Question

Construct an isosceles $△ABC$ when base $BC=6.8cm$ and $\angle A={90}^{\circ }$. The length of the other two sides is $a$. Then $10a=$ ? (approx.)

Answer 1

We know that in an isosceles triangle, two sides at equal.
So, when $AB=AC$, opposite angles will also be equal
$\Rightarrow \angle B=\angle C$
Since, $\angle A={90}^{\circ }$
$\Rightarrow \angle B=\angle C={45}^{\circ }$
Hence, follow the steps given below to construct the triangle.
1. Draw a line segment $BC=6.8cm$.
2. Both at $B$ and $C$ draw angles of ${45}^{\circ }$ towards each other.
3. Mark the point of intersection of the arms of the angles as $A$.
4. $ABC$ is the required triangle.

Measuring the length of the congruent sides, we get $a=4.8cm$

$\therefore 10a=48$