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17) The perimeter of a right triangle is 60 m. Its hypotenuse is 25 m. Find the area of the triangle.
Solution :
Length of the hypotenuse of the triangle = 25 m
Let the lengths of other base and height be x, y respectively.
Perimeter of the triangle = 60 m
⇒ Sum of length of all sides = 60
⇒ x + y + 25 = 60
⇒ x + y = 60 - 25
⇒ x + y = 35
⇒ y = 35 - x
By pythagoras theorem
⇒ (Base)² + (Height)² = (Hypotenuse)²
⇒ x² + y² = 25²
⇒ x² + (35 - x)² = 625
⇒ x² + 35² - 2(35)(x) + x² = 625
⇒ 2x² + 1225 - 70x = 625
⇒ 2x² - 70x + 1225 - 625 = 0
⇒ 2x² - 70x + 600 = 0
⇒ 2(x² - 35x + 300) = 0
⇒ x² - 35x + 300 = 0
⇒ x² - 15x - 20x + 300 = 0
⇒ x(x - 15) - 20(x - 15) = 0
⇒ (x - 20)(x - 15) = 0
⇒ x - 20 = 0 or x - 15 = 0
⇒ x = 20 or x = 15
i.e Base = x = 20 or 15 cm
⇒ Height y = (35 - x) = 15 or 20 cm
Area of the triangle = 1/2 * Base * Height
= 1/2 * 20 * 15
= 300/2 = 150 cm²
Hence area of the triangle is 150 cm²
18) Half of the perimeter of the rectangular garden is 36 m. If the length is 4m more than its width. Find the dimensions of the garden.
Solution :
Let the width of the garden be x m
Length of the garden = (x + 4) m
Half the perimeter = 36 m
⇒ 1/2 [ 2(Length + Breadth) ] = 36 m
⇒ Length + Breadth = 36
⇒ x + 4 + x = 36
⇒ 2x + 4 = 36
⇒ 2x = 32
⇒ x = 16
Breadth = x = 16 m
Length = x + 4 = 16 + 4 = 20 m
Hence, 20 m and 16 m are the dimensions of rectangle.
19) The denominator if the fraction is 1 more than the twice the numerator. If the sum of the fraction and reciprocal is 2 16/21, find the fraction.
Solution :
Let the numerator of the fraction be x
Denominator = 2x + 1
Fraction = x/(2x + 1)
Reciprocal of the fraction = (2x + 1)/x
Sum of fraction and its reciprocal = 2 16/21 = 58/21
⇒ x/(2x + 1) + (2x + 1)/x = 58/21
Taking LCM
⇒ [ x(x) + (2x + 1)² ] / [ (2x + 1)x ] = 58/21
⇒ ( x² + 4x² + 4x + 1 ) / (2x² + x) = 58/21
⇒ ( 5x² + 4x + 1 ) / (2x² + x) = 58/21
⇒ 21(5x² + 4x + 1) = 58(2x² + x)
⇒ 105x² + 84x + 21 = 116x² + 58x
⇒ 0 = 116x² - 105x² + 58x - 84x - 21
⇒ 11x² - 26x - 21 = 0
⇒ 11x² - 33x + 7x - 21 = 0
⇒ 11x(x - 3) + 7(x - 3) = 0
⇒ (11x + 7)(x - 3) = 0
⇒ 11x + 7 = 0 or x - 3 = 0
⇒ x = - 7/11 or x = 3
[ Neglecting - 7/11 ]
⇒ x = 3
Fraction = x/(2x + 1) = 3 / [ 2(3) + 1] = 3/7
Hence, 3/7 is the required fraction.
20) For what value of k of QE has real and real roots : (k + 1)x² - 2(3k + 1)x + 8k + 1 = 0
Solution :
(k + 1)x² - 2(3k + 1)x + 8k + 1 = 0
For QE to have real and equal roots b² - 4ac = 0
Comparing with ax² + bx + c = 0 we get,
- a = k + 1
- b = - 2(3k + 1)
- c = 8k + 1
b² - 4ac = 0
⇒ [ - 2(3k + 1) ]² - 4(k + 1)(8k + 1) = 0
⇒ [2(3k + 1)]² - 4[ k(8k + 1) + 1(8k + 1) ] = 0
⇒ (6k + 2)² - 4( 8k² + k + 8k + 1 )= 0
⇒ (6k)² + 2(6k)(2) + 2² - 32k² - 36k - 4 = 0
⇒ 36k² + 24k + 4 - 32k² - 36k - 4 = 0
⇒ 4k² - 12k = 0
⇒ 4k(k - 3) = 0
⇒ k - 3 = 0
⇒ k = 3
Hence the value of k is 3.
21) For what value of k of QE has real and distinct roots : kx² + 6x + 1 = 0
For QE to have real and distinct roots b² - 4ac > 0
Comparing with ax² + bx + c = 0 we get,
- a = k
- b = 6
- c = 1
b² - 4ac > 0
⇒ 6² - 4(k)(1) > 0
⇒ 36 - 4k > 0
⇒ 36 > 4k
⇒ 36/4 > k
⇒ 9 > k
⇒ k < 9
Hence, any value of k is such that k is lesser than 9.
22) If the roots of the equation (b - c)x² + (c - a)x + (a - b) = 0 are equal, then prove that 2b = a + c
(b - c)x² + (c - a)x + (a - b) = 0
For QE to have equal roots b² - 4ac = 0
Comparing with ax² + bx + c = 0 we get,
- a = (b - c)
- b = (c - a)
- c = (a - b)
b² - 4ac = 0
⇒ (c - a)² - 4(b - c)(a - b) = 0
⇒ c² + a² - 2ac - 4[ b(a - b) - c(a - b) ] = 0
⇒ c² + a² - 2ac - 4(ab - b² - ac + bc) = 0
⇒ c² + a² - 2ac - 4ab + 4b² + 4ac - 4bc = 0
⇒ c² + a² + 2ac - 4ab + 4b² - 4bc = 0
⇒ (a + c)² + 4b² - 4ab - 4bc = 0
⇒ (a + c)² + (2b)² - 4b(a + c) = 0
⇒ (a + c)² + (2b)² - 2(2b)(a + c) = 0
⇒ [ (a + c) - (2b) ]² = 0
⇒ a + c - 2b = 0
⇒ a + c = 2b
⇒ 2b = a + c