Area Problems
The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
Let the breadth of floor be 'b' m. Then, length of the floor is 'l = (b + 25)' Area of the rectangular floor = l x b = (b + 25) × b According to the question, (b + 15) (b + 8) = (b + 25) × b b 2 + 8 b + 15 b + 120 = b 2 + 25 b 2b = 120 b = 60 m. l = b + 25 = 60 + 25 = 85 m. Area of the floor = 85 × 60 = 5100 sq.m.
Let breadth = b meters. then, length = 3b/2 meters ? b x 3b/2 = 2/3 X 10000? b2 = (4 x 10000)/9? b = ( 2 X 100)/3 m ? Length = (3/2) x (2/3) x 100 m= 100 m
In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40
Let length = L and breadth = BLet , New breadth = ZThen, New length = ( 160 / 100) L.= 8L / 5? 8L / 5 x Z = LBor Z = 5B/8Decrease in breadth = (B-5B/8)= 3B/8? Decrease in percent = (3B/8 x1/B ) x 100 %= 371/2%
Let one diagonal be k.Then, other diagonal = (60k/100) = 3k/5 cmArea of rhombus =(1/2) x k x (3k/5) = (3/10) = 3/10 (square of longer diagonal)Hence, area of rhombus is 3/10 times.
EXAMIANS