Question

In the above figure, ABCD is a square and BCE is an equilateral triangle. Find the measure of $\angle$DEC.
(4 marks)

Answer 1

Given ABCD is a square and BCE is an equilateral triangle.

$\angle BCD={90}^{\circ }$ [Interior angle of a square]

$\angle BCE={60}^{\circ }$ [Interior angle of an equilateral triangle]

$\therefore \angle DCE={90}^{\circ }+{60}^{\circ }={150}^{\circ }$
(1 mark)

BC = CD [Sides of a square] ------ (1)

BC = CE [Sides of an equilateral triangle] ------ (2).

From (1) and (2),
DC = EC
(1 mark)

In $\Delta DCE$, DC = CE

$\Rightarrow \angle CDE=\angle CED$ --------- (3)
[Angles opposite equal sides]

Also, In $\Delta DCE$
$\angle DCE+\angle CDE+\angle CED={180}^{\circ }$ --------- (4) [Angle sum property]

From (3) and (4),
$2\angle DEC+\angle CED={180}^{\circ }$
(1 mark)

$\Rightarrow 2\angle DEC={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\angle DEC=\frac{1}{2}\left({30}^{\circ }\right)={15}^{\circ }$
(1 mark)