Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-1, -6),A(−1
2 answers:
Answer:
13 units
Step-by-step explanation:
Rectangle ABCD has coordinates A(-1,-6), B(-1,7), C(1,7), D(1,-6)
We want to find the length of AB.
We use the absolute value method to find the length of AB.
Since the first coordinates are the same for A(-1,-6) and B(-1,7).
The distance is given by;
The length of AB is 13 units.
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To find the perimeter of the triangle, you need to know the length of the hypotenuse (h). Use the Pythagorean theorem to find it.
.. h^2 = (4w)^2 + (15/2*w)^2
.. = w^2*(4^2 +(15/2)^2)
.. = w^2*(16 +56 1/4)
.. = w^2*(72 1/4)
Taking the square root, we find the hypotenuse to be
.. h = √(w^2*72.25) = 8.5w = (17/2)w
Now, the length of the triangle's perimeter is the sum of the lengths of the sides.
.. P = 4w +(15/2)w +(17/2)w
.. P = 20w
The area of the triangle is half the product of base and height:
.. A = (1/2)(4w)(15/2*w)
.. A = 15w^2
So, the ratio of perimeter to area is
.. P/A = (20w)/(15w^2) = 4/(3w) . . . . . . . . simplified expression for P/A
The problem is asking you to find (and graph) the values of w such that
.. P/A < 1
.. 4/(3w) < 1 . . . . . . substitute our value of P/A
.. 4/3 < w . . . . . . . . multiply by w
This is graphed with an open circle around the point 1 1/3, and a solid arrow to the right of that circle. The open circle shows that w=4/3 is NOT a solution, but all values greater than that are.
.. h^2 = (4w)^2 + (15/2*w)^2
.. = w^2*(4^2 +(15/2)^2)
.. = w^2*(16 +56 1/4)
.. = w^2*(72 1/4)
Taking the square root, we find the hypotenuse to be
.. h = √(w^2*72.25) = 8.5w = (17/2)w
Now, the length of the triangle's perimeter is the sum of the lengths of the sides.
.. P = 4w +(15/2)w +(17/2)w
.. P = 20w
The area of the triangle is half the product of base and height:
.. A = (1/2)(4w)(15/2*w)
.. A = 15w^2
So, the ratio of perimeter to area is
.. P/A = (20w)/(15w^2) = 4/(3w) . . . . . . . . simplified expression for P/A
The problem is asking you to find (and graph) the values of w such that
.. P/A < 1
.. 4/(3w) < 1 . . . . . . substitute our value of P/A
.. 4/3 < w . . . . . . . . multiply by w
This is graphed with an open circle around the point 1 1/3, and a solid arrow to the right of that circle. The open circle shows that w=4/3 is NOT a solution, but all values greater than that are.