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Can (x  2) be the remainder on division of a polynomial p(x) by(2x +3) ? Justify
Answer: No , (x  2) cannot be the remainder on division of a polynomial p(x) by(2x +3) Because degree of remainder must be less than degree of Divisor. Here degree of x2 and 2x3 is same that is 1 10th Math's Chapter 14 Statistics – Mean, Mode and Median of Grouped Data and Graphical Representation of Cumulative Frequency Distribution
10th Triangle (Similarity) : Problems and solution to excel in exam
Similar figures: “Two similar figures have the same shape but not necessarily the same sizes are called similar figures. “ This verifies that congruent figures are similar but the similar figures need not be congruent. Conditions for similarity of polygon: Two polygons of the same number of sides are similar, if (i) Their corresponding angles are equal and (ii) Their corresponding sides are in the same ratio (or proportion). Note: The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons. Equiangular triangles: If corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: “The ratio of any two corresponding sides in two equiangular triangles is always the same.” Q. The Basic Proportionality Theorem (now known as the Thales Theorem) : “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. “ [Prove it.] Q. The converse of The Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [Prove it by contradiction methods] Q. In a triangle ABC, E and F are point on AB and AC and EF  BC. Prove that AB/AE = AC/AF Q. Prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. Q. Prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. Q. In a triangle ABC, E and F are point on AB and AC Such that AE/EB = AF/FC and <AEF =<ACB. Prove that ABC is an isosceles Triangle. Q. In a trapezium ABCD , AB  DC and E and F are points on nonparallel sides AD and BC respectively such that EF is parallel to AB .Show that AE/ ED = BF /FC [join AC to intersect EF at G] Q. In a trapezium ABCD , AB  DC and its diagonals intersect each other at the point O. Show that AO/ BO = CO/DO Q. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/ BO = CO/DO . Show that ABCD is a trapezium. For solve and practice worksheet and CBSE Sample paper click on : Download File 10th maths Ch Similar Triangle Guess Paper SA1 Download File Q. Prove that no number of the type 4k+2 can be a perfect square.
Ans: If p is a prime factor of a perfect square, p2 must also be a factor of that perfect square. 4k+2 = 2(2k+1) 2 is a factor of 4k+2 but 2k+1 is odd and cannot have factor 2, so 4k+2 is not divisible by 4, and therefore cannot be a perfect square. Q. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Ans: We need to calculate the LCM to find the answer. 18 = 2x3x3 ;12= 2x2x3 ; LCM = 36 Q. show that only one out of n, n+2 or n+4 is divisible by 3 where n is positive integer. Solution: When n = 1, exactly one of 1, 1+2, and 1+4 is divisible by 3, namely 1+2, since 3 is divisible by 3, and the other two 1 and 5 are not. Suppose for n < k, only one out of n, n+2, n+4 is divisible by 3 For n = k, we consider k, k+2, k+4. By the induction hypothesis, only one of k1, k+1 and k+3 is divisible by 3. We look at the three possible cases. Case 1: k1 is the one which is divisible by 3. Then k1 = 3m, for some positive integer m. Then add 1 to both sides of k1 = 3m: k1+1=3m+1 k = 3m+1 then k is NOT divisible by 3 Now add 3 to both sides of k1 = 3m: k1+3=3m+3 k+2 = 3m+3 k+2 = 3(m+1) then k+2 IS divisible by 3 Now add 5 to both sides of k1 = 3m: k1+5=3m+5 k+4 = 3m+5 k+4 = 3m+3 + 2 = 3(m+1)+2 then k+4 is NOT divisible by 3. So, we have proved case 1 for n = k Case 2: k+1 is the one which is divisible by 3. Then k+1 = 3m, for some positive integer m. Then add 1 to both sides of k+1 = 3m: k+11=3m1 k = 3m1 then k is NOT divisible by 3 Now add 1 to both sides of k+1 = 3m: k+1+1=3m+1 k+2 = 3m+1 then k+2 is NOT divisible by 3 Now add 3 to both sides of k+1 = 3m: k+1+3=3m+3 k+4 = 3m+3 k+4 = 3(m+1) so k+4 IS divisible by 3. So, we have proved case 2. Case 3: k+3 is the one which is divisible by 3. Then k+3 = 3m, for some positive integer m. Then add 3 to both sides of k+3 = 3m: k+33=3m3 k = 3(m1) then k IS divisible by 3 Now add 1 to both sides of k+3 = 3m: k+31=3m1 k+2 = 3m1 then k+2 is NOT divisible by 3 Now add 1 to both sides of k+3 = 3m: k+3+1=3m+1 k+4 = 3m+1 then k+4 is NOT divisible by 3. So, we have proved case 3. Q.If the H C F of 210 and 55 is expressible in the form 210 × 5 + 55y then find y Answer: Let us first find the H C F of 210 and 55. Applying Euclid division lemna on 210 and 55, we get 210 = 55 × 3 + 45 ....(1) Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get 55 = 45 × 1 + 10 .... (2) Again, considering the divisor 45 and remainder 10 and applying division lemna, we get 45 = 4 × 10 + 5 .... (3) We now, consider the divisor 10 and remainder 5 and applying division lemna to get 10 = 5 × 2 + 0 .... (4) We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55. ∴ 5 = 210 × 5 + 55y ⇒ 55y = 5  1050 = 1045 ⇒ y = 19 Q. Finds the H.C.F. of 65 and 117 and express it in the form of 65m+117n. Answer: Among 65 and 117; 2117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 × 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 × 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 × 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52× 1 … Step 4 From Step 1: 52 = 117 – 65 × 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 × 1) ⇒13 = 65 × 2 – 117 ⇒13 = 65 × 2 + 117 × (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 Q.Find all positive integral values of n for which n2+96 is perfect square. Answer: Let n2 + 96 = x2 ⇒ x2 – n2 = 96 ⇒ (x – n) (x + n) = 96 ⇒ both x and n must be odd or both even on these condition the cases are x – n = 2, x + n = 48 x – n = 4, x + n = 24 x – n = 6, x + n = 16 x – n = 8, x + n = 12 and the solution of these equations can be given as x = 25, n = 23 x = 14, n = 10 x = 11, n = 5 x = 10, n = 2 So, the required values of n are 23, 10, 5, and 2. Q. Prove that one of every three consecutive integers is divisible by 3. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case – I when n = 3q In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3 Case  II When n = 3q + 1 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not divisible by 3 Case – III When n = 3q +2 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. But n and n+1 are not divisible by 3 Hence one of n, n + 1 and n + 2 is divisible by 3 More solved Questions:
X Triangles :Criteria for Similarity of Triangles Solved:
X Real Number Solved Questions F A  1 Download File 10th Real Number FA1 Test_paper Download File Class X Polynomial (MCQ's) Test PaperDownload File 10th Polynomial FA1 Test Paper Download File Factor and reminder theorem for class 10 Download Linear Equations in two Variables MCQ’ Test 1 Download File Linear Equations in two Variables Test paper2 Download File Linear Equations in two Variables Test paper3 Download File X_similar_triangle_guess_questions test Paper5 Download File X EBOOk Chapter Similar Triangle Download File X Maths: Similar Triangle:Optional Exercise Solved Download File 10th trigonometry_term1_number_worksheet Download File 10th trigonometry_term1_solved_questions Download File Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
Ans:Since 825>175, we apply division lemma to 825 and 175 to get825 = 175 x 4 + 125. Since r ≠ 0, we apply division lemma to 175 and 125 to get 175 = 125 x 1 + 50 Again applying division lemma to 125 and 50 we get, 125 = 50 x 2 + 25. Once again applying division lemma to 50 and 25 we get. 50 = 25 x 2 + 0. Since remainder has now become 0, this implies that HCF of 825 and 125 is 25. Q.2. Based on Showing that every positive integer is either of the given forms: Solved example: Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q. Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3. Q.Find H C F (26,91) if LCM(26,91) is 182 Sol: We know that LCM x HCF = Product of numbers. or 182 x HCF = 26 x 91 or HCF = 26 x 91 = 13 Hence HCF (26, 91) = 13. Q. prove that √5 is irrational. Solution: let us assume on the contrary that √5 is rational. That is we can find coprimes a and b b (≠0) such that √5 = a/b. Or √5b = a. Squaring both sides we get 5b2 = a2. This means 5 divides a2. Hence it follows that 5 divides a. So we can write a = 5c for some integer c. Putting this value of a we get 5b2 = (5c)2 Or 5b2 = 25c2 Or b2 = 5b2. It follows that 5 divides b2. Hence 5 divides b. Now a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprimes. This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational. Q. prove that product of three consecutive positive integers is divisible by 6? Ans: Let three consecutive positive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. Q. Express HCF of F 65 and 117 in the form of 65m +117n Ans: Among 65 and 117; 117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 x 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 x 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 x 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52 x 1 … Step 4 From Step 1: 52 = 117 – 65 x 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 x 1) x 1 ⇒13 = 65 x 2 – 117 ⇒13 = 65 x 2 + 117 x (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 1oth maths term1 
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