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Example 12 - A rectangular park is to be designed whose - Solving by quadratic formula - Equation to be formed

Example 12 - Chapter 4 Class 10 Quadratic Equations - Part 2
Example 12 - Chapter 4 Class 10 Quadratic Equations - Part 3 Example 12 - Chapter 4 Class 10 Quadratic Equations - Part 4

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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 = ( )/2 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.