Question
There is a rectangular paper, whose length and breadth
are in the ratio 6:5, respectively and it has a perimeter of 44 metres. If a triangular piece with height of 9 metres and base of 16 metres is cut out from the given paper then find the area (in m2) of the remaining paper.Solution
Let the length and breadth of the rectangular be ‘6x’ metres and ‘5x’ metres, respectively Then, perimeter of the paper = 2 × (length + breadth) = 2 × (6x + 5x) = 22 x metres So, 22x = 44 So, x = (44/22) = 2 So, area of the entire paper = length × breadth = (6 × 2) × (5 × 2) = 12 × 10 = 120 m2 Area of the triangular piece = (1/2) × base × height = (1/2) × 9 × 16 = 72 m2 So, area of the remaining paper = 120 – 72 = 48 m2
Equation 1: x² - 45x + 500 = 0
Equation 2: y² - 60y + 600 = 0
I. x² - 208 = 233
II. y² + 47 - 371 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 4x² - 12x + 9 = 0
Equation 2: 2y² + 10y + 12 = 0
i) p2+p=56 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
II) q2-17q+72=0
...In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer Â
I. 3x² –7x + 4 = 0�...
I. x2 + 24x + 143 = 0
II. y2 + 12y + 35 = 0
I. 195x² - 46x - 21 = 0
II. 209y² + 13y - 12 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 24x + 143 = 0
Equation 2: y² - 26y + 165 = 0
I. 5x² = 19x – 12
II. 5y² + 11y = 12
I. 5x2 – 18x + 16 = 0
II. 3y2 – 35y - 52 = 0
