what is the perimeter of a right triangle with an area of 78 and two sides of 12 & 13?

Question

show example of picture and steps

Answer 1

Pythagorean Theorem

c² - a² + b²

√c² = √(12)² + (13)²

√c² = √144 + 169

√c² = √313

c = 17.69180601

c = 17.69 rounded to two decimal places

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The perimeter =

12 + 13 + 17.69 = 42.69

The answer is the perimeter is 42.69

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Answer 2

For an area of 78; 12 & 13 must be the legs. The hypoyenuse is:
h=√(12^2+13^2)
h=√(144+169)
h=√313
h=17.69
p=12+13+17.69=42.69

Answer 3

p=42

The pythagorean theorem states that a²+b²=c². Therefore if you know the lengths of sides a(12) and b(13) then you square them a(144)+b(169)=c²(313). So to find the length of side C you take the square root of 313 which is 17.7 and you now have the lengths of all three sides. 12+13+17.7 or 42.7 the perimeter of the triangle.

Answer 4

A=(1/2)(base)(height).

We don't know whether 13 is the hypotenuse or not.

But if base=12 and height=13, A=(1/2)(12)(13) does equal 78, so 13 is not the hypotenuse.

(Hypotenuse)^2 = (12^2) + (13^2)
=144 + 169 = 313
Hypotenuse = 17.7

Perimeter=12+13+17.7=42.7

Answer 5

Hyp = sqrt( 12^2 + 13^2 ) = sqrt(313) = 17.69

Perimeter = 12 + 13 + 17.69 = 42.69