Word problem!!! AHH!!?

Question

Triangle One leg of a right triangle is two inches more than twice the other leg. The hypotenuse is 13 inches. Find the lengths of the three sides of the triangle. Enter the values without units and separated by semicolons.

Answer 1

just clarifying first answer...

let a = 1st leg
and b = 2nd leg
c = hypotenuse

a= 2 + 2b -->> one leg(a) is two inches(2+) more than twice the other leg(2b)

substitute a
[(2+2b)^2] + [b^2] = [13^2]
[4+4b+(4b^2)] + [b^2} = 169
5b^2 + 8b + 4-169 = 0
5b^2 + 8b - 165 = 0
A B C
using quadratic formula which is -B plus or minus the square root of B^2 - 4AC all over 2A

-8 plus/minus squareroot of 64 - [4(5)(-165)] all over 2x5

-8 plus/minus squareroot of 58 all over 10

[-8 + 58] / 10 = 5

or

[-8-58] /10 = -6.6

negative sides do not exist so one side (b) is 5

going back to a= 2 + 2b
a = 2+10
a = 12

therefore
a = 12,
b = 5
c = 13

Answer 2

Let the 3 sides be a, b and c (c be the hypotanuse)

given: a = 2b + 2

c = 13

from Pythagoras theorem; a^2 + b^2 = c^2

or (2b+2)^2 + b^2 = 13^2
or 4b^2 + 4 + 8b + b^2 = 169
or 5b^2 + 8b - 165 = 0
or 5b^2 +33b - 25b -165 = 0
or (b - 5)(5b + 33) = 0

or b = 5 , -13

Neglect -ve value of lenght,

b = 5 and thus a = 2b + 2 = 12

Ans: 5; 12 ; 13


Another way: A fmous Pythagorean Triplet is 5,12 and 13. Once u see that hypotanuse is 13, and its a rt triangle, plug in the values and see, easily u can see one side will be 5 and other is 12.

Answer 3

Hi. The area of the two legs (squared) has to equal 169 (13 times 13). x-2=2y x^2 +y^2=139. Any help?

Answer 4

Leg 1 =a
Leg 2=b
Hypoteneuse = 13

A=2b+2

Ok, now you can put it into the formula
(2b+2)^2+b^2=13^2

Solve for B.

Answer 5

let length of one leg be x
then other leg is 2+2x
x^2 +(x+2)^2=13^2
2x^2+4x-165=0
solve for x