Question

Difficulty: MediumRight Triangles and the Pythagorean Theorem

In the xyxy-plane, right triangle ABCABC has vertices at A(0,0)A(0, 0), B(k,0)B(k, 0), and C(0,6)C(0, 6), where kk is a positive constant. The right angle of the triangle is at vertex AA. If the length of the altitude from AA to side BCBC is 4.84.8, what is the value of kk?

  1. A
    3.6
  2. B
    4.5
  3. 8Answer
  4. D
    10

Answer

8
The correct value is 8. The area of right triangle ABCABC is given by 12×base×height\frac{1}{2} \times \text{base} \times \text{height}. Using the legs, the area is 12×k×6=3k\frac{1}{2} \times k \times 6 = 3k. Using the hypotenuse and the altitude, the area is 12×BC×4.8=2.4×BC\frac{1}{2} \times BC \times 4.8 = 2.4 \times BC. Equating the two expressions gives 3k=2.4×BC3k = 2.4 \times BC, which simplifies to BC=1.25kBC = 1.25k. Applying the Pythagorean theorem, (1.25k)2=k2+62    1.5625k2=k2+36    0.5625k2=36    k2=64(1.25k)^2 = k^2 + 6^2 \implies 1.5625k^2 = k^2 + 36 \implies 0.5625k^2 = 36 \implies k^2 = 64. Since kk must be positive, k=8k = 8.

Step-by-Step Solution

1
Express the area of right triangle ABCABC in two different ways using its base and height.
Area = 3k3k and Area = 2.4×BC2.4 \times BC.
The area of a right triangle can be calculated using the legs as the base and height, Area = 12×AB×AC=12×k×6=3k\frac{1}{2} \times AB \times AC = \frac{1}{2} \times k \times 6 = 3k. Alternatively, using the hypotenuse BCBC as the base and the altitude to the hypotenuse as the height, Area = 12×BC×4.8=2.4×BC\frac{1}{2} \times BC \times 4.8 = 2.4 \times BC.
2
Equate the two area expressions to find the length of the hypotenuse BCBC in terms of kk.
3k=2.4×BC    BC=1.25k3k = 2.4 \times BC \implies BC = 1.25k.
Since both expressions represent the area of the same triangle, they must be equal.
3
Apply the Pythagorean theorem to right triangle ABCABC to set up an equation for kk.
(1.25k)2=k2+62    1.5625k2=k2+36(1.25k)^2 = k^2 + 6^2 \implies 1.5625k^2 = k^2 + 36.
In right triangle ABCABC, the Pythagorean theorem states that BC2=AB2+AC2BC^2 = AB^2 + AC^2.
4
Solve the equation for kk.
0.5625k2=36    k2=64    k=80.5625k^2 = 36 \implies k^2 = 64 \implies k = 8 (since k>0k > 0).
Subtracting k2k^2 from both sides gives 0.5625k2=360.5625k^2 = 36. Dividing both sides by 0.56250.5625 yields k2=64k^2 = 64. Taking the square root gives k=8k = 8, since kk represents a positive coordinate.

Key Concept

Relating the area of a right triangle to its legs and hypotenuse altitude to set up proportions, followed by applying the Pythagorean theorem.
Estimated Time:1m 30s
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