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} $.

AcademicMathematicsNCERTClass 9


$BC=8\ cm, \angle B=45^o$ and $AB-AC=3.5\ cm$.

To do:

We have to construct a $\triangle ABC$.


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.

Updated on 10-Oct-2022 13:42:03