You did the hardest part correctly.
Once you have the corner coordinates, square the length of any side to get the area.
The length of the diagonal of your square is sqrt[7^2 + 7^2] = 7 sqrt 2, so the length of sides should be 7.
The area must be 49
Here is my question: If (2, -3) and
(-5, 4 are the endpoints of a diagonal of a square, what are the coordinates of the other two vertices?
I got (2,3) and (-5,-4). But the other question was:
What is the area of the square?
How would you find this? Do you just count the number of tickmarks in between the vertices and then multiply like you would a normal square, or is there another correct way?
Thank you! Any help is appreciated!
4 answers
Your other points are not correct, a simple case of graphing them would have shown you they don't form a square.
I got (2,4) and (-5,-3) as the other points.
Check: In a square the diagonals are equal and right-bisect each other.
For my points both are equal to 7√2 and their slopes are +1 and -1
the base length is 7, and the height of course is also 7, for an area of 49 square units
I got (2,4) and (-5,-3) as the other points.
Check: In a square the diagonals are equal and right-bisect each other.
For my points both are equal to 7√2 and their slopes are +1 and -1
the base length is 7, and the height of course is also 7, for an area of 49 square units
Ummm... I did graph them, and why would it be (-5, -3)?
That would make the square... slightly diagonal, wouldn't it. I said -4 because that is directly below 4.
That would make the square... slightly diagonal, wouldn't it. I said -4 because that is directly below 4.
Reiny is correct about the other two point coordinates. I should have used real graph paper, or verified the side lengths. All must be 7.
2 answers