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Example 12 - Chapter 4 Class 10 Quadratic Equations (Term 2)
Last updated at May 29, 2018 by Teachoo
Examples
Example 1 (i)
Important
Example 1 (ii)
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11
Example 12 Important You are here
Example 13 (i)
Example 13 (ii) Important
Example 13 (iii)
Example 14 (i)
Example 14 (ii)
Example 15 Important
Example 16
Example 17 Important
Example 18
Chapter 4 Class 10 Quadratic Equations
Serial order wise
- Ex 4.1
- Ex 4.2
- Ex 4.3
- Ex 4.4
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Examples
- Case Based Questions (MCQ)
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 = ( )/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
