Points (1, -1), (2, – 2), (4, – 5) and (-3, – 4) -Maths 9th

Points (1, -1), (2, – 2), (4, – 5) and (-3, – 4) -Maths 9th
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Answer :

(d) In points (1, -1), (2, -2) and (4, -5) x-coordinate is positive and y-coordinate is negative, So, they all lie in IV quadrant. In point (-3, – 4) x-coordinate is negative and y-coordinate is negative. So, it lies in III quadrant So, given points do not lie in the same quadrant.
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Related questions

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

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.

The points (- 5, 2) and (2, -5) lie in the -Maths 9th

Description : The points (- 5, 2) and (2, -5) lie in the -Maths 9th

Last Answer : (c) In point (-5,2), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant and in point (2, – 5), x- coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

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

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

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.

Points (1, -1), (2, – 2), (4, – 5) and (-3, – 4) -Maths 9th

Description : Points (1, -1), (2, – 2), (4, – 5) and (-3, – 4) -Maths 9th

Last Answer : (d) In points (1, -1), (2, -2) and (4, -5) x-coordinate is positive and y-coordinate is negative, So, they all lie in IV quadrant. In point (-3, – 4) x-coordinate is negative and y-coordinate is negative. So, it lies in III quadrant So, given points do not lie in the same quadrant.

The points (- 5, 2) and (2, -5) lie in the -Maths 9th

Description : The points (- 5, 2) and (2, -5) lie in the -Maths 9th

Last Answer : (c) In point (-5,2), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant and in point (2, – 5), x- coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.

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.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

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.

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

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

Two points with coordinates (3, 4) and (–5, 4) lie on a line parallel to which axis? Justify your answer. -Maths 9th

Description : Two points with coordinates (3, 4) and (–5, 4) lie on a line parallel to which axis? Justify your answer. -Maths 9th

Last Answer : Solution :- y-coordinate of both the points is 4. So, both points lie on the line y = 4 which is parallel to x-axis.

If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Description : If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Last Answer : Abscissa of P – Abscissa of Q = (–2) – (–3) = –2 + 3 = 1.

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

Description : 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

Last Answer : Solution :- The obtained triangle is an isosceles triangle.

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 :-

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...

Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

Find the ratio in which the x-axes divides the line joining the points (–2, 5) and (1, –9) ? -Maths 9th

Description : Find the ratio in which the x-axes divides the line joining the points (–2, 5) and (1, –9) ? -Maths 9th

Last Answer : Let the co-ordinates of the point of internal division A be (x, y). Then,\(x\) = \(rac{2 imes(-7)+3 imes8}{2+3}\) = \(rac{-14+24}{5}\) = \(rac{10}{5}\) = 2y = \(rac{2 imes4+3 imes9}{2+3}\) = \(rac{8+27}{5}\) = \(rac{35}{5}\) = 7∴ Co-ordinates of the point for internal division are (2, 7).

If the points A(a, –11), B(5, b), C(2, 15) and D(1, 1) are the vertices of a parallelogram ABCD, find the values of a and b. -Maths 9th

Description : If the points A(a, –11), B(5, b), C(2, 15) and D(1, 1) are the vertices of a parallelogram ABCD, find the values of a and b. -Maths 9th

Last Answer : Let the x-axis divide the line joining the points (-2, 5) and (1, -9) in the ratio k : 1. Let the point of division on x-axis is P Then,\(x\) = \(rac{k-2}{k+1}\), y = \(rac{-9k+ ... (rac{5}{9}\)k being positive, the division is internal. ∴ x-axis divides the given line internally in the ratio 5 : 9.

Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Description : Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Last Answer : Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, -2) Area of ΔABC = \(rac{1}{2}\) |{x1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}|= \(rac{1}{2}\) | \(x\)(1 + 2) + 2(-2 - y) + 3(y - 1) | = \(rac{1}{2 ... \(rac{3}{2}\)∴ Co-ordinates of A are \(\bigg(rac{7}{2},rac{13}{2}\bigg)\) or \(\bigg(rac{-3}{2},rac{3}{2}\bigg)\)

ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are mid-points -Maths 9th

Description : ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are mid-points -Maths 9th

Last Answer : (b) RhombusAB = \(\sqrt{(3-1)^2+(5-1)^2}\) = \(\sqrt{4+16}\) = \(\sqrt{20}\) = \(2\sqrt5\)BC = \(\sqrt{(1-5)^2+(1-3)^2}\) = \(\sqrt{16+4}\) = \(\sqrt{20}\) = \(2\sqrt5\)CD = \ ... = \(6\sqrt2\)Now, AB = BC = CD = AD ⇒ All sides are equal Also, AC ≠ BD ⇒ Diagonals are not equal. ⇒ ABCD is a rhombus.

The points A(2, 3), B(3, 5), C(7, 7) and D(5, 6) are such that: -Maths 9th

Description : The points A(2, 3), B(3, 5), C(7, 7) and D(5, 6) are such that: -Maths 9th

Last Answer : (d) 24 unitsAB ⊥ chord PQ ⇒ AB bisects chord PQ ⇒ PQ = 2PB. AB = \(\sqrt{(2-5)^2+(-3-1)^2}\) = \(\sqrt{(-3)^2+(-4)^2}\)= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5AP = radius of circle = 13 ∴ By Pythagoras' ... = \(\sqrt{AP^2-AB^2}\)= \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 units∴ PQ = 2 PB = 24 units.

In what ratio is the line joining the points (2, –3) and (5, 6) divided by the x-axis. -Maths 9th

Description : In what ratio is the line joining the points (2, –3) and (5, 6) divided by the x-axis. -Maths 9th

Last Answer : (b) (2, 1) (- 2, 1)Let PQRS be the required square and P(0, -1) and R(0, 3) be its two opposite vertices. Length of diagonal PR = \(\sqrt{(0-0)^2+(3+1)^2}\) = \(\sqrt{16}\) = 4∴ Length of each side = \( ... + 4 = 8 ⇒ a2 = 4 ⇒ a = 2. ∴ The other two vertices of the square are (+2, 1) and (-2, 1).

If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th

Description : If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th

Last Answer : (d) 1 : 16Gvien \(rac{AD}{AB}=rac{1}{4}\) ⇒ \(rac{AD}{DB}=rac{1}{3}\),i.e., D divides AB internally in the ratio 1 : 3. ∴ Co-odinates of D are\(\bigg(rac{1+12}{1+3},rac{5+18}{1+3}\bigg)\)i.e.,\(\ ... {Area of}\,\Delta{ADE}}{ ext{Area of}\,\Delta{ABC}}\) = \(rac{rac{15}{32}}{rac{15}{2}}\) = 1 : 16.

The line through the points (4, 3) and (2, 5) cuts off intercepts of lengths λ and μ on the axes. Which one of the following is correct ? -Maths 9th

Description : The line through the points (4, 3) and (2, 5) cuts off intercepts of lengths λ and μ on the axes. Which one of the following is correct ? -Maths 9th

Last Answer : (c) a, b, c are in H.P. only for all m As the points A(a, ma), B[b, (m + 1)b] and C[c, (m + 2)c] are collinear. Area of Δ ABC should be equal to zero.⇒ \(rac{1}{2}\)[x1(y2 - y3) + x2(y3 - y1) + ... - bc = 0 ⇒ ab + bc = 2ac ⇒ b = \(rac{2ac}{a+c}\)∴ a, b, c are harmonic progression (H.P.) for all m.

A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

Without plotting the points indicate the quadrant in which they lie, if : (i) ordinate is 5 and abscissa is – 3 (ii) abscissa is -5 and ordinate is – 3 -Maths 9th

Description : Without plotting the points indicate the quadrant in which they lie, if : (i) ordinate is 5 and abscissa is – 3 (ii) abscissa is -5 and ordinate is – 3 -Maths 9th

Last Answer : answer:

Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Description : Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Last Answer : answer:

Without plotting the points indicate the quadrant in which they will lie, if (i) the ordinate is 5 and abscissa is – 3 -Maths 9th

Description : Without plotting the points indicate the quadrant in which they will lie, if (i) the ordinate is 5 and abscissa is – 3 -Maths 9th

Last Answer : (i) In the point (−3,5) abscissa is negative and ordinate is positive, so it lies in the second quadrant. (ii) In the point (−5,−3) abscissa and ordinate both are negative, so it lies in the ... . (iv) In the point (3,5) abscissa and ordinate both are positive, so it lies in the first quadrant

If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Description : If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Last Answer : Since A and B lie on the circle having centre O. Therefore, OA = OB root(4−2)2+(3−3)2​=(root(x−2)2+(5−3)2​2=root(x−2)2+4​(x−2)2+4=4 (x−2)2=0 x−2=0 x=2 hope it helps thank you

Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Description : Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Last Answer : answer:

Points P (5, -3) is one of the two points of trisection of the line segment joining points A(7, -2) and B(1, -5) near to A. find the coordinates of the other point of trisection. -Maths 9th

Description : Points P (5, -3) is one of the two points of trisection of the line segment joining points A(7, -2) and B(1, -5) near to A. find the coordinates of the other point of trisection. -Maths 9th

Last Answer : answer:

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

Description : 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

Last Answer : As we can see from the figure, that two circles have two points in common. Two circles cannot intersect each other at more than two points. Let us assume that two circles cut each other at ... circle can pass. So, two circles if intersect each other will intersect at maximum two points.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Now, In ΔACD, R and S are the mid points of CD and DA respectively. , ... , PQRS is parallelogram. PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

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

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.