Last updated at Dec. 13, 2024 by Teachoo
Question 6 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
Examples
Example 1 (ii)
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6
Example 7
Example 8 Important
Example 9
Question 1
Question 2 Important
Question 3 Important
Question 4
Question 5
Question 6 Important You are here
Question 7 (i)
Question 7 (ii) Important
Question 7 (iii)
Question 7 (i)
Question 7 (ii)
Question 8 Important
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo