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Difficulty: HardPythagorean Theorem and Special Right Triangles

In the standard (x,y)(x, y) coordinate plane, a circle is centered at the origin (0,0)(0,0) and has a radius of 88. A horizontal chord ABAB lies entirely in the first and second quadrants at a distance of 44 units from the xx-axis. A point PP is located on the circle such that △ABP\triangle ABP is a right triangle. If the hypotenuse of △ABP\triangle ABP is a diameter of the circle, what is the area of △ABP\triangle ABP?

  1. A
    16316\sqrt{3}
  2. B
    3232
  3. 32332\sqrt{3}Answer
  4. D
    32532\sqrt{5}
  5. E
    48348\sqrt{3}

Answer

The area of the right triangle is 32332\sqrt{3}.
The correct answer is 32332\sqrt{3}. The horizontal chord ABAB has yy-coordinate 44, and its endpoints lie on the circle x2+y2=64x^2 + y^2 = 64. Solving for xx gives x=±43x = \pm 4\sqrt{3}, so the length of the chord is 838\sqrt{3}. Because the triangle is inscribed in the circle and is a right triangle, its hypotenuse must be a diameter of the circle (length 1616). Since AB<16AB < 16, ABAB is a leg, and the hypotenuse is one of the other sides (e.g., APAP). The remaining leg BPBP is found using the Pythagorean theorem: BP=162−(83)2=256−192=64=8BP = \sqrt{16^2 - (8\sqrt{3})^2} = \sqrt{256 - 192} = \sqrt{64} = 8. The area of the right triangle is 12×83×8=323\frac{1}{2} \times 8\sqrt{3} \times 8 = 32\sqrt{3}.

Step-by-Step Solution

1
Determine the length of chord ABAB.
The length of chord ABAB is 838\sqrt{3}.
The equation of the circle is x2+y2=64x^2 + y^2 = 64. Since the chord is horizontal and at a distance of 44 units from the xx-axis, its yy-coordinate is 44. Substituting y=4y = 4 gives x2+16=64⇒x2=48⇒x=±43x^2 + 16 = 64 \Rightarrow x^2 = 48 \Rightarrow x = \pm 4\sqrt{3}. The distance between A(−43,4)A(-4\sqrt{3}, 4) and B(43,4)B(4\sqrt{3}, 4) is 838\sqrt{3}.
2
Apply the rule for a right triangle inscribed in a circle to identify the hypotenuse.
The hypotenuse must be a diameter of length 1616, so the right angle is at BB (or AA).
Any right triangle inscribed in a circle must have a diameter as its hypotenuse. The diameter of this circle is 2×8=162 \times 8 = 16. Since the chord AB=83≈13.86AB = 8\sqrt{3} \approx 13.86 is shorter than the diameter, it cannot be the hypotenuse. Therefore, either APAP or BPBP is the hypotenuse (a diameter), making the angle opposite to it (either ∠ABP\angle ABP or ∠BAP\angle BAP) the 90∘90^\circ angle.
3
Calculate the length of the remaining leg of the right triangle.
The length of leg BPBP is 88.
Using the Pythagorean theorem for right triangle ABPABP with hypotenuse AP=16AP = 16 and leg AB=83AB = 8\sqrt{3}: AB2+BP2=AP2⇒(83)2+BP2=162⇒192+BP2=256⇒BP2=64⇒BP=8AB^2 + BP^2 = AP^2 \Rightarrow (8\sqrt{3})^2 + BP^2 = 16^2 \Rightarrow 192 + BP^2 = 256 \Rightarrow BP^2 = 64 \Rightarrow BP = 8.
4
Compute the area of right triangle ABPABP.
The area is 32332\sqrt{3}.
The area of a right triangle is 12×base×height=12×AB×BP=12×83×8=323\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BP = \frac{1}{2} \times 8\sqrt{3} \times 8 = 32\sqrt{3}.

Key Concept

Applying the Pythagorean theorem and Thales's theorem (inscribed right triangles) to solve multi-step geometric problems on the coordinate plane.
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Topics

Pythagorean Theorem and Special Right Triangles

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