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 area = (22/7) x 9 x 9 cm2New area = (22/7) x 7 x 7 cm2? Decrease = 22/7 x [(9)2 -(7)2] cm2=(22/7) x 16 x 2 cm2Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %= 39.5 %
Area of the room =(544 x 374) cm2size of largest square tile = H.C.F. of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) cm2? Least number of tiles required = (544 x 374) / (34 x 34) = 176
perimeter = total cost / cost per m = 10080 /20 = 504mside of the square = 504/4 = 126mbreadth of the pavement = 3mside of inner square = 126 - 6 = 120marea of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq mcost of pavement = 246*6*50 = Rs. 73800
Let the breadth of the given rectangle be x then length is 2x. thus area of the given rect is 2 x 2 after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)= 2 x 2 + 5 x - 25 since new area is 75 units greater than original area thus 2 x 2 + 75 = 2 x + 5 x - 25 5x=75+25 5x=100 therefore x=20 hence length of the rectangle is 40 cm.
EXAMIANS