Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th
by

1 Answer

Answer :

Solution  :- The obtained triangle is an isosceles triangle.
by

Related questions

Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

Description : Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

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

Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed. -Maths 9th

Description : Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed. -Maths 9th

Last Answer : answer:

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Last Answer : NEED ANSWER

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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

A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then, show that OA^2 + OC^2 = OB^2 + OD^2. -Maths 9th

Description : A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then, show that OA^2 + OC^2 = OB^2 + OD^2. -Maths 9th

Last Answer : answer:

If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Description : If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Last Answer : (b) 16π cm2.AB = OA ⇒ AB = OA = OB (radii of circle are equal) ⇒ ΔAOB is equilateral. ∴ If ‘r’ is the radius of the circle,then area of ΔAOB = \(rac{\sqrt3}{4}\)side2⇒ \(rac{\sqrt3}{4}\)(r)2 = 4√3 (given)⇒ r2 = 16 ⇒ r = 4∴ Area of circle = πr2 = 16π cm2.

On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Description : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Description : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Description : In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Last Answer : Solution :- As OD is perpendicular to AB ⇒ AC = AB (Perpendicular from the centre to the chord bisects the chord) ∴ AC = AB/2 = 8cm In right △OCA, OA2 = AC2 + OC2 (102) = 82 + OC2 OC2 = 100 - 64 OC2 = 36 ... = 6cm CD = OD - OC = 10 - 6 = 4cm [∴ OA = OD = 10cm (radii)]

In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th

Description : In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th

Last Answer : answer:

In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Description : In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Last Answer : In △OBC, OB = BC ⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal] ∠OBA is the exterior angle of △BOC So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles] Similarly, ∠AOD is the exterior angle of △AOC ∴ x = 2y + y = 3y

In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Description : In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Last Answer : In △OBC, OB = BC ⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal] ∠OBA is the exterior angle of △BOC So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles] Similarly, ∠AOD is the exterior angle of △AOC ∴ x = 2y + y = 3y

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Description : How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Last Answer : 2x + 1 = x - 3 2x-x = -3-1 ∴ x = - 4 ..(i) and it can be written as 1.x + 0. y = - 4 ..(ii) (i) Number line represent the all real values of x on the X ... the equation x + 4 = 0 represent a straight line parallel to Y-axis and infinitely many points lie on a line in the cartesian plane.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Description : How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Last Answer : 2x + 1 = x - 3 2x-x = -3-1 ∴ x = - 4 ..(i) and it can be written as 1.x + 0. y = - 4 ..(ii) (i) Number line represent the all real values of x on the X ... the equation x + 4 = 0 represent a straight line parallel to Y-axis and infinitely many points lie on a line in the cartesian plane.

Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Description : Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Last Answer : Solution :-

What is the name of the horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? -Maths 9th

Description : What is the name of the horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? -Maths 9th

Last Answer : The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Last Answer : Solution :-

Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Last Answer : answer:

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Last Answer : answer:

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Last Answer : answer:

In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@)` and `/_ ABO=20^(@)`. If `/_ OCB=(1)/(2) /_ A

Description : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@)` and `/_ ABO=20^(@)`. If `/_ OCB=(1)/(2) /_ A

Last Answer : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@) ... OCB=(1)/(2) /_ ACO,` then find `/_ BOC.`

The bearing of lines OA and OB are 16° 10' and 332° 18', the value of the included angle BOA is (A) 316° 10' (B) 158° 28' (C) 348° 08' (D) 43° 52'

Description : The bearing of lines OA and OB are 16° 10' and 332° 18', the value of the included angle BOA is (A) 316° 10' (B) 158° 28' (C) 348° 08' (D) 43° 52'

Last Answer : (D) 43° 52'

What are the angles and area of a triangle with vertices at -3 7 and 2 19 and 10 7 on the Cartesian plane?

Description : What are the angles and area of a triangle with vertices at -3 7 and 2 19 and 10 7 on the Cartesian plane?

Last Answer : The vertices (-3, 7) and (2, 19) and (10, 7) will form anisosceles triangle when plotted on the Cartesian plane with anglesof 67.38 degrees, 56.31 degrees and 56.31 degrees all rounded totwo decimal places and the area of the triangle works out as 78square units.

Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Description : Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Last Answer : In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we ... Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.

Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Description : Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Last Answer : In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we ... Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

Plot the points A(1,-3) and B(5,4). -Maths 9th

Description : Plot the points A(1,-3) and B(5,4). -Maths 9th

Last Answer : Solution :-

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER