Question

Construct ∆LMN, right angled at M,given that LN = 5cm and MN = 3cm

Answer 1

Here are steps of construction :
First of all draw a line segment MN = 3 cm
Using a compass draw an angle of 90° at M i.e. <NMQ =90°
(Using Pythagoras theorem we get LM = 4 cm )
Mark an arc on Ray MQ taking the radius of compass = 4 cm, name the point of intersection as L
Join L to N
∆LMN is the required triangle