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# Construct a triangle $ \mathrm{ABC} $ in which $ \mathrm{BC}=7 \mathrm{~cm}, \angle \mathrm{B}=75^{\circ} $ and $ \mathrm{AB}+\mathrm{AC}=13 \mathrm{~cm} $.

**Given:**

$BC=7\ cm, \angle B=75^o$ and $AB+AC=13\ cm$.

**To do:**

We have to construct a $\triangle ABC$.

**Solution:**

Steps of construction:

(i) Let us draw a line segment $BC$ of length $7\ cm$.

(ii) Now, construct an angle $CBX=75^o$ from point $B$.

(iii) Now, by taking a measure of $AB+AC=13\ cm$ with the compasses, let us draw an arc from point $B$ on the ray $BX$ and mark the intersection of the arc with ray $BX$ as point $D$.

(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 ray $BX$ as $A$

(v) Now, let us join $AC$. Therefore, $ABC$ is the required triangle.

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