Ex 4.3, 11 - Chapter 4 Class 10 Quadratic Equations

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Ex 4.3 ,11 Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. Let side of square 1 be x metres Perimeter of square 1 = 4 × Side = 4x Now, it is given that Difference of perimeter of squares is 24 m Perimeter of square 1 – Perimeter of square 2 = 24 4x – perimeter of square 2 = 24 4x – 24 = perimeter of square 2 Perimeter of square 2 = 4x – 24 Now, Perimeter of square 2 = 4x – 24 4 × (Side of square 2 ) = 4x – 24 Side of square 2 = (4𝑥 − 24)/4 = (4(𝑥 − 6))/4 = x – 6  Hence, Side of square 1 is x & Side of square 2 is x – 6 Also, given that Sum of area of square is 468 m2 Area of square 1 + Area of square 2 = 468 (Side of square 1)2 + (Side of square 2)2 = 468 𝑥2+(𝑥−6)2=468 x2 + x2 – 2 ×𝑥×6+6^2=468 x2 + x2 – 12 x + 36 = 468 x2 + x2 – 12x + 36 – 468 = 0 2x2 – 12x – 432 = 0 Dividing both sides by 2 (2𝑥2 − 12𝑥 − 432)/2=0/2 x2 – 6x – 216 = 0  Comparing equation with ax2 + bx + c = 0, Here a = 1, b = –6, c = – 216 We know that D = b2 – 4ac D = (–6)2 – 4 × 1 × (– 216) D = 36 + 4 × 216 D = 36 + 864 D = 900 So, the roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (−(−6) ± √900)/(2 × 1)  x = (−(−6) ± √900)/(2 × 1) x = (6 ± √900)/2 x = (6 ± √(9 × 100))/2 x = (6 ± √(3^2 × 〖10〗^2 ))/2 x = (6 ± √(3^2 ) × √(〖10〗^2 ))/2 x = (6 ± 3 × 10)/2 x = (6 ± 30)/2 Solving  So, x = 18 & x = – 12 Since x is side of square and x cannot be negative So, x = 18 is the solution ∴ Side of square 1 = x = 18 m & Side of square 2 = x – 6 = 18 – 6 = 12 m