Area Problems
Floor of a square room of side 10 m is to be completely covered with square tiles, each having length 50 cm. The smallest number of tiles needed is?
Area of square room = (10)2 = 100 sq m= 100 x (100)2 sq cm= 100 x 100 x 100 sq cmNow, area of tiles = (50)2 = 50 x 50 sq cm? Number of tiles needed = (Area of square room) / (Area of tile)= (100 x 100 x 100) / (50 x 50) = 400Hence, 400 tiles will be needed.
Let there be n sides of the polygon. Then it has n vertices. The total number of straight lines obtained by joining n vertices by talking 2 at a time is nC2 These nC2 lines also include n sides of polygon. Therefore, the number of diagonals formed is nC2 - n. Thus, nC2 - n = 44? [n(n - 1)/2] - n = 44? ( n2 - 3n) / 2 = 44? n2 - 3n = 88 ? n2 - 3n - 88 = 0 ?(n - 11) (n + 8) = 0 ? n = 11
Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%
Area of square = 40 x 40 = 1600 sq.cm Given that the areas of Square and Rectangle are equal => Area of rectangle = 1600 Sq.cm We know that, Area of rectangle = L x B Given L = 64 cm Breadth of rectangle = 1600/64 = 25 cm Perimeter of the rectangle = 2(L + B) = 2(64+25) = 178 cm.
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
Given that, l = 2b [Here l = length and b = breadth]Decrease in length = Half of the 10 cm = 10/2 = 5 cmIncrease in breadth = Half of the 10 cm = 10/2 = 5 cm Increase in the area = (70 + 5) = 75 sq cm According to the question, (l - 5) (b + 5) = lb + 75 ? (2b - 5) (b + 5) = 2b2 + 75 [since l = 2b]? 5b - 25 = 75 ? 5b = 100? b = 100/ 5 = 20? l = 2b = 2 x 20 = 40 cm
EXAMIANS