How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th
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You should study them thoroughly so that you won't find them difficult when you are preparing for competitive exams like Olympiad s,etc. Study other reference books along with ncert textbook. Hope my and will help you.
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How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Description : How should i study maths main chapters like Lines and Angle,Triangles class 9 ?? -Maths 9th

Last Answer : You should study them thoroughly so that you won't find them difficult when you are preparing for competitive exams like Olympiad s,etc. Study other reference books along with ncert textbook. Hope my and will help you.

Cbqs (case base study ) of chapter 9 Areas of Parallelograms and Triangles of maths class 9th -Maths 9th

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

Cbqs (case base study ) of chapter 7 Triangles of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 7 Triangles of maths class 9th -Maths 9th

Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1.Answers: 1. 2. 3

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 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)

MCQ Questions for Class 9 Maths Chapter 7 Triangles with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 7 Triangles with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 7 Triangles Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. These Class ... . These Triangles MCQ Questions will help you in practising more and more questions in less time.

MCQ Questions for Class 9 Maths Chapter 9 Areas of Parallelogram and Triangles with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 9 Areas of Parallelogram and Triangles with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 9 Areas of Parallelograms and Triangles Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking ... and Triangles MCQ Questions will help you in practising more and more questions in less time.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

Triangles Class 9th Formulas -Maths 9th

Description : Triangles Class 9th Formulas -Maths 9th

Last Answer : Linear equation in one variable: Linear equation of the form ax + b = 0, where a, b are real numbers such that a≠ 0. For example, 3x + 5 = 0. The letters used in an equation are called ... equation. More over every solution of the linear equation is a point on the graph of the linear equation.

Area of Parallelograms and Triangles Class 9th Formula -Maths 9th

Description : Area of Parallelograms and Triangles Class 9th Formula -Maths 9th

Last Answer : An angle is formed by two rays originating from same point. The rays making an angle are called arms of the angle and the point is called vertex of the angle. Right angle : An angle whose measure ... parallel to each other. Lines which are perpendicular to a given line are parallel to each other.

Cbqs (case base study ) of chapter 6 Lines and Angles of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 6 Lines and Angles of maths class 9th -Maths 9th

Last Answer : answer:

If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Description : If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Last Answer : hope it helps if the vertices of a triangle are (1,k),(4,−3)(−9,7) area = 15 sq.units. find the value of k. Area of △ 21 [x1 (y2 −y3 )+x2 (y3 −y1 )+x3 (y1 −y2 )]=15 21 [1(−3−7)+ ... k+3)]=15 21 [(−10+28−4k−9k−27)]=15 −10+28−4k−9k−27=30 −10+28−13k−27=30 −13k=30+10+27−28 −13k=39 k=1339 k=−3 thank u

What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

In how many chapters did Euclid divide his famous treatise ' The elements' ? -Maths 9th

Description : In how many chapters did Euclid divide his famous treatise ' The elements' ? -Maths 9th

Last Answer : Solution :- 13 chapters.

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

MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 6 Lines and Angles Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. ... Lines and Angles MCQ Questions will help you in practising more and more questions in less time.

case study questions class 9 maths surface area and volume -Maths 9th

Description : case study questions class 9 maths surface area and volume -Maths 9th

Last Answer : Q. Read the source or text given below and answer the following questions: A conical circus tent has to be made with a cloth that is 5m wide, whose height is 24m, and the radius of the base is 7m. ... 4. Find the Curved Surface Area Answers: 1. Curved Surface Area 2. 550 m 3. Rs.3850 4.110m²

Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Description : Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Description : Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

Description : In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

Last Answer : Solution :-

If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

If the corresponding angles of two triangles are equal, then they are always. State true or false and justify your answer. -Maths 9th

Description : If the corresponding angles of two triangles are equal, then they are always. State true or false and justify your answer. -Maths 9th

Last Answer : Solution :- False, because two equilateral triangles with sides 3 cm and 6 cm respectively have all angles equal, but the triangles are not congruent.

Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th

Description : Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th

Last Answer : Solution :-

ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Last Answer : Solution :-

Prove that the diagonal divides a parallelogram into two congruent triangles. -Maths 9th

Description : Prove that the diagonal divides a parallelogram into two congruent triangles. -Maths 9th

Last Answer : Solution :-

Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th

Description : Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th

Last Answer : Solution :-

Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Description : Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Last Answer : Reflexive : A ≅ A True Symmetric : if A ≅ B then B ≅ A True Transitive : if A ≅ B and B ≅ C, then A ≅ C True Therefore, the relation ‘≅’ is an equivalence relation.

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

Description : In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

Last Answer : (b) 7√3 cm2.AB = 6 cm, ∠C = 60º (∴ ∠A = ∠B = 60º) ∴ ΔABC is an equilateral triangle Area of ΔABC = \(rac{\sqrt3}{4}\) × (6)2 = 9√3 Area of (ΔADE + ΔBFG) = 2 x \(\bigg(rac{\sqrt3}{4} imes(2)^2\bigg)\) = 2√3 ∴ Area of pentagon = 9√3 - 2√3 = 7√3 cm2.

The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Description : The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

Important properties of triangles: -Maths 9th

Description : Important properties of triangles: -Maths 9th

Last Answer : answer:

Important Theorems on Triangles. -Maths 9th

Description : Important Theorems on Triangles. -Maths 9th

Last Answer : answer:

If the medians of two equilateral triangles are in the ratio 3 : 2, then what is the ratio of their sides? -Maths 9th

Description : If the medians of two equilateral triangles are in the ratio 3 : 2, then what is the ratio of their sides? -Maths 9th

Last Answer : answer:

Let ABCD be a cyclic quadrilateral. Show that the incentres of the triangles ABC, BCD, CDA and DAB form a rectangle. -Maths 9th

Description : Let ABCD be a cyclic quadrilateral. Show that the incentres of the triangles ABC, BCD, CDA and DAB form a rectangle. -Maths 9th

Last Answer : answer:

There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Description : There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Last Answer : answer:

What are the 4 congruence for congruent triangles ? -Maths 9th

Description : What are the 4 congruence for congruent triangles ? -Maths 9th

Last Answer : The four congruence are : 1. SSS congruence ( side-side-side). 2. SAS congruence (side-angle-side). 3. ASA congruence (angle-side-angle). 4. RHS congruence (Right-Hypotenuse-Side)