Construct a triangle $ \mathrm{ABC} $ in which $ \mathrm{BC}=8 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ} $ and $ \mathrm{AB}-\mathrm{AC}=3.5 \mathrm{~cm} $.
Given:
$BC=8\ cm, \angle B=45^o$ and $AB-AC=3.5\ cm$.
To do:
We have to construct a $\triangle ABC$.
Solution:
Steps of construction:
(i) Let us draw a line segment $BC$ of length $8\ cm$.
(ii) Now, construct an angle $XBC$ such that $\angle XBC=45^o$
(iii) Now, by taking a measure of $AB-AC=3.5\ cm$ with the compasses, let us draw an arc from point $B$ and mark it as Point $D$ on the ray $BX$.
(iv) Now, let us join $DC$. Then by taking compasses let us draw a perpendicular bisector of the line $DC$ and mark the intersection point of the bisector with the ray $BX$ as point $A$.
(v) Now, let us join $AC$. Therefore, $ABC$ is the required triangle.
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