Last updated at Feb. 25, 2017 by Teachoo

Transcript

Example 12 A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig.). Find its length and breadth. Given that area of rectangular park is 4m2 more than area of isosceles park So, Area of rectangle = 4 + Area of triangle x(x + 3) = 4 + 6x x2 + 3x = 4 + 6x x2 + 3x – 6x – 4 = 0 x2 – 3x – 4 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 1, b = – 3, c = – 4 We know that D = b2 – 4ac = (–3)2 – 4×1×(− 4) = 9 + 16 = 25 Hence , roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (− (− 3) ± √25)/(2 × 1) x = (+ 3 ± √52)/2 x = (3 ± 5)/2 Solving So, x = 4 & x = – 1 Since x is the breadth of the rectangle, x cannot be negative So, x = 4 is the solution Breadth = x = 4 m Length = x + 3 = 4 + 3 m = 7 m

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .