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Difficulty: Very hardQuadratic Equations and the Quadratic Formula

A right triangle has an area of 66 square units. The lengths of its legs, aa and bb, satisfy the quadratic equation a2−2.75ab+1.5b2=0a^2 - 2.75ab + 1.5b^2 = 0. If the length of the hypotenuse is an integer, what is the perimeter of the triangle?

  1. A
    9
  2. B
    10
  3. 12Answer
  4. D
    14
  5. E
    15

Answer

The perimeter of the triangle is 12.
The correct answer is 12. By solving the quadratic relationship between the legs, we find two possible ratios: one where one leg is twice the other, and one where one leg is 0.75 times the other. Using the area of 6, the first case yields non-integer side lengths, while the second case yields legs of length 3 and 4. This results in an integer hypotenuse of 5, giving a perimeter of 3 + 4 + 5 = 12.

Step-by-Step Solution

1
Express the relation a2−2.75ab+1.5b2=0a^2 - 2.75ab + 1.5b^2 = 0 as a quadratic in terms of the ratio r=abr = \frac{a}{b}.
The equation becomes r2−2.75r+1.5=0r^2 - 2.75r + 1.5 = 0.
To find the relationship between the two legs of the triangle by solving for their ratio.
2
Apply the quadratic formula to solve for rr.
r=2r = 2 or r=0.75r = 0.75, meaning either a=2ba = 2b or a=0.75ba = 0.75b.
To determine the two possible linear relationships between the legs of the right triangle.
3
Substitute each ratio case into the area formula Area=12ab=6\text{Area} = \frac{1}{2}ab = 6, which simplifies to ab=12ab = 12.
For a=2ba = 2b, we get b=6b = \sqrt{6} and a=26a = 2\sqrt{6}. For a=0.75ba = 0.75b, we get b=4b = 4 and a=3a = 3.
To calculate the actual leg lengths for both geometric cases.
4
Calculate the hypotenuse c=a2+b2c = \sqrt{a^2 + b^2} for both cases to check which yields an integer value.
The first case yields c=30c = \sqrt{30}, which is not an integer. The second case yields c=32+42=5c = \sqrt{3^2 + 4^2} = 5, which is an integer.
To satisfy the constraint that the hypotenuse must be an integer, identifying the correct leg lengths as 33 and 44.
5
Calculate the perimeter of the triangle for the valid case.
The perimeter is 3+4+5=123 + 4 + 5 = 12.
To find the final requested value.

Key Concept

Solving quadratic relationships and applying the quadratic formula in geometric constraints.
Estimated Time:3m 0s
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Topics

Quadratic Equations and the Quadratic Formula

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