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

Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th
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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-point formula.\(rac{x_1+x_2}{2}\) = 1 and \(rac{y_1+y_2}{2}\) = 2  ⇒ x1 + x2 = 2 and y1 + y2 = 4          ...(i)\(rac{x_1+x_3}{2}\) = 2 and \(rac{y_1+y_3}{2}\) = -1  ⇒ x1 + x3 = 2 and y1 + y3 = -2          ...(ii)\(rac{x_2+x_3}{2}\) = 0 and \(rac{y_2+y_3}{2}\) = -1  ⇒ x2 + x3 = 0 and y2 + y3 = -2          ...(iii)From (i), (ii) and (iii)x1 + x2 + x1 + x3 +x2 + x3 = 2 + 4 + 0 and y1 + y2 + y1 + y3 + y2 + y3 = 4 + (–2) + (–2)⇒ x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0Now, (x1 + x2 + x3) – (x1 + x2) = 3 – 2 ⇒ x3 = 1 and (y1 + y2 + y3) – (y1 + y2) = 0 – 4 ⇒ y3 = – 4 (x1 + x2 + x3) – (x1 + x3) = 3 – 4 ⇒ x2 = –1 and (y1 + y2 + y3) – (y1 + y3) = 0 – (–2) ⇒ y2 = 2 (x1 + x2 + x3) – (x2 + x3) = 3 – 0 ⇒ x1 = 3 and (y1 + y2 + y3) – (y2 + y3) = 0 – (–2) ⇒ y1 = 2∴ The vertices of the ΔPQR are P(3, 2), Q(–1, 2) and R(1, –4)Hence, co-ordinates of centroid of ΔPQR = \(\bigg(rac{3+(-1)+1}{3},rac{2+2+(-4)}{3}\bigg)\) = (1, 0).
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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 = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

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

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 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).

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 :-

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°