The diameters of three circles are in the ratio 3 : 5 : 6. If the sum of the circumferences of these circles be 308 cm; find the difference between the areas of the largest and the smallest of these circles.

Let the diameters of the three circles be 3a, 5a and 6a.

Radius of three circles = $\dfrac{3a}{2}$, $\dfrac{5a}{2}$ and $\dfrac{6a}{2}$

Circumference of a circle = 2πr

For the first circle,

r₁ = $\dfrac{3a}{2}$ cm

Circumference₁ = 2 x π x $\dfrac{3a}{2}$ = 3aπ cm

For the second circle,

r₂ = $\dfrac{5a}{2}$ cm

Circumference₂ = 2 x π x $\dfrac{5a}{2}$ = 5aπ cm

For the third circle,

r₂ = $\dfrac{6a}{2}$ cm

Circumference₂ = 2 x π x $\dfrac{6a}{2}$ = 6aπ cm

Total circumference = Circumference₁ + Circumference₂ + Circumference₃

⇒ 3aπ + 5aπ + 6aπ = 308

⇒ 14aπ = 308

⇒ 14 x a x $\dfrac{22}{7}$ = 308

⇒ 2 x a x 22 = 308

⇒ 44a = 308

⇒ a = $\dfrac{308}{44}$

⇒ a = 7 cm

Radius of three circles = $\dfrac{3}{2}$ x 7 cm, $\dfrac{5}{2}$ x 7 cm and $\dfrac{6}{2}$ x 7 cm

= 10.5 cm, 17.5 cm and 21 cm

Difference between the area of the largest and the smallest circles = π(21)² - π(10.5)²

= 441π - 110.25π cm²

= 330.75π cm²

= 330.75 x $\dfrac{22}{7}$ cm²

= 47.25 x 22 cm²

= 1039.5 cm²

Hence, the difference in the area = 1039.5 cm².