The side lengths of triangle are 6 units, 8 units and 10 units.
<u>SOLUTION: </u>
Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)
We know that, distance between two points is given by
Now,
Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".
"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6
Since we're solving for the length and width, let's give them each variables.
length = l = w+6
width = w
The next bit of information is "the area is 40 square inches"
Applying the formula for the area of a rectangle we can set up:
l x w = 40
replace "l", or length, with it's alternate value.
(w+6) x w = 40
distribute
+ 6w = 40
subtract 40 from both sides
+ 6w - 40 = 0
factor
(w - 4)(w + 10) = 0
solve for w
w= 4, or -10
So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.
We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.
l = w + 6
w=4
l = 4 + 6 = 10
The length is 10 in. and the width is 4 in. Hope this helps!