Construct a $∆ P Q R$ in which $P Q = 3.8 c m$ $, Q R = 4.2 c m$ and $P R = 3.5 c m$ .

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Solution

Step 1. Draw a line segment $P Q = 3.8 c m$

Step 2.Taking the radius of $4.2 c m$ and $Q$ as centre cut an arc.
Step 3.Taking the radius of $3.5 c m$ and $P$ as centre cut an arc which intersects the previous arc at point say point $R$ .
Step 4.Now join $P R$ and $Q R$ .
Hence $∆ P Q R$ is the required triangle.

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