Question

The sides of a triangle have lengths x, x+4, and 20. If the length of the longest side is 20, what values of x make the triangle acute?

Answer 1

12 < x < 16

What we know:

• The longest side of the triangle is 20, another side length is 'x + 4'
• Since the triangle is acute, we know that the value of 'x' will have to be greater than its right triangle value because whatever 'x' equals will result in a right triangle, not an acute one
• We must solve for the right triangle first in order to determine the acute triangle

Solving for 'x' in the right triangle:

1. a^2 + b^2 = c^2
2. (x+4)^2 + x^2 = 20^2
3. x^2 + x^2 + 8x + 16 = 400
4. 2x^2 + 8x - 384 = 0
5. simplify x^2 + 4x - 192 = 0
6. factor (x + 16) (x - 12) = 0
7. solve for each 'x' x + 16 = 0 & x - 12 = 0
8. x = -16 & x = 12
9. check the value of 'x' cannot be negative, so it is not -16. this means the value of 'x' is 12
10. the value of 12 for 'x' makes a right triangle, so that is not the answer. this means the the value of 'x' needs to be greater than 12

Checking your answer:

• original equation (x + 4)^2 + x^2 = 20^2
• with answer (12 + 4)^2 + (12)^2 = 20^2
• (16)^2 + (12)^2 = 20^2
• 256 + 144 = 400
• 400 = 400 this is true, so the answer of 'x = 12' is correct for making it into a right triangle

Solving for 'x' in the acute triangle:

1. x = 12 makes a right triangle
2. x > 12 minimum for an acute triangle
3. what is the maximum value for 'x' in the acute triangle? take the two other side lengths --> (x + 4) < 20 (longest side, so it will be the greater number)
4. simplify x < 16
5. x < 16 maximum for an acute triangle

Conclusion:

• if 'x = 12' makes a right triangle, then 'x > 12' and 'x < 16' are the values that will make it acute
• answer 12 < x < 16