Description : The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their -Maths 9th
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Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th
Description : The median of a triangle divides it into two -Maths 9th
Last Answer : (a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.
Description : Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th
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Description : Cbqs (case base study ) of chapter 9 Areas of Parallelograms and Triangles of maths class 9th -Maths 9th
Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1.Answers: 1. 2. 3
Description : Cbqs (case base study ) of chapter 13 surface areas and volume of maths class 9th -Maths 9th
Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1 . Dev was doing an experiment to find the radius r of a sphere.For this he took a cylindrical container with radius R = 7 cm and height 10 cm. He filled the container ... .14)*6²*8 = 301.44 m³ area of the floor = πR² = 3.14 (6)² = 113.04 m²
Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th
Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th
Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25
Description : Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th
Last Answer : Consider the equation 3x + 2y = 12 Now, square both sides: (3x + 2y)2 = 122 => 9x2 + 12xy + 4y2 = 144 =>9x2 + 4y2 = 144 – 12xy From the questions, xy = 6 So, 9x2 + 4y2 = 144 – 72 Thus, the value of 9x2 + 4y2 = 72
Description : Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th
Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th
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Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th
Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39
Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th
Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35
Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th
Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.
Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th
Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.
Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th
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Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th
Last Answer : According to question find the median of the data
Last Answer : The child does not understand, that data has to be arranged in ascending or descending order before finding the median.
Description : A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th
Last Answer : In order to find the median, data has to be arranged in ascending or descending order before finding the median.
Description : Find the median of the numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3. -Maths 9th
Last Answer : Arranging the data in ascending order, we get 3, 4, 4, 5, 6, 7, 7, 7, 12 Here, n = 9 Median = (9 +1)th/2 observation = 5th observation = 6.
Description : If AD is the median of DABC, then -Maths 9th
Description : Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th
Last Answer : (b) a = √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg ... \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = √2 .
Description : If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th
Last Answer : D=slid ht of BC D≅(20+4,20+2) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2=root42+32=5 hope it helps thank u
Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th
Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.
Description : The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th
Last Answer : (d) Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis. So, we put x = 0 in the given equation 2x+ 3y = 6, we get 2 x 0+ 3y = 6 ⇒ 3y = 6 y = 2. Hence, at the point (0, 2), the given linear equation cuts the Y-axis.
Description : G is the centroid of ΔABC with height h units. If a line DE parallel to BC cuts ΔABC at a height h/4 from BC, find the distance GG' in terms of AG -Maths 9th
Description : The line through the points (4, 3) and (2, 5) cuts off intercepts of lengths λ and μ on the axes. Which one of the following is correct ? -Maths 9th
Last Answer : (c) a, b, c are in H.P. only for all m As the points A(a, ma), B[b, (m + 1)b] and C[c, (m + 2)c] are collinear. Area of Δ ABC should be equal to zero.⇒ \(rac{1}{2}\)[x1(y2 - y3) + x2(y3 - y1) + ... - bc = 0 ⇒ ab + bc = 2ac ⇒ b = \(rac{2ac}{a+c}\)∴ a, b, c are harmonic progression (H.P.) for all m.
Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th
Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3
Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th
Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required
Description : How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th
Last Answer : l = √h2 + r2 = √(3.5)2 + (12)2 = √12.25 + 144 = √156.25 = 12.5 m Curved surface area = πrl = 22 / 7 × 3.5 × 12.5 = 137.5 m2 Area of cloth = 137.5 m2 Length of cloth required = C.S.A. / Width l = 137.5 / 5 = 27.5 m
Description : How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th
Last Answer : Slantheight5l = h2+r2 =(3.5)2+(12)2 total canvas required = πrl =π×12×12.5 =471 sq m
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Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th
Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.
Description : Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th
Last Answer : Volume of water discharged through the pipe = Volume increase of the tank First, consider the pipe. Volume discharge through pipe = length breadth speed time Let the time taken to fill the tank to 3 m depth be t. ... the tank V=150 100 3 cu. m ⇒ V=450 100 Therefore, 450t=450 100 ⇒ t=100 hours
Description : What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th
Last Answer : Lateral surface area of a prism = Perimeter of base Height ⇒ 840 = (5 + 5 + 8) Height ⇒ Height = 8401884018 = 46 cm. = Semi perimeter of the triangular base = 182182 = 9 cm ∴ Area of triangle = 9(9- ... 4 1 = 12 cm2 ∴ Required volume of prism = Area of base Height = (12 46) cm3 = 552cm3
Description : The volumes of two spheres are in the ratio 64 : 27. Find the difference of their surface areas, if the sum of their radii is 7 cm. -Maths 9th
Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th
Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9
Description : The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th
Last Answer : If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2 Ncert solutions class 9 chapter 13-6
Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th
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