Saved Bookmarks
| 1. |
A wire bent in the form of a circle of radius 42 cm is cut and again bent in the form of a square. Find the ratio of the areas of the regions enclosed by the circle and the square. |
|
Answer» Since, Wire is bent in the form of a circle of radius 42 cm. ∴ The wire length = the circumference of circle = 2π×42 = 2 × \(\frac{22}{7}\) ×42 = 44 × 6 = 264 cm. Given that, The wire again bent in form of a square. Let the side length of square is a cm. Therefore, The perimeter of square = length of the wire. ∴ 4a = 264 cm ⇒ a = \(\frac{264}{4}\) = 66 cm. Hence, The area of square = a2 = (66)2cm2. And the area of circle = πr2 = \(\frac{22}{7}\) × 42 × 42 = 22 × 6 × 42 cm2. Now, \(\frac{Area\,of\,circle}{Area\,of\,square}\) = \(\frac{22\times 6\times 42}{66\times 66}\) = \(\frac{22}{3\times 11}\) = \(\frac{14}{11}\) Hence, The ratio of areas of the regions enclosed by the circle and the squares is 14:11. |
|