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# Construct an isosceles triangle in which the lengths of each of its equal sides is $6.5\ cm$ and the angle between them is $110^{\circ}$.

**To do:**To construct an isosceles triangle in which the lengths of each of its equal sides are $6.5\ cm$ and the angle between them is $110^{\circ}$.

**Steps of construction:**

- Let us draw a line segment $QR=6.5\ cm$.
- At the point $Q$, let us draw a ray $QX$ making an angle of $110^{\circ}$ with $QR$ such that $\angle XQR=110^{\circ}$.
- Assuming $Q$ as the center, let us draw an arc of radius $6.5\ cm$ that cuts $QX$ at point $P$.
- Now let us join $PR$ to get the required triangle.

Thus, $\triangle PQR$ is the required triangle.

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