Shikhakaushal

By shikha kaushal

You can make model also on it

CBSE NCERT Solutions for Class 10 Mathematics Chapter 1 Back of Chapter Questions 1. Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Solution: (i) 135 and 225 Step 1: Since 225 is greater than 135, we can apply Euclid's division lemma to a = 225 and b = 135 to find q and r such that 225 = 135q + r, 0 ≤ r < 135 So, dividing 225 by 135 we get 1 as the quotient and 90 as remainder. i. e 225 = (135 × 1) + 90 Step 2: Remainder r is 90 and is not equal to 0, we apply Euclid's division lemma to b = 135 and r = 90 to find whole numbers q and r such that 135 = 90 × q + r, 0 ≤ r < 90 So, dividing 135 by 90 we get 1 as the quotient and 45 as remainder. 𝑖𝑖. 𝑒𝑒 135 = (90 × 1) + 45 Step 3: Again, remainder r is 45 and is not equal to 0, so we apply Euclid's division lemma to b = 90 and r = 45 to find q and r  such that 90 = 45 × q + r,   0 ≤ r < 45 So, dividing 90 by 45 we get 2 as the quoti

Revision Notes on Circle The equation of a   circle   with its center at C(x 0 , y 0 ) and radius r is: (x – x 0 ) 2  + (y – y 0 ) 2  = r 2     If x 0  = y 0  = 0 (i.e. the centre of the circle is at origin) then  equation of the circle   reduce to x 2  + y 2  = r 2 . If r = 0 then the circle represents a point or a point circle. The equation x 2  + y 2  + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius √(g 2 +f 2 -c). Equation of the circle with points P(x 1 , y 1 ) and Q(x 2 , y 2 ) as extremities of a diameter is (x – x 1 ) (x – x 2 ) + (y – y 1 )(y – y 2 ) = 0.  For general circle, the equation of the chord is x 1 x + y 1 y + g(x 1  + x) + f(y 1  +y) + c = 0 For circle x 2  + y 2  = a 2 , the equation of the chord is x 1 x + y 1 y = a 2   The equation of the  chord  AB (A ≡ (R cos α, R sin α); B ≡ (R cos β, R sin β)) of the circle x 2  + y 2  = R 2  is given by x cos ((α + β )/2) + y sin ((α - β )/2) = a cos (