can anyone please solve questions no. 2,3,6. tomorrow is my maths exam
Answers
2. Let the area of square - 1 be A₁ and the area of square - 2 be A₂
given that the sum of two areas = 468 cm²
⇒ A₁ + A₂ = 468 cm²
let the side of square - 1 be S₁ and the side of square -2 be S₂
we know that area of square = S²
⇒ (S₁)² + (S₂)² = 468 cm²
we know that perimeter of square is given by 4S
given that the difference of the Perimeter of square -1 and square - 2 is 24
⇒ 4S₁ - 4S₂ = 24
⇒ S₁ - S₂ = 6
squaring on both sides of above equation we get :
⇒ (S₁ - S₂)² = 36
⇒ (S₁)² + (S₂)² - 2S₁S₂ = 36
but we know that (S₁)² + (S₂)² = 468
⇒ 468 - 2S₁S₂ = 36
⇒ 2S₁S₂ = 432
Now let us consider the equation (S₁)² + (S₂)² = 468
let us add 2S₁S₂ on both sides of the above equation :
⇒ (S₁)² + (S₂)² + 2S₁S₂ = 468 + 2S₁S₂
⇒ (S₁ + S₂)² = 468 + 432
⇒ (S₁ + S₂)² = 900
⇒ (S₁ + S₂) = 30
and we know that (S₁ - S₂) = 6
adding both equations (S₁ - S₂) = 6 and (S₁ + S₂) = 30 we get :
⇒ (S₁ + S₂) + (S₁ - S₂) = 30 + 6
⇒ 2S₁ = 36
⇒ S₁ = 18
substituting S₁ = 18 in equation (S₁ - S₂) = 6
we get 18 - S₂ = 6
⇒ S₂ = 12
So the sides of two squares are 18 and 12
3. 1/(a + b + x) = 1/a + 1/b + 1/x
⇒ 1/(a + b + x) - 1/x = 1/a + 1/b
taking LCM on both sides we get :
⇒ (x - a - b - x)/x(a + b +x) = (a + b)/ab
⇒ -(a + b)/x(a + b +x) = (a + b)/ab
(a +b) on both sides get cancelled
⇒ -1/x(a + b +x) = 1/ab
⇒ -ab = xa + xb + x²
⇒ x² + ax +bx + ab = 0
⇒ x(x + a) + b(x + a) = 0
⇒ (x + a)(x + b) = 0
⇒ x = a or x = b
6. if a quadratic equation Ax² + Bx + C = 0 has equal roots then B² - 4AC = 0
in the above question we can notice that B = 2mnc , A = (1 + m²)n² , C = c² - a²
substituting the above values in B² - 4AC = 0 we get
⇒ 4m²n²c² - 4n²(1 + m²)(c² - a²) = 0
4n² is taken common and made zero
⇒ m²c² - (c² - a² + m²c² - m²a²) = 0
⇒ m²c² - c² + a² - m²c² + m²a² = 0
⇒ m²a² + a² = c²
⇒ a²(m² + 1) = c²