such that $BD=1$cm.

Step $4:$ Join $DC$.

Step $5:$ Draw a perpendicular bisector of DC such that, it intersects ray $BX$ at point $A$.

Step $6:$ Join $AC$.

Thus, $ABC$ is the required triangle.