Question
Construct an isosceles triangle in which: Base $DE = 6 \ cm$ and $\angle F = 45^\circ $
✓
Answer
In isosceles $\triangle DEF,$
Base $DE = 6\ cm$
$\angle F = 45^\circ ....($given$)$
$\Rightarrow \angle D = \angle E ....(\text{DEF}$ is isosceles triangle$)$
Now, $\angle D + \angle E + \angle F = 180^\circ $
$2\angle D + 45 = 180^\circ $
$2\angle D = 135^\circ $
$\Rightarrow \angle D = \angle E = 67.5^\circ $
Steps:
$1.$ Draw $DE = 6\ cm.$
$2.$ Construct $\angle DEP = 67.5^\circ $ and $\angle EDQ = 67.5^\circ $
$3$. Ray $EP$ and ray $DQ$ meets at $F.$
Thus, $\text{DEF}$ is the required triangle.
Base $DE = 6\ cm$
$\angle F = 45^\circ ....($given$)$
$\Rightarrow \angle D = \angle E ....(\text{DEF}$ is isosceles triangle$)$
Now, $\angle D + \angle E + \angle F = 180^\circ $
$2\angle D + 45 = 180^\circ $
$2\angle D = 135^\circ $
$\Rightarrow \angle D = \angle E = 67.5^\circ $
Steps:
$1.$ Draw $DE = 6\ cm.$
$2.$ Construct $\angle DEP = 67.5^\circ $ and $\angle EDQ = 67.5^\circ $
$3$. Ray $EP$ and ray $DQ$ meets at $F.$
Thus, $\text{DEF}$ is the required triangle.
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