summary of chapter 9 plzzz?

summary of chapter 9 plzzz?
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what is the setting of chapter 9
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I want contact number of study ranker who give answers of our question on whatsapp I am having that number earlier but plzzz it's my request please provide me your number I am PREMIUM MEMBER -Maths 9th

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I want contact number of study ranker who give answers of our question on whatsapp I am having that number earlier but plzzz it's my request please provide me your number I am PREMIUM MEMBER -Maths 9th

Description : I want contact number of study ranker who give answers of our question on whatsapp I am having that number earlier but plzzz it's my request please provide me your number I am PREMIUM MEMBER -Maths 9th

Last Answer : Hi there, You can ask your doubts here. Just post your question by clicking on Ask Question button select the subject and enter the chapter name in tag. Thanks.

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NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.3 -Maths 9th

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Description : NCERT Solutions for class 9 Maths Chapter 1 Exercise 1.1 -Maths 9th

Last Answer : In this chapter, though our main topic is Heron’s (or Hero’s) formula given by Heron to calculate area of triangle whose sides are a, b, c as given under Yet, we list all formulas that were used at one point or other in this chapter.

NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.2 -Maths 9th

Last Answer : Solids Bodies occupying space are called solids. A solid has three dimensions viz., length, breadth and thickness or depth or height. The space occupied by a solid body is called its volume .Volume ... a right-angled triangle VOA is revolved about OV, it generates a cone shown in figure alongside.

NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.4 -Maths 9th

Last Answer : 1. Is zero a rational number ? Can you write it in the form p/q , where p and q are integers and q ≠ 0? Yes, zero is a rational number. We can write it in the form (p/q). That is zero. 2. ... such as -1, -2 are non-whole numbers] (iii) False statement [∵ Rational number (1/2) is not a whole number]

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Description : NCERT Solutions for class 9 Maths Chapter 1 Number System Exercise 1.5 . -Maths 9th

Last Answer : 1. State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number. (ii) Every point on the number line is of the form √m , where m is a natural ... 5 ⇒ OC = √5 With O as centre and OC as radius, draw an arc intersecting OX at D. Since

NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.1 -Maths 9th

Last Answer : 1. Write the following in decimal form and say what kind of decimal expansion each has: 4. Express 0.99999 in the form p/q . Are you surprised by your answer? With your teacher ... ) 0.767076700767000767 (iii) 0.78080078008000780 9. Classify the following numbers as rational or irrational:

NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2 -Maths 9th

Last Answer : 1. Visualise 3.765 on the number line, using successive magnification. 3.765 lies between 3 and 4. Let us divide the interval (3, 4) into 10 equal parts. Since, 3.765 lies between 3.7 and 3. ... 10 equal parts. Now, the point corresponding to 3.765 is clearly located, as shown in Fig. (iii) above.

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Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.4 -Maths 9th

Last Answer : Classify the following numbers as rational or irrational: 2: Simplify each of the following expressions: 3. Recall, π is defined as the ratio of the circumference (say c) of a circle to ... is no contradiction in saying that π is irrational . 4. Rationalise the denominators of the following:

NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.5 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.5 -Maths 9th

Last Answer : 1. Which of the following expressions are polynomials in one variable and which are not ? State reasons for your answer. 2: Write the co-efficients of x2 in each of the following: (i) 2 + x2 + x The ... It is a quadratic polynomial. (vii) 7x3 ∵ The degree of 7x3 is 3. ∴ It is a cubic polynomial.

NCERT Solutions for class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.2 -Maths 9th

Last Answer : For x + 1 = 0, we have x = -1. ∴ The zero of x + 1 is -1. (i) p(x) = x3 + x2 + x + 1 ∴ p(-1) = (-1)3 + (-1)2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0 i.e. when p(x) is divided by (x + 1), then the remainder is zero. ∴ ... ] = (y - 1)[2y(y + 1) + 1(y + 1)] = (y - 1)[(y + 1)(2y + 1)] = (y - 1)(y + 1)(2y + 1)

NCERT Solutions for class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.1 -Maths 9th

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NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.1 -Maths 9th

Last Answer : (i) (x + 4)(x + 10): Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10) = x2 + 14x + 40 (ii) (x + 8)(x - 10): Here, a = 8 and b = ( ... have 64m3 - 343n3 = (4m)3 - (7n)3 = (4m - 7n)[(4m)2 + (4m)(7n) + (7n)2] = (4m - 7n)(16m2 + 28mn + 49n2)

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.2 -Maths 9th

Last Answer : 1. How will you describe the position of a table lamp on your study table to another person ? To describe the position of a table lamp placed on the table, let us consider the table lamp as P and the table as a plane ... the point A(4, 3). (ii) A unique cross street as shown by the point B(3, 4).

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3 -Maths 9th

Last Answer : 1: Draw the graph of each of the following linear equations in two variables: (i) x + y = 4 (ii) x - y = 2 (iii) y = 3x (iv) 3 = 2x + y Solution: (i)x + y= 4 ⇒ y = 4 - x If ... graph paper and joining them, we get a straight line PQ. Thus, PQ is the required graph of the linear equation y = 5x + 3.

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.4 -Maths 9th

Last Answer : 1. Give the geometric representations of y = 3 as an equation (i) in one variable (ii) in two variables (i) y = 3 [An equation in one variable] ∵ y = 3 is an equation in one variable, i.e. y only. ∴ It has ... equation in two variables] We can write 2x + 9 = 0 as 2x + 0y + 9 = 0 or 2x = -9 + 0y

NCERT Solutions for class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Exercise 5.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Exercise 5.2 -Maths 9th

Last Answer : 1. How would you rewrite Euclid's fifth postulate so that it would be easier to understand ? We can have: Two distinct intersecting lines cannot be parallel to the same line. 2. Does Euclid's fifth postulate imply ... other side also. ∴ The lines m' and n' never meet, i.e., they are parallel.

NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.1 -Maths 9th

Last Answer : 1. Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. (ii) There are an infinite number of lines which pass ... statement is true for all things and in all parts of Universe. So, it is a Universal truth'.

NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.2 -Maths 9th

Last Answer : 1. In the following figure, find the values of x and y and then show that AB || CD. In the figure, we have CD and PQ intersect at Y. ∴ y = 130º ... Perpendiculars to the parallel lines are parallel. (ii) According to the laws of reflection, angle of incidence = angle of reflection

NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 6 Lines and Angles Exercise 6.3 -Maths 9th

Last Answer : 1. In the adjoining figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠ SPR = 135º and ∠ PQT = 110º, find ∠ PRQ. ∵ TQR is a straight line, ∴ ∠ TQP + ∠ PQR = 180º [ ... + ∠ T ⇒ (1/2)∠ P= ∠ T i.e. (1/2) ∠ QPR = ∠ QTR or ∠ QTR = (1/2) ∠ QPR

NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.1 -Maths 9th

Last Answer : 1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ΔABC ∠ ΔABD. What can you say about BC and BD ? In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD ... [Given] ∴ CM = (1/2) DC = (1/2) AB ⇒ CM = (1/2) AB

NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.2 -Maths 9th

Last Answer : 1. In an isosceles triangle ABC, with AB = AC, the bisectors of B and ∠C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects ∠ A (i) In ΔABC, we have AB = AC ... 180° or x =(180o/3) = 60° ∴ ∠A= ∠B = ∠C = 60° Thus, the angles of an equilateral triangle are 60° each.

NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.3 -Maths 9th

Last Answer : 1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that (i) ΔABD ≌ ... [common] ∴ Using RHS criteria, ΔABP ≌ ΔACP ∴ Their corresponding parts are congruent. ⇒ ∠B= ∠C

NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.4 -Maths 9th

Last Answer : 1. Show that in a right angled triangle, the hypotenuse is the longest side Let us consider ΔABC such that ∠B = 90º ∴ ∠A + ∠B + ∠C = 180º ∴ [∠A + ∠C] + ∠B = 180º ⇒ ∠A + ∠C = 90º ⇒ ... on the line l. Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.5 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 7 Triangles Exercise 7.5 -Maths 9th

Last Answer : 1. ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC. Let us consider a ΔABC. Draw l' the perpendicular bisector of AB. Draw m' the ... figure (i) and 300 equilateral triangles in the figure (ii). ∴ The figure (ii) has more triangles.

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 -Maths 9th

Last Answer : 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Let the angles of the quadrilateral be 3x, 5x, 9x and 13x. ∵ Sum of all the angles of ... ABCD is a square, and diagonals of a square bisect the opposite angles. ∴ BD bisects ∠B as well as ∠D.

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 -Maths 9th

Last Answer : 1. ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that: (i) SR || AC and SR = (1/2) AC (ii) PQ = SR (iii) PQRS is a parallelogram. We ... MA =(1/2)AB (2) From (1) and (2), we have CM = MA = (1/2) AB

NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1 -Maths 9th

Last Answer : 1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. The figures (i), (iii), and (v) lie on the same base and between the same parallels.

NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2 -Maths 9th

Last Answer : 1. In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. We have AE ⊥ DC and AB = 16 cm ∵ AB = CD [opp. sides of parallelogram ABCD] ∴ ... sow wheat in (ΔPAQ) and pulses in [(ΔAPS) + (ΔQAR)] OR wheat in [(ΔAPS) + (ΔQAR)] and pulses in (ΔPAQ).

NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 -Maths 9th

Last Answer : 1. In the figure, E is any point on median AD of a DABC. Show that ar (ABE) = ar (ACE). We have a ΔABC such that AD is a median. ∴ ar (ΔABD) = ar (ΔADC) ...(1) [∵ A median divides ... from both sides, we get [ar (ΔABD) - ar (ΔAOB)] = [ar (ΔABC) - ar (ΔAOB)] ⇒ ar (ΔAOD) = ar (ΔBOC)

NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4 -Maths 9th

Last Answer : 1. Parallelogram ABCD and rectangle ABEF are on the same base AB have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. We have a parallelogram ABCD and rectangle ABEF such that ar ( ... (BCED) = ar (ABMN) + ar (ACFG) [from (5)