Area Problems
One diagonal of a rhombus is 60% of the other diagonal. Then, area of the rhombus is how many times the square of the length of the larger diagonal?
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.
let the side of the square be x meterslength of two sides = 2x metersdiagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0 . 59 x 2 x * 100 % = 30% (approx)
Area of the room=(544 * 374) size of largest square tile= H.C.F of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) c m 2 Number of tiles required== [(544 x 374) / (34 x 34)] = 176
Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %
Let the length, breadth and height of the room be l, b and h respectively As per question Cost of 2(l + b) x h = Rs. 48 ? Required cost = cost of 2 (2l + 2b) x 2h= cost of 4 [2(l + b) x h ]= 4 x Rs. 48= Rs. 192
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