algebra help please?

Question

The hypotenuse of a right triangle is 15, and one leg is 3 more than the other leg. Find the lengths of the legs of the triangle. show work and give explanations for each step.

how do i slove this problem???
thanks

Answer 1

Let the legs be of length x and x+3. Pythagoras' theorem then tells us that
x^2 + (x+3)^2 = 15^2
<=> x^2 + x^2 + 6x + 9 - 225 = 0
<=> 2x^2 + 6x - 216 = 0
<=> x^2 + 3x - 108 = 0
<=> (x+12)(x-9) = 0
<=> x = -12 or 9
But x > 0, so we must have x = 9. So the legs are 9 and 12 units long.

Answer 2

Well, I'm not really sure how you would "slove" the problem, but I can tell you how to solve it! Heh, heh...

Anyways, the equation you want is:

leg^2 + leg^2 = hypotenuse^2

This is what you know about your legs:
x = one leg
x + 3 = another leg

Plug in what you have:
leg^2 + leg^2 = hypotenuse^2
(x)^2 + (x+3)^2 = 15^2
(x^2) + (x^2 + 6x + 9) = 225
2x^2 + 6x + 9 = 225
2x^2 + 6x - 216 = 0

Look familiar? You have a quadratic now, so what you need to do it factor it and then solve for your zeros. Try it and then compare it with the answer below:

Factored: 2 (x + 12) (x - 9)

Now you solve it and get your solution set:
x = {-12, 9}
Look, you can cross off the -12 because you cannot have a negative length!

So x = 9. Oh, look, that means one of the legs is 9. The other one must be 12 because 9 + 3 = 2.

Answer:
9 and 12

Now to check:
9^2 + 12^2 = 15^2
81 + 144 = 225
225 = 225
....../
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YES! It works! That's supposed to be a check mork up there. Hopefully you got more out of this than just an answer. You have seen the full process.

Answer 3

The square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

H^2=L^2+M^2
Your problem also says that one side is 3 units longer than the other one.
Picking L to be the longer:
L=M+3
L^2 = M^2+6*M +9
15^2=2*M^2 +6*M +9
2*M^2 +6*M - 216=0
using the quadratic formula
A=2, B=6, C=-216
M=(-B+sqrt(B^2-4AC))/2A
M=(-6+sqrt(6^2-4*2*(-216)))/(2*2)
M=9
note that I have ignored the negative solution for M since a triangle can not have a side with a negative length.
L=M+3
L=12
Putting the numbers back in the Pythagorean theorem's equation shows that 9,12 and 15 are indeed three lengths of line segments that would form a right triangle.
9^2+12^2=81+144=225=15^2

Answer 4

Given: Hyp = 15
Let: X = shorter leg
X + 3 = longer leg

Use the pythagorean theorem: c^2 = a^2 + b^2
where c = hypotenuse and a and b the legs.
Thus, c = 15; a = X; b = X + 3.
Substitute values/terms:
c^2 = a^2 + b^2
15^2 = X^2 + (X+3)^2 [simplify by squaring binomial]
225 = X^2 + X^2 + 6X + 9 [combine like terms]
225 = 2X^2 + 6X + 9 [move 225 to other side with the opposite sign and equate with zero]
0 = 2X^2 + 6X + 9 - 225
0 = 2X^2 + 6X - 216 [factor trinomial by 2]
0 = 2(X^2 + 3X - 108) [factor trinomial by FOIL]
0 = 2(X-9)(X+12) [equate each binomial to zero]
0=X-9; 0=X+12 [solve for x]
X=9; X=-12 [X is equal to 9!]

**the legs will be 9 and 12!

To check,
15^2 = 9^2 + 12^2
225 = 81 + 144
225 = 225

Answer 5

The equation for a right triangle is A^2 + B^2 = C^2 when A and B are the legs of the triangle and C is the hypotenuse.

So C = 15
A is a leg.
and B is 3 times A or 3A

So the equation is A^2 + (3A)^2 = 15^2
solving this
A^2 + 9A^2 =225
10A^2 = 225
A^2 = 225/10
A = sqrt(225/10)
B = 3 * sqrt(225/10)

Answer 6

one leg is X, other leg is 3+X.

The formula that connects all three legs is a^2 + b^2 = c^2, where c is the hypotenuse, and a and b are the two legs.

x^2 + 9+6x+x^2 = 225
--> 2x^2 + 6x -216 = 0
--> x^2 + 3x - 108 = 0
Use the quadratic formula, or factor to find X.

Answer 7

let x = short leg

x^2 + (x+3)^2 = 15^2
x^2 + x^2 + 3x + 3x + 9 = 225
2x^2 + 6x - 216 = 0
(2x+6) (x-9) = 0
x = -12
or
x = 9

since x cant be negative, x = 9
x+3 = 12

9 - 12 - 15 triangle

Answer 8

let x be one of the leg
the other is x + 3

a^2 + b^2 = c^2

x^2 + (x+3)^2 = 15^2
x^2 + x^2 + 6x + 9 = 225
2x^ + 6x - 216=0

x^2 + 3x - 108 = 0

factor
(x+12)(x-9)

x = -12 or 9

since you can't have a nagetive distance. 9 is your answer

the legs are 9 and 12

Answer 9

x² + (x+3)² = 15²
x² + x²+6x+9 = 225
2x² + 6x - 216 = 0
(2x+24)(x-9) = 0
x= {-12, 9}