Solve Problems Requiring the Use of Quadratic Equation?

Question

A small local park is shaped like a right triangle. The hypotenuse of the triangle is 10 yards long. One leg of the triangular park is 2 yards shorter than the other leg. Find the length of the other two sides of this park. PLEASE HELP!

Answer 1

[05]
Let one of the remaining legs be x yards and the length of the last leg is x-2 yds
According to Pythogorean theorem,
x^2+(x-2)^2=10^2
x^2+x^2-4x+4=100
2x^2-4x+4-100=0
2x^2-4x-96=0
x^2-2x-48=0
(x-8)(x+6)=0
x=8 or -6
Ignoring the negative value of x,we get x=8 yard
Therefore the length of the other two legs of the right angled triangles are 8 yards and 8-2=6 yards

Answer 2

The most basic problem.

Let one leg be x and the other be (x + 2)

By Pythagoras theorem,
10^2 = x^2 + (x + 2)^2
100 = x^2 + x^2 + 4x + 4
2x^2 + 4x + 4 = 100
x^2 + 2x + 2 = 50
x^2 + 2x - 48 = 0
(x + 8)(x - 6) = 0
x = -8 or x = 6

But x cannot be -8 as it is a length.

So, x = 6
-> x + 2 = 8

The other two sides of the park are of length 6 yards and 8 yards,

Answer 3

So, the one leg = x
The other leg = x - 2

Using pythagorean theorem:

(x)^2 + (x - 2)^2 = 10^2
x^2 + (x^2 - 4x + 4) = 100
2x^2 - 4x + 4 = 100
(divide everything by 2)
x^2 - 2x + 2 = 50
x^2 - 2x - 48 = 0
(x - 8)(x + 6) = 0
x = 8 or -6

x = 8 yard

The lengths of the two legs are 8 yards and 6 yards.

Answer 4

Set up the following equation using the pythagorean theorem:
x^2+y^2=100 (10^2)
you know that one length is 2 shorter than the other so x-2 can equal other length, so plug this into y and simplify
x^2+(x-2)^2=100
x^2+x^2-4x+4=100
2x^2-4x+4=100
2x^2-4x-96
2(x^2-2x-480
2(x-8)(x+6) (factoring)
x=8 or -6, 8 can only work because it is positive, therefore one leg is 8 and the other is x-2 or 6 (both in yards).

Answer 5

One leg=x
The other leg is 2 shorter, or x-2
The hypotenuse is 10
For right-angled triangles. the Pythagorean Theorem
says H^2=s(1)^2 + s(2)^2
10^2=x^2+(x-2)^2
100=x^2+x^2-4x+4
2x^2-4x=96
2x^2-4x-96=0
Dividing by 2,
x^2-2x-48=0
(x-8)(x+6)=0
For (x-8)(x+6) to multiply out to zero,
Either x-8=0, so x=8
Or x+6=0, so x=-6
-6 isn't a practical answer
x=8
x-2=6
Sides are 6yds. and 8 yds.

Answer 6

Pythagorean Theorem:
a^2+b^2=c^2
a=x
b=x-2
c=10

x^2+(x-2)^2=10^2
x^2+x^2-4x+4=100
2x^2-4x-96=0
2(x^2-2x-48)=0

The quadratic formula:
x=-b+-sqrtb^2-4ac/2a
a=1,b=-2,c=-48
x=2+-sqrt(-2)^2-4(1*-48)/2(1)
x=(2+-sqrt4+192)/2
x=(2+-sqrt196)/2
x=(2+-14)/2
x=-6 or 8
Ignore the -6 because you do not want a negative answer for this problem.

8-2=6

The legs are 6 and 8 yards.

Answer 7

hi

c^2 + (c-2)^2 = 10^2

c^2 + c^2 - 4 c + 4 = 100

c^2 - 2 c - 48 = 0
c^2 - 2 c + 1 = 49
(c-1)^2 = 7^2

c (positive) = 8 yards
c-2 = 6 yards

greetings