Question

From the given figure, find the sum of x, y and z.


(4 marks)

Answer 1

From the figure, we can say that BE is parallel to DC.

⇒ $\angle EBC+\angle z={180}^{\circ }$ (Co-interior angles)

⇒ ${70}^{\circ }+\angle z={180}^{\circ }$

⇒ $\angle z=(180-70{)}^{\circ }={110}^{\circ }$
(1 mark)

Also, $\angle ABE=\angle z={110}^{\circ }$ (corresponding angles)

$\angle x=\angle ABD-\angle ABE$

$={150}^{\circ }-{110}^{\circ }={40}^{\circ }$
(1 mark)

Since AE || BD, y = x as they are alternate angles.

$\angle y={40}^{\circ }$
(1 mark)

$\therefore$ The required sum $=x+y+z={40}^{\circ }+{40}^{\circ }+{110}^{\circ }={190}^{\circ }$
(1 mark)