Area Problems
Area of rectangular field is 3584 m 2 and the length and the breadth are in the ratio 7 : 2, respectively. What is the perimeter of the rectangle?
Let length of the rectangular field = 7k m and breadth of the rectangular field = 2k mAccording to the question,Area of a rectangular field = Length x Breadth? 3584 = 7k x 2k ? 14 x k2 = 3584 ? k2 = 3584/14 = 256? k2 = 256 = 16 m? Length of rectangular field = 7k = 7 x 16 = 112 mAnd breadth of rectangular field = 2 x 16 = 32 m? Perimeter of rectangle = 2(Length x Breadth)= 2(112 + 32) = 2 x 144 = 288 m
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
Let a = 13, b = 14 and c = 15. Then, s = 1 2 a + b + c =21 (s- a) = 8, (s - b) = 7 and (s - c) = 6. Area = s s - a s - b s - c = 21 × 8 × 7 × 6 = 84 sq.cm
According to the question,l/[2(l + b)] = 5/18? 10l + 10b = 18l = 10b? l/b = 10/8 = 5/4? l : b = 5 : 4Hence, ratio of length and breadth of a rectangle is 5 : 4
Area of the room =(544 x 374) cm2size of largest square tile = H.C.F. of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) cm2? Least number of tiles required = (544 x 374) / (34 x 34) = 176
EXAMIANS