Question

In ΔPQR, \(∠P\), \(∠Q\) and \(∠R\) are in geometric progression in the same order. \(∠Q\) = \(60°\). What is the perimeter of the triangle if the height of the triangle is \(6\) cm

• A. \(24\sqrt{3}\;cm\)
• B. 24 cm
• C. \(18\sqrt{3}\;cm\)
• D. 18 cm
• E. \(12\sqrt{3}\;cm\)

Answer 1

The Correct Option is C

To find the perimeter of the triangle \(\Delta PQR\), where \(\angle P\), \(\angle Q\), and \(\angle R\) are in geometric progression, and \(\angle Q = 60^\circ\), let's solve this step by step. 

1. If the angles \(\angle P\), \(\angle Q\), and \(\angle R\) form a geometric progression, we can express them as:
   • \(\angle P = a\),
   • \(\angle Q = ar\),
   • \(\angle R = ar^2\).
2. Given: \(\angle Q = 60^\circ\). So, \(ar = 60^\circ\).
3. The sum of angles in a triangle is \(180^\circ\).
4. Thus, we have: \[ a + ar + ar^2 = 180^\circ \] Substitute \(ar = 60^\circ\): \[ a + 60 + ar^2 = 180 \] So, \[ a + ar^2 = 120 \]
5. Since \(a\), \(ar\), \(ar^2\) are in a geometric progression, we solve for \(r\) via the relation: \[ 60^2 = a \times ar^2 \] Or, \[ a \times ar = (ar)^2 \implies a^2 \times r^2 = \left(\frac{60}{r}\right)^2 \] Therefore, with simplifications and initially knowing \(ar = 60\), it suffices to use \(a + ar^2 = 120\) to determine: Using \( ar = 60 \): \[ a + ar^2 = 120 \] \[ a\left(1 + r\right)r = 120 \] But given only these conditions, it is more precise to use known \(ar = 60\) directly to determine \(r\) as \(1\) solving practical resolve, often equates \(a\) and proportionally via basic conditions or direct geometric sums and end at expected these specific calculations.
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