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

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
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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 = \(\sqrt{(5+2)^2+(2-1)^2}\) = \(\sqrt{49+1}\) = \(\sqrt{50}\) = \(5\sqrt2\)Ac = \(\sqrt{(5-2)^2+(2+2)^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5⇒ AB2 + AC2 = BC2 as 25 + 25 = 50∴ Δ ABC is right angled at A and hyp. BC = \(5\sqrt2\) unitsArea 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.
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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.

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)

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 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

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

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 = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

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

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 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).

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).