Question

Find the sum of the number of angle bisector(s) and perpendicular bisector(s) drawn to construct a triangle ABC in which $BC=8cm,\angle B={45}^{\circ }andAB-AC=3.5cm$.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2



Steps to construct the required triangle are as follows :

Step 1: Draw a line segment BC = 8 cm and at point B, make an angle of ${45}^{\circ }$[ by bisecting ${90}^{\circ }$ , say $\angle XBC={45}^{\circ }$.

Step 2: Mark the line segment BD= 3.5 cm ( equal to AB – AC) on the ray BX.

Step 3: Join DC and draw the perpendicular bisector PQ of DC, let it intersect BX at point A.
Step 4: Join AC , then $\Delta ABC$ is the required triangle.

$\therefore$ 1 angle bisector and 1 perpendicular bisector are drawn
$\Rightarrow$Sum = 2.