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

Open in App

Solution

The rough sketch of the required ΔDEF is as follows.

The steps of construction are as follows.

(i)Draw a line segment DE of length 5 cm.

(ii) At point D, draw a ray DX making an angle of 90° with DE.

(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.

(iv) Join F to E. ΔDEF is the required triangle.

Suggest Corrections

Similar questions

△ D E F

D E = 5 c m, D F = 3 c m

m ∠ E D F = 90^{o}

Δ

∠

90^{\circ}

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}

∆

∠

°

Join BYJU'S Learning Program