The hypotenuse of a right triangle measures 20m. If one leg of the triangle is 4m longer than the other leg, find the lengths of the two legs

Question

So I have a test on Thursday a math exam and I just don't understand this question I need help please help me with his question 

Answer 1

You can use the Pythagorean theorem since you know the hypotenuse.

 

a^2 + b^2 = c^2 You have a hypotenuse of 20m.

 

a^2 + b^2 = (20m)^2

 

Let x = one leg, then the other leg is x + 4m.

 

Substitute into the Pythagorean theorem.

 

x^2 + (x + 4m)^2 = (20m)^2

 

square everything

 

x^2 + x^2 + 8mx + 16m^2 = 400m^2

 

Simplify left side

 

2x^2 + 8mx + 16m^2 = 400m^2

 

Factor out a 2 on the left side

 

2(x^2 + 4mx + 8m^2) = 400m^2

 

Divide both sides by 2

 

x^2 + 4mx + 8m^2 = 200m^2

 

Set equation to 0 by subtracting both sides by 200m^2

 

x^2 + 4mx + 8m^2 - 200m^2 = 0

 

Simplify left side

 

x^2 + 4mx - 192m^2 = 0

 

Factor left side. Two factors of 192 whose difference makes a positive 4

 

(x + 16m)(x - 12m) = 0

 

Set x - 12m to zero to get a positive result. x + 16m will give a negative result. Only positive results are needed in measurement problems

 

x - 12m = 0 add 12m to both sides

 

x = 12m

 

x + 4m = 16m

 

Your legs are 12m and 16m

 

Check using the Pythagorean Theorem

 

(12m)^2 + (16m)^2 = (20m)^2

 

144m^2 + 256m^2 = 400m^2

 

400m^2 = 400m^2

 

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Answer 2

x = length of one side

 

x + 4 = length of other side

 

20 m = length of hypotenuse

 

Pythagorean theorem:

 

x^2 + (x + 4)^2 = (20)^2 = 400

 

x^2 + (x^2 + 8x + 16) = 400

 

2x^2 + 8x - 384 = 0

 

Divide through by 2:

 

x^2 + 4x - 192 = 0

 

x = (1/2){-4 ± √[(4)^2 -4(-192)]} = (1/2)(-4 ± 28)

 

Use positive square root:

 

x = (1/2)(-4 + 28) = 12

 

The lengths of the sides are 12 m and 16 m.