The hypotenuse of a right triangle?

Question

The hypotenuse of a right triangle is 2 more than one of the sides. The other side has length 8m. What is the perimeter of the triangle?

Answer 1

C^2 = A^2 + B^2

let one side be A
and the other be the 8 meters.

if the hypotenuse is 2 more than A; C = A+2
so the equation is

(A+2)^2 = A^2 + 8^2

A^2 + 4A + 4 = A^2 + 8^2
4A +4 = 64
4A = 60
A =15
so the triangle is a 8,15,17 triangle. This is an identity. you should memorize all triangle identities or at least have a cheat sheet next to you when doing homework so you can identify them quickly. If you see that one of the sides is an 8 you should have immediately seen if the 3,4,5 triangle or the 8,15,17 triangle would make the statement true.

Answer 2

On problems like these, it is very helpful to draw a picture, so you can see what you are working with. I drew a right triangle, and then labeled each side according to the information you gave me.

Ok, I labeled one side x. Now, you said the hypotenuse is 2 more than one of the sides. Therefore, I have x + 2 for the hypotenuse. And I have 8 for the third side. We now have an equation. Remember, when doing these problems, the basic formula is a^2 + b^2 = c^2 (c^2 is the hypotenuse).

Here is the equation:

x^2 + 8^2 = (x + 2)^2

We can now solve for x.

x^2 + 64 = x^2 + 4x + 4

(Subtract x^2 from both sides)
64 = 4x + 4

(Subtract 4 from both sides)
60 = 4x

(Now we can divide the 4)
60/4 = 4x/4
15 = x

Since x = 15, one side is 15, one side is 8, and the hypotenuse is 17. Hope this helps!

Answer 3

Let the other side be x and the hypotenuse will be x+2

So, the square of the two sides should equal the square of the hypotenuse (Pythagorean)

8^2 + x^2 = (x+2)^2
8^2 + x^2 = (x+2)(x+2)
8^2 + x^2 = x^2 + 4x + 4
64 - 4 = x^2 - x^2 + 4x
60 = 4x
x = 15

So... the sides are 8, 15 and 17 m long. So, the perimeter is 40m.

Answer 4

side a = x
side b = 8
side c (hypotenuse) = x + 2 correct?

so, x^2 + 8^2 = (x+2)^2
x^2 + 64 = x^2 + 4x + 4
60 = x^2 + 4x - x^2
4x = 60
x = 15

hence, perimeter = 15 + 8 + (15 + 2)
= 40 m

Answer 5

Let x denote side 1
Let y denote side 2
Let h denote hypotenuse

Side 1, x, is equal to 8.
Side 2 = y.
hypotenuse, h, is equal to y + 2.

x²+y² =h²
8²+y² = (y+2)²
64 + y² = y² + 4y + 4
4y = 60
y = 15.

h = y + 2 = 15 + 2 = 17.

The perimetre of the triangle is therefore 8 + 15 + 17 = 40 metres.

Answer 6

A^2 + B^2 = C^2
8^2 + x^2 = (x+2)^2
64 +x^2 = x^2 +4x +4
the x^2 cancels out so...
64 = 4x + 4
60 = 4x
15 = x

thus the sides are 8, 15, and 17, so the perimiter is 40m

Answer 7

The perimeter is 40. The triangle is an 8,15,17
right triangle.
Let's work it out algebraically:
We use Pythagorean theorem:
64 + b² = (b+2)².
64 = 4b + 4
4b = 60
b = 15.
h = 15+2=17.

Answer 8

Pythagorean triple
8, 15, 17

8 + 15 + 17 = 40 m