What is the perimeter and area of a right triangle with a hypotenuse of 26 and a leg of 24?

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Answer 1 · 2018-05-23T14:52:29

$Perimeter = 60$ $"Perimeter" = 60$ and $Area = 120$ $"Area" = 120$

Assign the sides $b = 24 and c = 26$ $b = 24 and c = 26$

Use the Pythagorean theorem,

$c^{2} = a^{2} + b^{2}$ $c^2 = a^2 + b^2$

,to find the length of side a:

$a = \sqrt{26^{2} - 24^{2}}$ $a = sqrt(26^2-24^2)$

$a = 10$ $a = 10$

The perimeter is:

$Perimeter = a + b + c$ $"Perimeter" =a +b +c$

$Perimeter = 10 + 24 + 26$ $"Perimeter" =10 +24 +26$

$Perimeter = 60$ $"Perimeter" = 60$

The area is:

$Area = \frac{1}{2} b h$ $"Area" = 1/2bh$

If we select side b as the base, then side a is the height:

$Area = \frac{1}{2} (24) (10)$ $"Area" = 1/2(24)(10)$

$Area = 120$ $"Area" = 120$