Question

Construct a triangle ABC having its perimeter 12 cm and base angles ${50}^{\circ }$ and ${70}^{\circ }$

Answer 1

Perimeter of $△$ABC = 12 cm

AB + BC + CA = 12 cm

$\angle B={50}^{\circ }\text{and}\angle C={70}^{\circ }.$

We are required to construct $△$ ABC.

Steps of construction:

Step 1. Draw a ray PX and cut off a line segment PQ=12 cm from it.

Step 2. At P, construct $\angle YPB={25}^{\circ }$ with the help of protractor i. e $[\frac{1}{2}\times {50}^{\circ }]$

Step 3. At Q, construct $\angle ZQP={35}^{\circ }i.e.[\frac{1}{2}\times {70}^{\circ }]$

Step 4. Draw perpendicular bisectors of AP intersecting PQ at B.

Step 5. Draw perpendicular bisectors of AQ intersecting PQ at C.

Step 6. Join AB and AC.

Then, $△$ABC is the required triangle.