Heron's Formula Class 9th Formula -Maths 9th

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Answer :

• Two figures are said to be on same base and between the same parallels, if they have a common side (base) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. •  Figures on the same base and between the same parallels : In fig.(i) triangles ABC and DBC have a common side, BC. So we can say that Δ’s ABC and DBC are on the same base BC. Similarly, in figs.(ii), (iii), (iv) this condition can be checked. Area Axioms: The concept of area possesses properties similar to the concept of the length of a line segment. So, it is natural to think that there may be some analogy between the two concepts: (a) The concept of length (b) The concept of area. Area of a figure is a number associated with the part of the plane enclosed by the figure . • Two congruent figures have equal area but the converse need not to be true. • A diagonal of a || gm  divides it into two triangles of equal area. ar (ΔABD) = ar (ΔCDB)  • The area of a || gm is the product  of  its  base  and corresponding altitude, ar (|| gm  ABCD) = Base × Altitude • Parallelograms on the same base and between the same parallels are equal in area. • Triangles on the same base and between the same parallels are equal in area. • The area of a triangle is half the product of any of its sides and the corresponding altitude. • In given fig., ar(△ABC) = ½  (BC × AL) • The area of a trapezium is half the product of its height and the sum of parallel sides. • The area of a rhombus is half the product of the length of its diagonals. • Triangles having equal areas and having one side of the triangle, equal to one side of the other, have their corresponding altitudes equal. Note:  Diagonals of a parallelogram divide it into four triangles of equal area. • If the diagonals AC and BD of quad. ABCD intersect at O and separate the quad. Into four triangles of equal area, then the quad. ABCD is a parallelogram. • A median of a triangle divides it into two triangles of equal area.

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Last Answer : Need answer

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Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

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Last Answer : Clearly, the mean score of two sections A and B is same

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Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : The class mark of the class 90-120 is -Maths 9th

Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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Last Answer : NEED ANSWER

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