Gunjo Pachauri

Solution Here Given, Length=32cmBreadth=21cmRequired length of wooden strip = Perimeter of photograph=2(length +breadth)=2(32+21)=2×53cm=106cmRequired Length of wooden strip = 106cm

Solution L = 2B – 1, Diagonal = 17. D² = L²+ B²⇒ 289 = (2B –1)²+ B²⇒ B = 8 & L = 15.Also 8, 15, 17 is a Pythagorean triplet.So answer can be reached directly.

Solution L. H. S. = (3x + 2)²= (3x)² + 2² + 2 × 2 × 3x= 9x² + 4 + 12xR.H. S. = 3x² + 6x + 4Therefore, L. H. S. ≠ R. H. S.Hence, correct statement is (3x + 2)² = 9x² + 4 + 12x

Solution Let Sum of zeroes (α + β) = S = -8 …[Given]Product of zeroes (αβ) = P = 12 …[Given]Quadratic polynomial is x² – Sx + P= x² – (-8)x + 12= x²+ 8x + 12= x² + 6x + 2x + 12= x(x + 6) + 2(x + 6)= (x + 2)(x +

Solution We have to solve these two equations5x + 4y = 20x + 2y = 4Let’s we pick the second equation,x = 4 – 2yNow substituting the value of x in the other equation.5(4 – 2y) + 4y = 2020 – 10y + 4y = 20-6y = 0y = 0Finding out the value of x

Solution The 2nd and the 7th terms are (a + d) and (a + 6d) respectively. The ratio of these terms is 1/3. Solving this ratio, we get 2a = 3d. The 5th term is (a + 4d) = 11. Substituting for a, we get a = 3 and d = 2. Therefore, the 15th

Solution The 2nd and the 6th terms are (a + d) and (a + 5d) respectively. The ratio of these terms is 2/5. Solving this ratio, we get 3a = 5d. The 8th term is (a + 7d) = 26. Substituting for a, we get a = 5 and d = 3. Therefore, the 10th

Solution Value of a² + b² + 3ab , at a = 1 and b = – 2= (1)2 + (–2)2 + 3 (1)(–2)= 1 + 4 – 6= 5 – 6= – 1

Solution Given,a = 5 + 2√6b = 1/a= 1/(5 + 2√6)Rationalizing the denominator, we get;b = [1/(5 + 2√6)] [(5 – 2√6)/(5 – 2√6)]= (5 – 2√6)/ [(5)2 – (2√6)2]= (5 – 2√6)/ (25 – 24)= 5 – 2√6Now,a² + b²= (a + b)² – 2ab= (5 + 2√6 + 5 – 2√6)2 – 2(5

Solution Let the breadth of floor be x metre. :. Length = (x + 20) metre :. Area of the floor =(x+ 20) x sq. metre According to question. (x + 10) (x + 5) = x (x + 20) ⇒x² + 15x + 50 = x² + 20x ⇒ 20x = 15x + 50