Geometric progression, anyone?

Question

Given the lengths of sides of a triangle are in a geometric progression. The length of the longest side is 36cm. If the perimeter of the triangle is 76cm, find the positive value of the common ratio.
That's all that's given in the ques. I tried assuming it as an isoceles but couldnt find too. Can anyone think of any idea please. Thank you!

Answer 1

Recall that a geometric progression goes as follows:

For a = the first term and r being the ratio, our sequence would be

{a, ar, ar^2, ar^3, ar^4, ..... }

Given that the longest side is 36cm, a geometric sequence containing 36 and two other values would be

{36, 36r, 36r^2}

The perimeter of a triangle is just the sum of the three sides. Since the three sides are of that geometric sequence,

P = 36 + 36r + 36r^2

76 = 36 + 36r + 36r^2

Let's reorder the terms in terms of descending r power.

36r^2 + 36r + 36 = 76

Moving 76 to the left hand side,

36r^2 + 36r - 40 = 0

Dividing both sides by 4,

9r^2 + 9r - 1 = 0

And we solve this using the quadratic formula.

r = [-9 +/- sqrt (81 - 4(9)(-1))] / [18]
r = [-9 +/- sqrt (81 + 36)] / [18]
r = [-9 +/- sqrt (117)] / [18]

This gives us two solutions:

r = { [-9 + sqrt (117)] / [18]} , { [-9 - sqrt (117)] / [18] }

However, we can discard the second solution because we only want the positive solution as per the question.

r = [-9 + sqrt (117)] / [18]

Answer 2

.

Let the 3 sides of the triangle be a, ar , ar^2

ar^2 = 36
a = 36 / r^2

a + ar + ar^2 = 76
a = 76 / ( 1 + r + r^2 )

36 / r^2 = 76 / ( 1 + r + r^2 )

36 ( 1 + r + r^2 ) = 76 * r^2

9 + 9r + 9r^2 = 19r^2
10r^2 - 9r - 9 = 0

r = (9 ± √81 + 360) / 20
r = (9 ± 21) / 20
r = (9 + 21) / 20 [ The ratio is said to be positive ]
r = 30/20
r = 1.5

a = 36 / (1.5)^2
a = 16

Hence, the 3 sides are 16 (a), 24 (ar) and 36 (ar^2).

.

Answer 3

since the sides are in geometric procession

the sides are a,ar,ar^2

the longest side is 36 there fore ar^2= 36

a = 36/r^2

given a+ar+ar^2 = 76

substitute a in this equation

36/r^2 + 36/r +36 = 76

36 + 36r=40r^2
40r^2 -36r +36=0
if we solve this equation
we get the answer
but
here
the roots are imaginary since B^2 - 4AC <0