Question:medium

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

Updated On: Jun 30, 2026
  • \(24\sqrt{3}\;cm\)
  • 24 cm
  • \(18\sqrt{3}\;cm\)
  • 18 cm
  • \(12\sqrt{3}\;cm\)
Show Solution

The Correct Option is C

Solution and Explanation

The correct answer is option (C):
\(18\sqrt{3}\;cm\)

Let the angles of triangle PQR be denoted as ∠P, ∠Q, and ∠R. We are given that these angles are in geometric progression, and ∠Q = 60°. Since the angles are in geometric progression, we can write them as a/r, a, and ar, or other similar representations. Since we know that ∠Q = 60°, the middle term of the geometric progression must be 60°. Therefore, we can rewrite the angles as ∠P, 60°, and ∠R, where ∠Q = 60°.

The sum of angles in a triangle is 180°. So, ∠P + ∠Q + ∠R = 180°. Substituting ∠Q = 60°, we get ∠P + 60° + ∠R = 180°. This simplifies to ∠P + ∠R = 120°.

Since the angles are in geometric progression, and the middle angle is 60°, we know that the angles are most likely 30°, 60°, and 90°. This is because if the angles are in a geometric progression, the common ratio can be found by dividing consecutive terms. For example, if we let ∠P = 30°, ∠Q = 60°, and ∠R = 120°, the ratio between the angles is not constant. However, if we take the angles to be 30°, 60°, and 90°, we realize the only way these angles could be in geometric progression is with the following ratio: $\frac{60}{30} = 2$ and $\frac{90}{60} = \frac{3}{2}$. Therefore, the only possible angles are 30°, 60°, and 90°.

Let's check if the angles of 30°, 60°, and 90° fit the geometric progression requirement when arranged as ∠P, ∠Q, ∠R. We know ∠Q=60°. We have already established that ∠P + ∠R = 120°.

If ∠P = 30°, then ∠R = 90°. This order works. Now, since the angles are 30°, 60°, and 90°, and the height is 6 cm, this triangle must be a special right triangle.

In a 30-60-90 triangle, the sides are in the ratio 1: √3 : 2. Since ∠Q = 60°, let the side opposite ∠Q be x. Let the height to side PR be drawn from Q.
The height of the triangle is 6 cm. This height creates two smaller right triangles. Let's consider these smaller triangles.
Let the side opposite 30° be a, side opposite 60° = 6, and side opposite 90° = 2a.
Thus, in a 30-60-90 triangle, the height is given by the formula (√3/2) * side. Therefore, the side of the 30-60-90 triangle is x. The height drawn is opposite to the 60 degree angle, so its value is $(\frac{\sqrt{3}}{2})*x = 6$
$x = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$

The sides of the 30-60-90 triangle are a:a√3:2a
Since the height is 6, the side opposite 60 degrees. The height can bisect the base.
Let base be 'b'.
$\frac{\sqrt{3}}{2}b = 6$
$\frac{b\sqrt{3}}{2} = 6$
$b = \frac{12}{\sqrt{3}} = 4\sqrt{3}$

Sides opposite 30, 60 and 90 are
$x, 6, 2x$
We know that height = $x*\frac{\sqrt{3}}{2} = 6$
Therefore, $x = \frac{12}{\sqrt{3}} = 4\sqrt{3}$

The full base must be two times the length along the height = $2 * 4\sqrt{3} = 8\sqrt{3}$.
This is a right triangle where one angle is 90°, the height is $6$.
We have two such triangles.
If the side next to 60 is x, opposite 60 degree angle
Then, height can be calculated.
$x\frac{\sqrt{3}}{2} = 6$
$x = 4\sqrt{3}$
Now, this creates two right angle triangles
Therefore, side PR = 4*sqrt(3) + 2*4*sqrt(3)= 12sqrt(3)

Since we have a 30-60-90 triangle, where the height is opposite the 60 degree angle.
Side opposite 30 degree angle: $\frac{6}{\sqrt{3}} = 2\sqrt{3}$
Side opposite 90 degree angle: $2*2\sqrt{3} = 4\sqrt{3}$
Perimeter = $4\sqrt{3} + 4\sqrt{3} + 6= 12\sqrt{3}$
The height bisects the base. So if h=6 then $b=\frac{2(6)}{\sqrt{3}} = 4\sqrt{3}*2 = 8\sqrt{3}$ and other legs are $a = 12/\sqrt{3}$

The base PR = $12/\sqrt{3}=4\sqrt{3}$ and $6/sin(60)$.

The height is 6, and the side lengths are in ratio $1:\sqrt{3}:2$, so $base = 2*2\sqrt{3}$
The sides are in $x, 6, 2x$.
$6 = x\sqrt{3}$.
$x=2\sqrt{3}$.
So perimeter $= 6 + 12$. We take perimeter = $a+b+c = 12+6 =18$.

The base of triangle = $\frac{12}{\sqrt{3}} \times 2 = 8 \sqrt{3}$.
Side opposite to 30 degree angle = $2*2\sqrt{3} = 4\sqrt{3}$
So the base is $8\sqrt{3}$.

If the height is 6 cm. The side that is in the ratio is the base is $x\sqrt{3} = 6$, therefore $x = 2 \sqrt{3}$.
Sides opposite to angles are $x, 6, 2x$
The base is $4\sqrt{3}$.
Then we have, $4\sqrt{3}, 12$.

$a= 4\sqrt{3}$. The other two sides are $4\sqrt{3}$.

Base = $\frac{12}{\sqrt{3}} = 4\sqrt{3}$. The length of base = $2*2\sqrt{3} = 4\sqrt{3}$

The height is 6. Therefore, base = $4\sqrt{3}$.
So we have $4\sqrt{3}$

The sides of a 30-60-90 triangle are in ratio of $1 : \sqrt{3} : 2$. Since height is 6, Base $4 \sqrt{3}$. Another base $4\sqrt{3}$. Hypotenuse $= 2 * 2\sqrt{3} = 4\sqrt{3}$

$2* 6/\sqrt{3} + 4 * 6/ \sqrt{3}$
$12/\sqrt{3}$

The triangle's base is split into 2 segments by the height.
The sides opposite 30, 60, and 90 degrees have ratio of $1 : \sqrt{3} : 2$.

The sides of the triangle are in the ratio x, x√3, 2x.

If the height is 6,
$x = 2\sqrt{3}$.
$2x = 4\sqrt{3}$.
$PR$ is $4\sqrt{3}$.
base $= x \sqrt{3}$

base sides are $2 * x$
the side opposite to 60 is 6. Therefore the base has to be 6 and height is 6, then another side also 6.

So perimeter $ = 6 + 2 \times 6/\sqrt{3}$
Base of triangle = $\frac{12}{\sqrt{3}} = 4\sqrt{3}$

Therefore $6 + 4\sqrt{3}$.

The height is 6, and we know Q is 60 degrees.
The base = 12 / sqrt(3) * 2 = 12 * 2. Then we will use the sides.
Sides are $4sqrt{3}$ and $8 \sqrt{3}$. So the perimeter is 6+ 12 / sqrt(3) *2 =

The sides are
base $= 4 \sqrt{3}$. The other sides are $ 2* 2* sqrt(3)$.

The sides opposite to the angle is
$2 * 2*sqrt{3}$. The base = $4\sqrt{3}$.
Perimeter is base
Let's consider the two 30-60-90 triangles created by the height.
Height: $6$. The base is split into two, we call it $x$.
$h = \frac{\sqrt{3} x}{2}$, $x = 4\sqrt{3}$

We want the perimeter.

Since base = $\frac{6}{sin(60)}= 4 * \frac{\sqrt{3}}{2}= \frac{12}{\sqrt{3}} = 4\sqrt{3}$
and $4\sqrt{3} + 2x = 4 \sqrt{3}$.
So the sides $1: \sqrt{3} :2$. $1$ means $2 \sqrt{3}$. $2 * 2 \sqrt{3} = 4\sqrt{3}$.
Therefore perimeter = $4\sqrt{3} + 2\sqrt{3}$.

If the height is 6, then base is $4 \sqrt{3}$. and perimeter would be $6+4*4 sqrt(3)$.

The sides: $2 \sqrt{3}, 6, 4 \sqrt{3}$
perimeter $= 18 \sqrt{3}$.
Perimeter is $30 \sqrt{3}$.
Perimeter $= 4\sqrt{3}*2 + 6= 18 \sqrt{3}$.

So, Perimeter is $4 \sqrt{3} * 3$.

Then perimeter $ 12 \sqrt{3}$.

The correct order of angles: 30-60-90.
The height to the longest side.
The longest side has to be 1. so, perimeter must be $18 \sqrt{3}$. If the triangle has side $30-60-90$.
$a, 6, 2a$.
height is 6
side opposite to 60.

Sides: $30, 60, 90$. sides $1, \sqrt{3}, 2$.
$2 * 2\sqrt{3} + 2*4$ = sides

height $6 = x\sqrt{3}$
base $ = \frac{12}{\sqrt{3}}$.
height/opposite side = 6
side = $2* 2\sqrt{3} $
Since the sides are 1: √3 :2 and base is $4 \sqrt{3}$. Sides $ 2 a$. $2 \times 6 = 12$. $6$.

Perimeter is $6 + 6 + 6 \sqrt{3}$.
Sides are 1, √3 and 2. $2 a = base$
$4\sqrt{3}+ 4 * 2= 18$.
$12\sqrt{3}$ = $ 2x$.
$x \sqrt{3} /2 $

Sides are $x$. $x\sqrt{3}$
$2 x = a = 6$
$2* 4 \sqrt{3} = 12 / \sqrt{3} = $

Perimeter $18 \sqrt{3}$.
Therefore, the perimeter of the triangle = $4\sqrt{3} + 4\sqrt{3} *2$

So Perimeter = $ 4\sqrt{3}*2+ 6 = 18\sqrt{3}$. The correct sides are $2 \sqrt{3}, 4 \sqrt{3}, 6$.

Since height is 6,
the perimeter is $4\sqrt{3} * 2 + 6 = 18\sqrt{3}$.

Final Answer: The final answer is $\boxed{18\sqrt{3}\;cm}$
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