Area Problems
The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
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 length of the longer diagonal = d cmThen, length of other diagonal = 0.8 x d cm Area of rhombus = (1/2) x d x 0.8 x d = 2/5 d2= 2/5 d2Area of square of the length of the longer diagonal = d2So the area of the rhombus is 2/5 times the square of the length of the longer diagonal.
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
EXAMIANS