Construct $△ D E F$ with $E F = 4.5 c m, D E = 5 c m$ and $D F = 3 c m$

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Solution

Given that
$E F = 4.5 c m$
$D E = 5 c m$
And, $D F = 5 c m$
step of construction
(i) Draw a line segment $D E = 5 c m$
(ii) With $E$ as centre and radius $4.5 c m$ draw an arc.
(iii) With $D$ as centre and radius $4 c m$ draw an arc cutting the previous arc at $F$
(iv) join $E F$ and $D F$ then $Δ D E F$ is the required triangle

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