A circle has center O, and points T and S lie on the circle. From a point P outside the...

1. TRANSLATE the problem information

• Given information:
  • Circle with center O and radius \(\mathrm{135\,mm}\)
  • Points T and S lie on the circle
  • Point P is outside the circle
  • PT and PS are tangent to the circle at T and S respectively
  • Area of quadrilateral OTPS = \(\mathrm{24{,}300\,square\,mm}\)
• What we need: Distance OP

2. INFER key geometric relationships

• Since PT and PS are tangents to the circle:
  • \(\mathrm{OT \perp PT}\) and \(\mathrm{OS \perp PS}\) (tangent perpendicular to radius)
  • \(\mathrm{PT = PS}\) (tangent segments from external point are equal)
  • This creates two congruent right triangles: OTP and OSP

3. INFER the area decomposition strategy

• The quadrilateral OTPS can be split into triangles OTP and OSP
• Each triangle has legs: radius = \(\mathrm{135\,mm}\) and tangent length = \(\mathrm{t}\)
• Area of each triangle = \(\mathrm{\frac{1}{2} \times 135 \times t}\)
• Total area = \(\mathrm{2 \times \frac{1}{2} \times 135 \times t = 135t}\)

4. SIMPLIFY to find the tangent length

• Set up equation: \(\mathrm{135t = 24{,}300}\)
• Solve for t: \(\mathrm{t = 24{,}300 \div 135 = 180\,mm}\)

5. INFER how to find distance OP

• In right triangle OTP: OP is the hypotenuse
• We know both legs: \(\mathrm{OT = 135\,mm}\) and \(\mathrm{TP = 180\,mm}\)
• Use Pythagorean theorem: \(\mathrm{OP^2 = OT^2 + TP^2}\)

6. SIMPLIFY the final calculation

• \(\mathrm{OP^2 = 135^2 + 180^2}\)
  \(\mathrm{OP^2 = 18{,}225 + 32{,}400}\)
  \(\mathrm{OP^2 = 50{,}625}\)
• \(\mathrm{OP = \sqrt{50{,}625} = 225\,mm}\) (use calculator)

Answer: D. 225

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students struggle to visualize that the quadrilateral OTPS decomposes into two right triangles, or they attempt to find the area using incorrect formulas like treating OTPS as a rectangle or trapezoid. Without recognizing the tangent perpendicularity creating right triangles, they cannot set up the correct area equation \(\mathrm{135t = 24{,}300}\).

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{135t = 24{,}300}\) and find \(\mathrm{t = 180}\), but make arithmetic errors when computing \(\mathrm{OP^2 = 135^2 + 180^2}\). They might calculate \(\mathrm{135^2 = 18{,}225}\) correctly but err on \(\mathrm{180^2 = 32{,}400}\), or make mistakes adding these values or taking the final square root.

This may lead them to select Choice B (180) if they confuse the tangent length with the distance OP, or Choice C (210) from calculation errors.

The Bottom Line:

This problem requires strong spatial reasoning to decompose the quadrilateral and recognize the right triangle relationships, combined with careful arithmetic in the final calculations. Students who miss either the geometric insight or make computational errors will struggle to reach the correct answer.