Find the value of x and find the value of polygon JKLM
1 answer:
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Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.
Givens
a = x
b = x + 4
c = 20
Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2
Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20 Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0
2x^2 + 8x - 4 = 0 Divide through by 2
x^2 + 4x - 2 = 0
Use the quadratic formula
a = 1
b = 4
c = - 2
From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6) This is an extraneous root. No line can be minus.
x1 = + 0.4495
x2 = x + 4 = 4.4495
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.
Givens
a = x
b = x + 4
c = 20
Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2
Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20 Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0
2x^2 + 8x - 4 = 0 Divide through by 2
x^2 + 4x - 2 = 0
Use the quadratic formula
a = 1
b = 4
c = - 2
From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6) This is an extraneous root. No line can be minus.
x1 = + 0.4495
x2 = x + 4 = 4.4495