Question

The 5 interior angles of a Pentagon are x;x+30;2x+40;3x+20&2x. The sum of the interior angles is 540. Calculate the size of each of the angles.

Answer 1

Step-by-step explanation:

Sum of all the angles of a pentagon = 540°

$x+(x+30) + (2x+40)+ (3x+20) + 2x=540^{o}$

$x+x+30+2x+40+3x+20+2x=540^{o}$

$x+x+2x+3x+2x+30+40+20=540^{o}$

$9x+90=540^{o}$

$9x=(540 - 90)^{o}$

$9x = 450^{o}$

$x=\frac{450^{o}}{9}$

∴ $x=50^o$

∴ the five interior angles =

x = 50°

(x + 30) = 80°

(2x + 40) = 140°

(3x + 20) = 170°

2x = 100°