All categories

3 years ago

Find the area of an octagon whose perimeter is 120 cm.

Mathematics

1 answer:

stira [4]3 years ago

5 0

Answer:

The area of an octagon whose perimeter is 120 cm is 1086.4 $cm^{2}$

Step-by-step explanation:

An octagon is a polygon with eight sides. If the lengths of all the sides and the measurement of all the angles are equal, the octagon is called a regular octagon.

There is a predefined set of formulas for the calculation of perimeter, and area of a regular octagon.

The perimeter of an Octagon is given by

$P=8a$

and the area of an Octagon is given by

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

We know that the perimeter is 120 cm, solving for side length (a) in the perimeter formula we get

$120=8a\\\frac{8a}{8}=\frac{120}{8}\\a=15$

Now, we calculate the area

$A=2a^{2}(1+\sqrt{2})\\A=2(15)^{2}(1+\sqrt{2})\\A=450\left(1+\sqrt{2}\right)\\A\approx 1086.4 \:cm^{2}$

You might be interested in

For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec

telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

$4sin(x^2) = 0$

$sin(x^2) = 0$

$x^2 = n\pi$ where n = 0, 1, 2,3, 4, 5,...

$x = \sqrt(n\pi)$

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

$f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0$

$xcos(x^2) = 0$

$x = 0$ or

$cos(x^2) = 0$

$x^2 = \frac{\pi}{2} + n\pi$ where n = 0, 1, 2, 3

$x = \sqrt{\pi(n+1/2)}$

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25, $y = f(1.25) = 4sin(1.25^2) = 4$

x = 2.17 , $y = f(2.17) = 4sin(2.17^2) = -4$

x = 2.8, $y = f(2.8) = 4sin(2.8^2) = 4$

For the inflection point, we can take the 2nd derivative and set it to 0

$f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0$

$cos(x^2) = 2x^2sin(x^2)$

$tan(x^2) = \frac{1}{2x^2}$

We can solve this numerically to get the inflection points are at

x = 0.81, $y = f(0.81) = 4sin(0.81^2) = 2.44$

x = 1.81, $y = f(1.81) = 4sin(1.81^2) = -0.54$

x = 2.52, $y = f(2.52) = 4sin(2.52^2) = 0.27$

3 0

3 years ago

Need help plz..........

boyakko [2]

Subtraction property of equality
2/3y = -6
2/3y × 3/2 = -6 × 3/2
multiplication property of equality
y = -9

8 0

2 years ago

What is the volume of a cylinder with height 4 and radius 2? Use 3.14 for pi.

Over [174]

Answer:

50.27

v=3.14 (2 squared) 4

5 0

3 years ago

What is the additive of p when p = -19.57

scoray [572]

The additive inverse of a number is just the opposite of the given number. For example; 2 and the additive inverse is -2.

If we apply this to the number we are given, we get 19.57

Best of Luck!

5 0

3 years ago

The distance d traveled at a constant speed r is directly proportional to the time t spent traveling. a car moving at a constant

svet-max [94.6K]

150/2.5=x/6 is the proportion

3 0

2 years ago

Read 2 more answers