Q1. Complete the Following Statements: (i) Probability of an event E + Probability of the event ‘not E’ = 1 (ii) The Probability of an event that can not happen is 0 such an event is called an impossible event. (iii) The probability of an event that is certain to happen is 1 such an event is…
Q.1 State which pairs of tringles in fig. 6.34 are similar. write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic. (i) Solution: In Δ ABC and Δ PQR ∠A =∠P =60° ∠B =∠Q =80° ∠C=∠R =40° ∴ Δ ABC ∼ Δ PQR…
Q.1 In fig.6.17,(i)and(ii),DE || BC. find EC in(i)and AD in (ii) (i) Solution: According to figure (i) in ΔABC DE || BC { given } By B.P.T. [ In a triangle if a line is parallel to one side divides remaining to sides in equal ratio ] AD / BD = AE / EC 1.5/3…
Arithmetic progressions: Basic concepts Sequence: Some numbers arranged in a definite order, according to a definite rule, are said to form a sequence. ex (I) 1, 2, 3,……… (II) 100, 70, 40, 10 ……… (III) -1·0, -1·5, -2·0, -2·5, ……. Arithmatic progressions (AP): An arithmetic progression is a list of numbers in which each term…
Q.1 Find the area of the triangle whose vertices are : (i) (2,3),(-1,0),(2,-4) (ii) (-5,-1),(3,-5),(5,2) Solution: (i) (2,3),(-1,0),(2,-4) Let The co-ordinate of vertices of the triangle are A (2, 3), B (-1, 0), C(2, -4) x1 = 2 ; y1 = 3 x2 = -1 ; y2 = 0 x3 = 2 ; y3 =…
Quadratic formula / Shreedharacharya sutra: x=-b±√b²-4ac/2a For Finding Nature of roots: First we find Discriminant (D) : b²- 4ac There are three different cases Case I : b²-4ac<0 Nature of Roots : Imaginary Roots/ No Real roots Case II : b² – 4ac = 0 Nature of Roots : Real and two different roots Case III…
Q1. Find the roots of the following quadratic equations by factorization Method: (I) x² -3x-10 =0 by factorization method: ⇒ x² – 3x – 10 = 0 ⇒ x² – ( 5 – 2 )x – 10 = 0 ⇒ x² – 5x + 2x – 10 = 0 ⇒ x( x – 5 ) +…
Q.1. Check whether the following are quadratic equations ÷ (i) (x + 1)2 = 2 (x – 3) Solution: (x + 1)2 = 2 (x – 3) [By Applying ( a + b )² = a²+2ab+b2] ⇒ x²+2x+1= 2x – 6 ⇒ x²+2x+1 – 2x + 6 = 0 ⇒ x² + 7 = 0…
Important formula for this Exercise: Sum of n terms of AP= Sn (i) Sn = n/2[2a+(n-1)d] (ii) Sn = n/2[a + an ] (iii) Sn = n/2[a + l ] (iv) Sum of first n positive integers = Sn = n(n+1)/2 Q.1 Find the sum of the Following APs; (i) 2, 7,12,…………to 10 terms a = first term = 2…
Q1 Fill in the blanks in the following table, given that a is the first term d, the common difference and an the nth term of the AP: (i) a = 7 d = 3 n = 8 an = ? Solution: nth term of AP an = a + (n-1)d a8 = 7 +…