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murzikaleks [220]
3 years ago
15

How to calulate the area of a square with a perimeter of 60 inches​

Mathematics
2 answers:
liberstina [14]3 years ago
7 0

Answer: 225 in.²

Step-by-step explanation: First, we want to find the length of one side. Since we know that a square has 4 sides and the perimeter is 60 inches, we can set up the following equation to find the length of one side.

Let's use l as our variable to represent length.

4l = 60

To get l by itself, we need to divide by 4 on the left side of the equation. Since we divided by 4 on the left side, we must also divide by 4 on the right side. On the left side, the 4's cancel and we are left with l. On the right side, we have 60 over 4 which simplifies to 15.

This means that a side of the square is 15 inches.

Now, we are asking ourselves what is the area of a square with a side length of 15 inches.

To find the area of a square, remember that a square is a type of rectangle so we can use the following formula.

<em>Area = length × width</em>

However, since the length and width of a square are always equal, we use a special version of this formula to find the area of a square.

Instead of <em>area = length × width</em>, we say that the area of a square equals side × side or S².

Since the sides of the square each have a length of 15 inches, the area of the square is equal to (15 in.)² or (15 in) (15 in) which equals 225 in.²

Therefore, the area of the square is 225 in.².

olasank [31]3 years ago
6 0

Answer:

<h2>A = 225 in²</h2>

Step-by-step explanation:

The formula of an area of a square:

A=s^2

<em>s</em><em> - side</em>

The formula of a perimeter of a square:

P=4s

<em>s</em><em> - side</em>

<em />

We have a perimeter

P=60\ in

Substitute:

4s=60             <em>divide both sides by 4</em>

\dfrac{4s}{4}=\dfrac{60}{4}\\\\s=15\ in

Put it to the formula of an area of a square:

A=15^2=225\ in^2

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a. The equation of the tangent at (1,3) is y = -x/3 + 10/3

b. The equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

<h3>a. How to find the equation of the tangent at (1, 3)?</h3>

Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.

So, x² + y² = 10

d(x² + y²)/dx = d10/dx

dx²/dx + dy²/dx = 0

2x + 2ydy/dx = 0

2ydy/dx = -2x

dy/dx = -2x/2y

dy/dx = -x/y

At (1,3), dy/dx = -1/3

Using the equation of a straight line in slope-point form, we have

m = (y - y₁)/(x - x₁) where

  • m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
  • (x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

-1/3 = (y - 3)/(x - 1)

-(x - 1) = 3(y - 3)

-x + 1 = 3y - 9

3y = -x + 1 + 9

3y = -x + 10

3y + x = 10

y = -x/3 + 10/3

So, the equation of the tangent at (1,3) is y = -x/3 + 10/3

<h3>b. The equation of the normal at the point (1, 3)</h3>

Since the tangent and normal line are perpendicular at the point, for two perpendicular line,

mm' = -1 where

  • m = gradient of tangent = -1/3 and
  • m' = gradient of normal

So, m' = -1/m

= -1/(-1/3)

= 3

Using the equation of a straight line in slope-point form, we have

m' = (y - y₁)/(x - x₁) where

  • m' = gradient of normal at (1, 3)  and (
  • x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

3 = (y - 3)/(x - 1)

3(x - 1) = (y - 3)

3x - 3 = y - 3

y = 3x - 3 + 3

y = 3x + 0

y = 3x

So, the equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

Learn more about equation of tangent and normal here:

brainly.com/question/7252502

#SPJ1

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