Saved Bookmarks
| 1. |
enamplesExample 4: In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularflower bed is the point of intersection O of thediagonals of the square lawn, find the sum of theareas of the lawn and the flower beds.56mFig. 12.15 |
|
Answer» Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)Diagonal of a square (AC) =√2×side of a square.Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]Let OA = OB = r m (ràdius of sector)Area of sector OAB = (90°/360°) πr² Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²Area of sector OAB = 1232 m²Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²Area of flower bed AB = 448 m²Similarly, area of the other flower bed CD = 448 m²Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²Hence, the sum of the areas of the lawns and the flower beds are 4032 m². Like if you find it useful! |
|