Question

Write the steps to construct $△ABC$, in which $BC=5.2cm$, $\angle ACB={45}^{\circ }$ and perimeter of $△ABC$ is $10cm$.

Answer 1

$\begin{array}{c}BC=5.2cmP=10cm\\ AB+BC+CA+10\\ \Rightarrow AB+CA=4.8and\angle C={45}^{\circ }\end{array}$

$\to$ Draw a line segment BC = 5.2 cm

$\to$ Draw $\angle C={45}^{\circ }$

$\to$ with centre C and radius 4.8 cm, makes an AX which intersect CX at D

$\to$ join A to C

$\to$ Draw a perpendicular bisector at segment DB it intersect the

line segment CD at point A.

$\to$ Join C to A.