The sides of two regular octagons are 3 in. And 6 in. respectively. Find the ratio of their perimeters and their areas

Mathematics

1 answer:

photoshop1234 [79]2 years ago

6 0

The ratio of their perimeters is $\frac{1}{2}$ and the ratio of their areas is $\frac{1}{4}$

Step-by-step explanation:

A regular n-side polygon has

n equal sides
n equal angles
The measure of each interior angle = $\frac{(n-2).180}{n}$
Its perimeter = n × a, where a is the length of its side

V.I.Note: all regular polygons have same number of sides are similar, because their interior angles are equal in measures and their sides are proportion

∵ There are to regular octagons

- Regular octagon has 8 equal sides and 8 equal angles

∴ The two octagon are similar

∵ Their sides are 3 inches and 6 inches

∴ The constant ratio between their sides = $\frac{3}{6}$

- Divide the two terms of the ratio by 3 to simplify it

∴ The constant ratio between their sides = $\frac{1}{2}$

In similar figures the ratio between their perimeters is equal to the constant ratio between their sides, and the ratio between their areas is equal to square the constant ratio between their sides

∵ The two octagons are similar

∵ The constant ratio between their sides = $\frac{1}{2}$

∵ The ratio between their perimeters = the constant ratio between

their sides

∴ The ratio between their perimeters = $\frac{1}{2}$

∵ The ratio between their areas = (constant ratio)²

∴ The ratio between their areas = $(\frac{1}{2})^{2}=\frac{1}{4}$