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4 years ago

Find the circumference given the area = 50.3 m². Use 3.14 for π as necessary.

Mathematics

2 answers:

Marat540 [252]4 years ago

3 0

The circumference of a circle with an area of 50.3 m² is 25.12 m.

<h2>Further Explanation </h2><h3>Area </h3>

Area is a measure of how much space is occupied by a given shape.
Area of a substance is determined by the type of shape in question.

For example;

Area of a rectangle is given by; Length multiplied by width
Area of a triangle = 1/2 x base x height
Area of a circle = πr². where r is the radius of a circle,
Area of a square = S², Where s is the side of the square.etc.

<h3>Perimeter </h3>

Perimeter is defined as the distance along a two dimension shape. Perimeter of different shapes is given by different formulas

For example;

The perimeter of a rectangle = 2(length+width)
The perimeter of a triangle = a+b+c; where a, b and c are the sides of the triangle. etc.
The Circumference of a circle = 2πr , where r is the radius of the circle

In this case;

The Area of a circle = 50.3 m²

π = 3.14

But; Area of a circle = πr²

Therefore;

3.14r²= 50.3 m²

r² = 50.3/3.14

=16.019

r = √16.019

= 4.0023

≈ 4.00

But;

Circumference of a circle is given by 2πr

Thus;

Circumference = 2 × 3.14 × 4.00

= 25.12 m

Keywords; Perimeter, Area, Area of a circle, Circumference of a circle

<h3>Learn more about:</h3>

Perimeter:brainly.com/question/1322653
Area: brainly.com/question/1322653
Area of a circle: brainly.com/question/9404782
Circumference of a circle: brainly.com/question/9461882

Level: Middle school

Subject; Mathematics

Topic: Area and Perimeter

Sub-topic: Area and circumference of a circle

Karo-lina-s [1.5K]4 years ago

3 0

The answer would be 25.14

C=2 √ π•50.3

Convert into a fraction

C=2 √ π•503/10

Calculate the product

C=2 √ 503 π/10

Use radical rules

C=2 √ 503 π/√ 10

Calculate the product

C= 25.14

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============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

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$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

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