Area Problems
The ratio between the length and breadth of a rectangular field is 5:4. If the breadth is 20 meters less than the length, the perimeters of the field is?
Let length of rectangular field = 5y,so width = 4y.From question5y - 4y = 20m? x = 20m ? Length = (5 x 20)m = 100 mBreadth = (4 x 20) m = 80 m ? Perimeter = 2 (100 + 80 ) m = 360 m
Let l = 4k and b = 9kArea of rectangle = l x b144 = 4k x 9k ? k2 = 144/36 ? k2 = 4? k = 2? l = 8 cm and b = 18 cmPerimeter of rectangle = 2(l + b)= 2(8 + 18)= 2 x 26= 52 cm
Let lateral side = (5y) cm and base = (4y) cm ? perimeter = 5y + 5y + 4y = 14 ?y = 1So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm? Required Area = ? (7 x 2 x 2 x 3) cm2=2?21 cm2
Original breadth of rectangle = 720/30 = 24 cmNow , area of rectangle = (5/4) x 720 = 900 cm2? New length of rectangle = 900/24 = 37.5 cm? New perimeter of rectangle = 2(l+ b)= 2(37.5 + 24 ) = 2 x 61.5 = 123 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
EXAMIANS