Examples with solutions for Area of the Square: Finding Area based off Perimeter and Vice Versa

Exercise #1

A square has a side length of

(3125)232 (\sqrt{3}\cdot\sqrt{12}-5)^2\cdot3^2 .

Calculate its perimeter.

Video Solution

Step-by-Step Solution

To calculate the perimeter of a square, we need to first determine the length of one side and then use the formula for the perimeter of a square, which is given by 4×side length 4 \times \text{side length} .

The side length is given as (3125)232 (\sqrt{3}\cdot\sqrt{12}-5)^2\cdot3^2 . Let's simplify this expression step by step:

  • First, simplify the square roots: 3 \sqrt{3} and 12 \sqrt{12} .

    • 12=4×3=4×3=2×3 \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4}\times \sqrt{3} = 2 \times \sqrt{3}
    • Thus, 3×12=3×2×3=2×3=6 \sqrt{3} \times \sqrt{12} = \sqrt{3} \times 2 \times \sqrt{3} = 2 \times 3 = 6 .
  • Next, substitute back into the expression: (65)232 (6 - 5)^2 \cdot 3^2 .
  • Simplify inside the parentheses: 65=1 6 - 5 = 1 , so we have 1232 1^2 \cdot 3^2 .
  • Calculate the squares: 12=1 1^2 = 1 and 32=9 3^2 = 9 .
  • Multiply the results: 1×9=9 1 \times 9 = 9 .

The side length of the square is therefore 9 9 .

Thus, the perimeter of the square is:

  • 4×9=36 4 \times 9 = 36 .

Therefore, the perimeter of the square is 36 36 .

Answer

36

Exercise #2

Given the area of the square ABCD is

3216 3^2\sqrt{16}

Find the perimeter.

Video Solution

Step-by-Step Solution

To find the perimeter of the square ABCD, we first need to determine the side length of the square using the given area. The area of a square is calculated using the formula: Area=s2 \text{Area} = s^2 , where s s is the side length of the square.

According to the problem, the area of the square is given by the expression 3216 3^2\sqrt{16} .

Let's simplify this expression:

  • First, calculate 32 3^2 , which is 9 9 .

  • Next, calculate 16 \sqrt{16} , which is 4 4 .

Now, multiply these results: 9×4=36 9 \times 4 = 36 .

Thus, the area of the square is 36 36 .

Since the area is s2=36 s^2 = 36 , we can solve for s s :

  • Find the square root of both sides: s=36 s = \sqrt{36} .

  • This gives s=6 s = 6 .

Now that we have the side length s=6 s = 6 , we can find the perimeter. The perimeter P P of a square is given by:

P=4s P = 4s .

Substituting the side length, we get:

  • P=4×6=24 P = 4 \times 6 = 24 .

The solution to the question is: 24 24

Answer

24 24

Exercise #3

Given a square whose area is

(3212)+23 (3^2-1^2)+2^3

What is the perimeter of this square?

Video Solution

Step-by-Step Solution

To solve this problem, we need to find the perimeter of a square given its area. The area of the square is given by the expression (3212)+23 (3^2-1^2)+2^3 .

Let us evaluate the expression to find the area:

  • Calculate 32 3^2 , which is 9 9 .
  • Calculate 12 1^2 , which is 1 1 .
  • Subtract to find 3212=91=8 3^2 - 1^2 = 9 - 1 = 8 .
  • Calculate 23 2^3 , which is 8 8 .
  • Add the results: 8+8=16 8 + 8 = 16 .

Therefore, the area of the square is 16 16 .

In general, the area of a square is given by the formula s2 s^2 , where s s is the side length of the square. To find the side length, we solve the equation:

  • s2=16 s^2 = 16 .
  • Taking the square root of both sides, we find s=16=4 s = \sqrt{16} = 4 .

The perimeter P P of a square with side length s s is given by the formula:

  • P=4s P = 4s .

Thus, substituting the value of s s :

  • P=4×4=16 P = 4 \times 4 = 16 .

Therefore, the perimeter of the square is 16 16 .

Answer

16 16

Exercise #4

Given 2 squares. One side of the squares is smaller by 2 than the other. The value of the area of the large square is greater than the value of the perimeter of the small square by 20.

Find the length of the small square

XXXX-2X-2X-2

Video Solution

Answer

4