Question

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

Answer 1

Let the diameter of the three circles be 3d, 5d, and 6d respectively.

Now

π x 3d + π  x 5d + π  x 6d = 308
                                 14πd = 308
                                       d = 7

radius of the smallest circle = `21/2` = 10.5

                                    Area = π  x (10 .5)²
                                            = 346.5

radius of the largest circle = `42/2` = 21

                                  Area = π  x ( 21 )²
                                          = 1386
                         difference = 1386 - 346.5
                                          = 1039.5 cm²  

Answer 2

Let the diameter of the first circle, d₁ = 3x cm

Then diameter of the second circle, d₂ = 5x cm

Diameter of the third circle, d₃ = 6x cm

Now, sum of circumference of 3 circles = 308 cm

⇒ πd₁+ πd₂ + πd₃ = 308

⇒ π[d₁ + d₂ + d₃] = 308

⇒ `22/7`[3x + 5x + 6x ] = 308

⇒ `22/7` × 14x = 308

⇒ 44x = 308

⇒ x = 7

Now, diameter of smallest circle, d₁ = 3x = 3 × 7 = 21 cm radius of smallest circle, r₁ = `d_1/2 = (21/2)` cm

area of smallest circle = πr₁² = `22/7 xx (21/2)^2` = 346.5cm²

Diameter of the largest circle, d₂ = 6x = 6 × 7 = 42 cm

Now, radius of the largest circle, r₂ = `(d_2/2)` = 21 cm

area of largest circle = πr₂² = `22/7 xx (21)^2` = 1386 cm²

Difference between the areas = area of largest circle − area of smallest circle = 1386 − 346.5 = 1039.5 cm²