The side lengths of triangle are 6 units, 8 units and 10 units.
<u>SOLUTION:
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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,
A + b = 15 ..... eq 1
a – b = 12 ...... eq 2
So a = 15 – b
Substitute in eq 2 :
15 – b – b = 12
15 – 2b = 12
–2b = 12 – 15
b = 1.5
So a = 15 – 3/2 = 13.5
4ab = 4 x 13.5 x 1.5 = 81
Take 16x divide it by 2 and square it. 16/2= 8. 8^2= 64. So, the first step will be to add 64 to both sides.
Answer:
Step-by-step explanation:
A^2 + b^2 = c^2
a^2 + (12)^2 = (13)^2
a^2 + 144 = 169
a^2 = 169 - 144
a^2 = 25
Sqrt (a^2) = Sqrt(25)
a = 5
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