Question

Find the similarity ratio and the ratio of perimeters for two regular octagons with areas of 18in2 and 50in2
1. 3 : 5; 9 : 25
2. 9 : 25; 9 : 25
3. 3 : 5; 3: 5
4. 9 : 25; 3 : 5

Answer 1

Similarity ratio will be ratio of sides of two octagon

Now area are in square inches and sides are inches, so side ratio will be given by square roots of area

So first simplify 18: 50. For that we divide ratio by greatest common factor they have. Greatest common factor in 18 and 50 is 2. So divide 18 and 50 by 2

$\frac{18}{2} : \frac{50}{2}$

9 : 25

Now square root it

$\sqrt{9} : \sqrt{25}$

3 : 5

so similarity ratio will be 3 : 5

Similarly units for perimeter is in inches and that of area is in $in^{2}$

so ratio of perimeter will be given by square roots of that of area

so ratio of perimeter will be

$\sqrt{9} : \sqrt{25}$

3 : 5

So ratio of perimeter will be 3 : 5

So we got similarity ratio as 3 : 5 and perimeter ratio as 3 : 5

so choice (3) is the right answer here

Answer 2

Answer:

3. 3 : 5; 3: 5

Step-by-step explanation:

First, let's find the leght of the sides of each octagon.

$A= 18in^{2}$

The area of an octagon is defined by

$A=2(1+\sqrt{2})l^{2}$

Replacing the area

$18=2(1+\sqrt{2})l^{2}\\\frac{18}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{18}{3.4} } \approx 2.3 \ in$

Therefore, the side of the first octagon is 1.6 inches long.

Its perimeter is: $P=8(2.3in)=18.4in$

$A=50 in^{2}$

$50=2(1+\sqrt{2})l^{2}\\\frac{50}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{50}{3.4} } \approx 3.8 \ in$

Therefore, the side of the second octagon is 3.8 inches long.

Its perimeter is $P=8(3.8in)=30.4in$.

Now, let's divide to find each ratio:

$\frac{3.8}{2.3} \approx 1.65$ (the ratio between sides).

$\frac{30.4}{18.4} \approx 1.65$ (the ratio between perimeters).

Therefore, the closest ratio is 3. 3 : 5; 3: 5