Lines and Angles Class 9th Formula . -Maths 9th

1 Answer

Answer :

The system used for describing the position of a point in a plane with reference to two mutually perpendicular lines is called cartesian or rectangular coordinate system. Coordinate axes: Let us draw a horizontal line on a graph paper and draw a vertical line intersecting the horizontal line. The horizontal line is called  x -axis and vertical line is called  y -axis. The two lines taken together are called coordinate axes. Origin: The point of intersection of x-axis and y -axis are called origin. Quadrant: The coordinate axis divide the plane of graph paper into 4-regions called quadrant. Coordinate of a point:  Let A(x, y) be the point in a plane. x=distance of point from y-axis. y=distance of point from x-axis. Convention of sign: I  quadrant →x > 0, y > 0 II  quadrant →x < 0, y > 0 III quadrant →x < 0, y < 0 IV quadrant →x > 0, y < 0

Related questions

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'.

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

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

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.

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

Last Answer : answer:

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Last Answer : Solution :-

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

Description : If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Last Answer : answer:

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.

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.

Description : Number Systems Class 9th Formula -Maths 9th

Last Answer : Rational Numbers : Rational Numbers are numbers , which can be expressed in the form p/q, q≠0; where p and q are integers . The collection of all rationals are represented by Q. ∴ Q = {p/q ; p,q ... number of rational numbers x2 , x1, x3,... between any two rational numbers a and b so that a

Description : Polynomials Class 9th Formula -Maths 9th

Last Answer : An algebraic expression is the combination of constants and variable connected by the four basic operations (+, -, , ). For example : 2x , x2y , xy/3, 3 etc. Types of Algebraic expression : Polynomial in one variable : An ... b2) (x) a3 + b3 + c3 - 3abc = (a+b+c)(a2+b2+c2-ab - bc - ca)

Description : Coordinate Geometry Class 9th Formula -Maths 9th

Last Answer : Here, the traingle ABC is isosceles So,Angle A willbe 90°.And AB=BC therefore their angles will be equal So..90+x+x=180° 2x=90° X=45° Then angle B and angle C = 45°

Description : Quadrilaterals Class 9th Formula -Maths 9th

Last Answer : A point is that which has no part. A line is breadth less length. The ends of lines are points. A straight line is a line which lies evenly with the points on itself. A surface is that ... lines, if produced indefinitely, meet on that side on which the sum of angles are less than two right angles

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.

Description : Constructions Class 9th Formula -Maths 9th

Last Answer : If A, B, C, D are four points in a plane such that no three of them are collinear and the line segments AB, BC, CD and DA do not intersect except at their end points, the ... the intercepts made by them on one transversal are equal, then the intercepts on any other transversal are also equal.

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

Last 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 ... quad. ABCD is a parallelogram. A median of a triangle divides it into two triangles of equal area.

Description : NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 -Maths 9th

Last Answer : 1. A traffic signal board, indicating SCHOOL AHEAD', is an equilateral triangle with side a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board? 2 ... x = 30 - 12 - 12 cm ⇒ x = 6 cm Now, semi-perimeter = (30/2)cm = 15 cm

Description : NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 -Maths 9th

Last Answer : 1. A park, in the shape of a quadrilateral ABCD, has - C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? Solution: Let us join B and D, such that ΔBCD is ... The given field is in the form of a trapezium ABCD such that parallel sides AB = 10 m and DC = 25 m

Description : MCQ Questions for Class 9 Maths Chapter 12 Heron's Formula with answers Heron`s Formula -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 12 Heron's Formula Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. ... Heron's Formula MCQ Questions will help you in practising more and more questions in less time.

Description : Cbqs (case base study ) of chapter 12 Heron's Formula of maths class 9th -Maths 9th

Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1.A triangular park ABC has sides 120m, 80m and 50m (see Fig. 12.7). A gardener Dhania has to put a fence all around it and also plant grass ... used for wheat, potatoes, and onions? (1 hectare = 10000 m2 ) More questions to be added shortly...

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Description : 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

Description : 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

Description : 4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O. To show that, AC = BD AO = OC and ∠AOB = 90° Proof, In ΔABC and ΔBAD, AB = BA (Common) ∠ABC = ∠BAD = ... = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

Description : Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Last Answer : each interior opposite angles are 55

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

Description : If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Last Answer : Let the two adjacent angles be x° and 2x° . In a parallelogram, sum of the adjacent angles are 180°. ∴ x + 2x = 180° ⇒ 3x = 180° ⇒ x = 60° Thus , the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Description : If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Last Answer : Quadrilateral will be Rhombus .

Description : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Last Answer : Since opposite angles are equal in a parallelogram . Therefore , in parallelogram ABCD , we have ∠A = ∠C ⇒ 1 / 2 ∠A = 1 / 2 ∠C ⇒ ∠1 = ∠2 ---- i) [∵ AX and CY are bisectors of ∠A and ∠C ... intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal . ∴ AX | | CY .

Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

Description : In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Last Answer : We know that opposite angles of a paralleloram are equal . ∴ ∠P = ∠R ⇒ 3x - 4 = 56 - 3x ⇒ 6x = 60 ⇒ x = 10 Thus, ∠P = ∠R = 3 10 - 4 = 26° Also, ∠P + ∠Q = 180° [ ... Hence, the four angles of the parallelogram PQRS are 26°, 154°,26° and 154°. We should care our earth to have good eco balance.

Description : Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Last Answer : Given : In a circle C(O,r), chord AB = chord CD. To Prove : ∠AOB = ∠COD. Proof : In△AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] Chord AB = Chord CD [given] ⇒ △AOB ≅ △COD [by SSS congruence axiom] ⇒ ∠AOB = ∠COD. [c.p.c.t.]

Description : If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Last Answer : Given : In a circle C(O,r) , ∠AOB = ∠COD To Prove : Chord AB = Chord CD . Proof : In △AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] ∠AOB = ∠COD [given] ⇒ △AOB ≅ △COD [by SAS congruence axiom] ⇒ Chord AB = Chord CD [c.p.c.t]

Description : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Description : Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is -Maths 9th

Last Answer : (b) Given, the ratio of angles of a triangle is 2 : 4 : 3. Let the angles of a triangle be ∠A, ∠B and ∠C. ∠A = 2x, ∠B = 4x ∠C = 3x , ∠A+∠B+ ∠C= 180° [sum of all the angles of a triangle is 180°] 2x ... ∠B = 4x = 4 x 20° = 80° ∠C = 3x = 3 x 20° = 60° Hence, the smallest angle of a triangle is 40°.