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
Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2) = 2(15 + 2) = 34 cm
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
Cross section area = 1/2 x ( a + b ) x d where a and b are the parallel sides, d is the perpendicular distance between them.? 1/2 x ( a + b ) x d = 640? d = (640 x 2) / 16 = 80m
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
EXAMIANS