Construct $△ D E F$ such that $D E = 5 c m, D F = 3 c m$ and $m ∠ E D F = 90^{o}$ .

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Solution

Draw a line segment $D E$ of length $5 c m .$
Make a perpendicular at point $D$ and extend the line $D X .$
Make an arc at a distance $3 c m$ from point $D$ intersecting $D X$ at $F .$
Join $F$ to $E .$

$△ D E F$ is the required triangle.

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Similar questions

Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.

Question 1
Construct $Δ$ DEF such that DE = 5 cm, DF = 3 cm and $∠$ EDF = $90^{\circ}$

△ D E F

D E = 5

D F = 3

∠ E D F = 90^{\circ}

△ D E F

E F = 4.5 c m, D E = 5 c m

D F = 3 c m

Δ

∠

90^{\circ}

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