The formula for the measure of an interior angle of a regular n-sided polygon is:
$ \text{Angle} = \frac{(n-2) \times 180^\circ}{n} $For a regular pentagon, n = 5.
Substituting n = 5 into the formula:
$ \text{Angle}_{\text{pentagon}} = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ $For a regular octagon, n = 8.
Substituting n = 8 into the formula:
$ \text{Angle}_{\text{octagon}} = \frac{(8-2) \times 180^\circ}{8} = \frac{6 \times 180^\circ}{8} = \frac{1080^\circ}{8} = 135^\circ $The question asks for the ratio of the measure of an angle of a regular pentagon to that of a regular octagon.
Ratio = $ \frac{\text{Angle}_{\text{pentagon}}}{\text{Angle}_{\text{octagon}}} = \frac{108^\circ}{135^\circ} $
To simplify the ratio, find the greatest common divisor (GCD) of 108 and 135, which is 27.
Divide both numbers by the GCD:
$ \frac{108 \div 27}{135 \div 27} = \frac{4}{5} $The ratio is 4:5.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.