Area Problems
The perimeter of an isosceles triangle is 26 cm while equal sides together measure 20 cm. The third side and each of the equal sides are respectively.
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
AB = 60 m, BC = 40 m and AC = 80 m ? s = (60 + 40 + 80 ) / 2 m = 90 m (s-a) = 90 - 60 = 30 m, (s-b) = 90 - 40 = 50 m and (s-c) = 90 - 80 = 10 m? Area of ? ABC = ?s(s-a)(s-b)(s-c) = ?90 x 30 x 50 x 10 m2= 300?15 m2? Area of parallelogram ABCD = 2 x area of ? ABC = 600?15 m2
Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
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
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