Question

In an isosceles triangle \( PQR \) with \( PQ = PR \), PT and PZ trisect \( QR \) such that \( QT = TZ = ZR \). If \( \angle TPZ = \angle PQR \) and the area of triangle \( PQR \) is equal to \( \frac{27\sqrt{7}}{4} \), then find \( PR \).

• A. 3
• B. 8
• C. 9
• D. 6

Hint:

In problems involving isosceles triangles, remember that the height divides the base into two equal parts and helps in calculating the area. Use these properties to solve such problems effectively.

Answer 1

The Correct Option is D

- We are given that \( QT = TZ = ZR = n \) and \( QR = 3n \), so the base \( QR \) is divided into three equal parts.
- \( \triangle APQT \equiv \triangle APRZ \) by SAS (Side-Angle-Side), hence \( PT = PZ \).
- \( \angle PTZ = \angle PQT + \angle ZQT \), and since \( \angle PQT = \angle ZPT \), we have: \[ \angle PTZ = \angle PQT + \angle ZQPT. \] - Given \( \angle PQT = \angle ZTPZ \) and \( \angle PZT = \angle ZPTZ \), the triangles are congruent. Hence, \( PQ = QZ = 2n \) and \( PQ = PR = 2n \).

Step 1: Height of the Isosceles Triangle The height \( h \) of the isosceles triangle can be computed using: \[ h = \frac{1}{4} \sqrt{4a^2 - b^2}. \] Substituting the values: \[ h = \frac{1}{4} \sqrt{16n^2 - 9n^2} = \frac{1}{4} \sqrt{7n^2}. \] Thus, the height of the triangle is \( \frac{\sqrt{7}n}{2} \).

Step 2: Area of Triangle PQR The area of triangle \( PQR \) is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. \] Substituting the base as \( 4n \) (since \( QR = 3n \) and \( n = 3 \)): \[ \text{Area} = \frac{1}{2} \times 4n \times \frac{\sqrt{7}n}{2}. \] Simplifying: \[ \text{Area} = \frac{27\sqrt{7}}{4}. \]

Step 3: Final Calculation for \( PR \) From the calculation, we have: \[ n = 3 \quad \text{and} \quad AC = 2n = 2(3) = 6. \] Thus, the value of \( PR = 6 \).
Final Answer: The correct answer is (d) 6.