Question

Construct $\Delta$ DEF such that DE = 5 cm, DF = 3 cm and $\angle$ EDF = ${90}^{\circ }$. $\left[3\text{Marks}\right]$

Answer 1

To Construct: $\Delta$ DEF where DE = 5 cm, DF = 3 cm and $\angle$ EDF = ${90}^{\circ }$
Steps of construction:

$\left[1\text{Mark}\right]$

(a) Draw a line segment DF = 3 cm.
(b) At point D, draw an angle of 90° with the help of compass i.e., $\angle$XDF = ${90}^{\circ }$ $\left[1\text{Mark}\right]$
(c) Taking D as center, draw an arc of radius 5 cm, which cuts DX at the point E.
(d) Join EF.
It is the required right angled triangle DEF. $\left[1\text{Mark}\right]$