In a triangle ABC,D is the midpoint of BC and E is the midpoint of AD. If BE is produced meets AC in F. Then prove that AF =1/3AC.
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Find the ratio in which line y = x divides the line segment joining the points (6, 3) and (1, 6) Step 1: Let's assume the ratio in which the line divides the segment AB is 𝒌: 𝟏 Step 2: Find the coordinates of the point of intersection P as x = (𝒌+𝟔)/( k + 1); y = (𝟔𝒌 – 3)/(k + 1) Step 3: The point P lies on the line Y = X, so you equated the x-coordinate and the y-coordinate of point P So, P(x, y) lies on y = x k + 6 = 6k –3 k = 𝟗 /𝟓 Ratio is 9 : 5 More related material Class 10 chapter Co-Ordinate
sum of integers between 100 and 200 (i) divisible by 9 (ii) not divisible by 9 (CBSE 2024) Class 1015/8/2023 Find the sum of integers between 100 and 200 which are (i) divisible by 9 (ii) not divisible by 9 (5 mark question CBSE BOARD 2023 - 24) You can apply formula of nth term aₙ = a + (n - 1)d to find number of terms(n) then , Apply formula of sum of nth term Sn= n/2 (a + l)d to find Sum of n terms(Sn) Thus, you may find result for question of board paper 2024 the sum of integers between 100 and 200 which are (i) divisible by 9 (ii) not divisible by 9 (a) The integers between 100 and 200 that are divisible by 9 are 108, 117, 126,.......,198 Here a = 108,d = 9 ,l = 198aₙ = a + (n - 1)d So, 198 = 108 + (n - 1)9 n = 11 then , Sn = n/2[ a + l ] So, S = 11/2[108+198] = 1683 (b) All integers between 100 and 200 are 101, 102, 103,...........,199. a = 101 , d =1 , l = 199 aₙ = a + (n - 1)d 199 = 101 + (n - 1)1 n = 99 Now, S = n/2[ a + l ] = 99/2[101 + 199] = 14850 For final answer Sum of the integers between 100 and 200 that are not divisible by 9 = (Sum of the integers between 100 and 200) - (sum of the integers between 100 and 200 that are divisible by 9) = 14850 - 1683 = 13167 3.You can now try to solve these Related Questions and put your solution in comment box 1. Find the sum of all integers between 50 and 500, which are divisible by 7.(CBSE 2024) 2. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ? Also, find their sum.(CBSE 2024) 3. Solve the equation: - 4 + (-1) + 2 +...+ x = 437 (CBSE 2024) For self-assessment use study material available free for use by Jsunil Sir at www.Jsuniltutorial.in
1. Class 10th Arithmetic progression Solved Question Papers 2. Class 10th Arithmetic Progression CBSE Test Paper 3.Also Check: NCERT Solutions for Class 10 Maths Chapter 5 Be Happy and Enjoy Your Learning ! In the given figure, each one of PA, QB and RC is perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that 1/x + 1/y= 1/z. by Jsunil sir in ∆QBC and ∆ PAC , ∠B = ∠A (90°) ∠C = ∠C (angles) ∆QBC ~ ∆PAC BQ/AP = BC/AC z/x = b/(a+b)-------(i) In and ∆QBA and ∆ ARC ∠B= ∠C ( 90°) ∠A = ∠A (common) in ∆ABQ ~∆ARC BO/RC = AB/AC z/y = a/(a+b)------(ii) From equation (i) and (ii), z(1/x +1/y) = 1 So, (1/x +1/y) = 1/z Key prove, 1x,1y,1z if, pa, qb, rc, perpendicular, ac, Class,10
In Fig OB is the perpendicular bisector of the line segment DE, FA ⊥ OB and FE intersects OB at the point C. Prove that 1/OA + 1/OB = 2/OC Solution by Jsunil
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