(0, –1) and (0, 3) are the two opposite vertices of a square. The other two vertices are: -Maths 9th

(0, –1) and (0, 3) are the two opposite vertices of a square. The other two vertices are: -Maths 9th
by

1 Answer

Answer :

(c) (15, 19)Let Δ(a, b) be the fourth vertex of the parallelogram ABCD. The diagonals of a parallelogram bisect each other at point O (say), so the diagonals AC and BD have the same mid-point.
by

Related questions

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).

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).

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...

If (0, 0) and (2, 0) are the two vertices of a triangle whose centroid is (1, 1), then the area of the triangle is: -Maths 9th

Description : If (0, 0) and (2, 0) are the two vertices of a triangle whose centroid is (1, 1), then the area of the triangle is: -Maths 9th

Last Answer : (b) \(\bigg(rac{2\sqrt{13}+20\sqrt2}{\sqrt{13}+\sqrt{17}+5\sqrt2},rac{8\sqrt{13}-6\sqrt{17}}{\sqrt{13}+\sqrt{17}+5\sqrt2}\bigg)\)Let A(x1, y1), B(x2, y2), C(x3, y3) be the vertices of ΔABC the ... +6)^2}\) = \(\sqrt{4+196}\) = \(\sqrt{200}=10\sqrt{2}\)∴ Co-ordinates of incentre of Δ ABC are

Find the co-ordinates of the circumcentre of the triangle whose vertices are (3, 0), (–1, –6) and (4, –1). Also find its circum-radius. -Maths 9th

Description : Find the co-ordinates of the circumcentre of the triangle whose vertices are (3, 0), (–1, –6) and (4, –1). Also find its circum-radius. -Maths 9th

Last Answer : Let A ≡ (2, - 2), B ≡ (-2, 1), C ≡ (5, 2 ). Then,AB = \(\sqrt{(-2-2)^2+(1+2)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5BC = \(\sqrt{(5+2)^2+(2-1)^2}\) = \(\sqrt{49+1}\) = \(\sqrt{50}\) = \( ... of ΔABC = \(rac{1}{2}\) x base x height = \(rac{1}{2}\) x AB x AC = \(rac{1}{2}\)x 5 x 5 = 12.5 sq. units.

Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th

Description : Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th

Last Answer : Let A(1, 2), B(0, -1) and C(2, -1) be the mid-points of the sides PQ, QR and RP of the triangle PQR. Let the co-ordinates of P, Q and R be (x1, y1), (x2, y2) and (x3 , y3) respectively. Then, by the mid- ... ordinates of centroid of ΔPQR = \(\bigg(rac{3+(-1)+1}{3},rac{2+2+(-4)}{3}\bigg)\) = (1, 0).

Find the area of the quadrilateral whose vertices are (3, 4), (0, 5), (2, –1) and (3, –2). -Maths 9th

Description : Find the area of the quadrilateral whose vertices are (3, 4), (0, 5), (2, –1) and (3, –2). -Maths 9th

Last Answer : Let A(-36, 7), B(20, 7) and C(0, -8) be the vertices of the given triangle.Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25b = AC =\(\sqrt{(0-36)^2+(-8-7 ... (rac{-900+780}{120},rac{175+273-448}{120}\bigg)\) = \(\bigg(rac{-120}{120},rac{0}{120}\bigg)\) = (-1, 0)

Without using Pythagoras’ theorem, show that the points A (0, 4), B(1, 2) and C(3, 3) are the vertices of a right angle triangle. -Maths 9th

Description : Without using Pythagoras’ theorem, show that the points A (0, 4), B(1, 2) and C(3, 3) are the vertices of a right angle triangle. -Maths 9th

Last Answer : Slope (m) = \(rac{(y_2-y_1)}{(x_2-x_1)}\) = \(rac{6-2}{5-1}\) = \(rac{4}{4}\) = 1Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction. tan θ = 1 ⇒ θ = tan –11 = 45º.

What is the perimeter of the triangle with the vertices A(–4, 2), B(0, –1) and C(3, 3) ? -Maths 9th

Description : What is the perimeter of the triangle with the vertices A(–4, 2), B(0, –1) and C(3, 3) ? -Maths 9th

Last Answer : The two given lines are ax + by + c1 = 0 and ax + by + c2 = 0. Any line parallel to these two lines and midway between them is ax + by + c = 0 ...(i) Putting x = 0, y = \(-rac{c}{b}\) is ... c1 = c - c2 ⇒ c = \(rac{c_1+c_2}{2}\)∴ Required equation is ax + by + \(rac{c_1+c_2}{2}\) = 0.

The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if -Maths 9th

Description : The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if -Maths 9th

Last Answer : (c) \(rac{b+k}{f+h}\)Let the slope of the lin passing through the points (-k, h) and (b, - f) be m1. Then m1 = \(rac{-f-h}{b+k}\) = \(-\bigg(rac{f+h}{b+k}\bigg)\)\(\bigg[Slope = rac{y_2-y_1}{x_2-x_1}\bigg]\) ... \(-rac{1}{m_1}\)= \(rac{-1}{-\big(rac{f+h}{b+k}\big)}\) = \(\bigg(rac{b+k}{f+h}\bigg)\)

The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Description : The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Last Answer : (b) 3x - y = 0 Given lines are 3x - y - 3 = 0 and 3x - y + 5 = 0. Line parallel to the given lines can be written as 3x - y + c = 0 ...(i) Let us taken a point, say ... 5c + 15 = - 3c + 15 ⇒ 8c = 0 ⇒ c = 0. Substituting c = 0 in (i), the required equation is 3x - y = 0.

The orthocentre of a triangle whose vertices are (0, 0), (3, 0) and (0, 4) is -Maths 9th

Description : The orthocentre of a triangle whose vertices are (0, 0), (3, 0) and (0, 4) is -Maths 9th

Last Answer : (c) 60ºSince p is the length of perpendicular from origin on the straight line ax + by - p = 0.p = \(rac{|a.0+b.0-p|}{\sqrt{a^2+b^2}}\)⇒ 1 = \(\sqrt{a^2+b^2}\) ⇒ 1 ... 60° + y sin 60° = p Hence required angle is 60°, which is the angle between the perpendicular and the positive direction of x-axis.

If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th

Description : If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th

Last Answer : D=slid ht of BC D≅(20+4​,20+2​) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2​=root42+32​=5 hope it helps thank u

Find the area of quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4) -Maths 9th

Description : Find the area of quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4) -Maths 9th

Last Answer : answer:

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

Description : The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

Last Answer : Let A(x1, y1) = (3, 4), B(x2, y2) ≡ (0, 5), C(x3, y3) ≡ (2, -1)and D(x4, y4) ≡ (3, -2) be the vertices of quadrilateral ABCD.Area of quad. ABCD = \(rac{1}{2}\) |{(x1 y2 - x2 y1) + (x2y3 - x3y2) + (x3y4 - x4y3) ... ) + (12 + 6)}|= \(rac{1}{2}\) |{15 - 11 + 0 + 18}| = \(rac{1}{2}\)x 22 = 11 sq. units.

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Description : Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Last Answer : Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance 'a'on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance 'a' on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Last Answer : Given, length of a rectangle = 5 units and breadth of a rectangle = 3 units One vertex is at origin i.e., (0, 0) and one of the other vertices lies in III quadrant. So, the length of the rectangle is 5 ... negative,direction of y-axis and then vertex is C(0, -3). The fourth vertex B is (-5, - 3).

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Last Answer : Given, length of a rectangle = 5 units and breadth of a rectangle = 3 units One vertex is at origin i.e., (0, 0) and one of the other vertices lies in III quadrant. So, the length of the rectangle is 5 ... negative,direction of y-axis and then vertex is C(0, -3). The fourth vertex B is (-5, - 3).

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

Last Answer : Solution :-

Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Last Answer : Solution :-

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Description : If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Last Answer : Solution :- Let, ABCD be a cyclic quadrilateral such that its diagonal AC and BD are the diameters of the circle though the vertices A,B,C and D. As angle in a semi-circle is 900 ∴ ∠ABC = 900 and ∠ADC = 900 ∠DAB = 900 ... Hence, ABCD is a rectangle.

An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted? -Maths 9th

Description : An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted? -Maths 9th

Last Answer : (c) 33.33%When an equilateral triangle is cut from its three vertices to form a regular hexagon then out of the 9 equilateral triangles that form ΔABC, three triangle, ΔADE, ΔFCG,ΔIHB are cut off and 6 remain in the ... to get the hexagon.∴ Area wasted = \(\bigg(rac{1}{3} imes100\bigg)\)% = 33.33%

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

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:

ABCD is a trapezium where AB and CD are non-parallel sides. If the vertices A, B, C and D are concyclic, then -Maths 9th

Description : ABCD is a trapezium where AB and CD are non-parallel sides. If the vertices A, B, C and D are concyclic, then -Maths 9th

Last Answer : answer:

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).

Prove that the points (1, –1) ((-1/2),(1/2)) and (1, 2) are the vertices of an isosceles triangle. -Maths 9th

Description : Prove that the points (1, –1) ((-1/2),(1/2)) and (1, 2) are the vertices of an isosceles triangle. -Maths 9th

Last Answer : (x, y) is equidistant from the points (2, 1) and (1, –2) ⇒ Distance between (x, y) and (2, 1) = Distance between (x, y) and (1, –2)⇒ \(\sqrt{(x-2)^2+(y-1)^2}\) = \(\sqrt{(x-1)^2+(y+2)^2}\)⇒ x2 – 4x + 4 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 4y + 4⇒ – 4x + 2x – 2y – 4y = 0 ⇒ –2x – 6y = 0 ⇒ x + 3y = 0

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 area of a triangle whose vertices are (1, 3), (2, 4) and (5, 6). -Maths 9th

Description : Find the area of a triangle whose vertices are (1, 3), (2, 4) and (5, 6). -Maths 9th

Last Answer : Let OR = \(x\) units (i) ΔQOR ~ ΔPAR⇒ \(rac{PA}{AR}\) = \(rac{QO}{OR}\) ⇒ \(rac{6}{x+4}\) = \(rac{3}{x}\)⇒ \(rac{x}{x+4}\) = \(rac{3}{6}\) ⇒ \(rac{x}{x+4}\) = \(rac{1}{2}\)⇒ 2\(x\) = \(x\) + 4 ⇒ \(x ... = \(rac{1}{2}\) (OQ + AP) x OA = \(rac{1}{2}\) (3+6) x 4 = \(rac{1}{2}\) x 9 x 4 = 18 sq. units.

(–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Description : (–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Last Answer : 3\(x\) - 2y + 5 = 0 ⇒ -2y = -3\(x\) - 5 ⇒ y = \(rac{3}{2}\)\(x\) + \(rac{5}{2}\)On comparing with y = m\(x\) + c, we see that slope of given line = \(rac{3}{2}\)As the required line is perpendicular to the given line, ... - 4)⇒ 3(y - 5) = - 2\(x\) + 8 ⇒ 3y - 15 = -2\(x\) + 8 ⇒ 3y + 2\(x\) - 23 = 0

The coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and (2, –2). -Maths 9th

Description : The coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and (2, –2). -Maths 9th

Last Answer : (a) bx = ayGiven, AM = BM ⇒ AM2 = BM2 ⇒ [x – (a + b)]2 + [y – (b – a)]2 = [x – (a – b)]2 + (y –(a + b))2

Name the quadrilateral ABCD, the coordinates of whose vertices are A(3, 5), B(1, 1), C(5, 3) and D(7, 7). -Maths 9th

Description : Name the quadrilateral ABCD, the coordinates of whose vertices are A(3, 5), B(1, 1), C(5, 3) and D(7, 7). -Maths 9th

Last Answer : (b) Equilateral, \(2\sqrt3a^2\) sq. unitsLet A(a, a), B(-a, -a) and C \((-\sqrt3a,\sqrt3a)\) be the vertices of ΔABC. Then,AB = \(\sqrt{(a+a)^2+(a+a)^2}\) = \(\sqrt{4a^2+4a^2}\) = \(\sqrt{8a^2}\) = \( ... 4}\) (side)2 = \(rac{\sqrt3}{4}\) x (\(2\sqrt2a\))2= \(rac{\sqrt3}{4}\) x 8a2 = \(2\sqrt3a^2\).

The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Description : The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Last Answer : (a) (5, 2)Let A(8, 6) , B(8, -2) and C(2, -2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. Then, PA2 = PB2 = PC2 Now, PA2 = PB2 ⇒ (x - 8)2 + (y - 6)2 = (x ... 4 = x2 - 4x + 4 + y2 + 4y + 4 ⇒ 12x = 60 ⇒ x = 5. ∴ Co-ordinates of the circumcentre are (5, 2).

If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Description : If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Description : Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Last Answer : (b) a = √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg ... \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = √2 .

Find the area of triangle ABC whose vertices are A (-5, 7), B (-4, -5) and C (4, 5). -Maths 9th

Description : Find the area of triangle ABC whose vertices are A (-5, 7), B (-4, -5) and C (4, 5). -Maths 9th

Last Answer : the answer is 56.55u because height is 8.7 base is 13 cm

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

a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

Description : a square is inscribed in an isosceles triangle so that the square and the triangle have one angle common. show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse. -Maths 9th

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

Let the opposite angular points of a square be (3, 4) and (1, – 1), Find the co-ordinates of the remaining angular points. -Maths 9th

Description : Let the opposite angular points of a square be (3, 4) and (1, – 1), Find the co-ordinates of the remaining angular points. -Maths 9th

Last Answer : P ≡ (-3, 2), Q ≡ (-5, -5), R ≡ (2, -3), S ≡ (4, 4)∴ PQ = \(\sqrt{(-5+3)^2+(-5-2)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)QR = \(\sqrt{(2+5)^2+(-3+5)^2}\) = \(\sqrt{49+4}\) = \ ... of rhombus = \(rac{1}{2}\) x (Product of length of diagonals) = \(rac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.

One root of x^2 + kx – 8 = 0 is the square of the other, then the value of k is : -Maths 9th

Description : One root of x^2 + kx – 8 = 0 is the square of the other, then the value of k is : -Maths 9th

Last Answer : answer:

The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a: A(-3,-5) and(a)Square (b) Rhombus (c) Rectangle (d) Trapezium

Description : The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a: A(-3,-5) and(a)Square (b) Rhombus (c) Rectangle (d) Trapezium

Last Answer : (c) Rectangle

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