PLEASE HELP WITH QUESTION. I am taking a test right now with little time left.
1 answer:
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First, we need to find the length of the side of the square.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.