Area Problems
The perimeter of an isosceles triangle is equal to 14 cm . The lateral side is to the base in the ratio 5 : 4 the area of the triangle is?
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 circumference = 2?r New circumference = (150 /100) x 2 ?r = 3?r 2?R = 3?r? R = 3r/2 Original area = ?r2New area = ?R2= ?9r2 / 4 = 9?r2/4Increase in area = (9?r2/4 ) - (?r2)= (5/4) ?r2Req. increase per cent = [{(5/4) ?r2} / {?r2}] x 100 = 125 %
Let Length = 5y meters and breadth = 3y meters.Then, perimeter = 2 x (5y + 3y ) m = 16y meters ...(i)But perimeter = Total Cost / Rate = 3000 / 7.50 m = 400 m ...(ii)from eqs. (i) and (ii)16y = 400? y = 25? Length - Breadth = (5 x 25 - 3 x 25 ) m = 2 x 25 m = 50 m
EXAMIANS