The median of a triangle divides it into two -Maths 9th

1 Answer

Answer :

(a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.

Related questions

Description : The median of a triangle divides it into two -Maths 9th

Last Answer : (a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.

Description : Prove that median of a triangle divides it into two triangles of equal area. -Maths 9th

Last Answer : Solution :-

Description : Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th

Last Answer : Solution :-

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

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 .

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

Description : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

Description : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

Description : Prove that the diagonal divides a parallelogram into two congruent triangles. -Maths 9th

Last Answer : Solution :-

Description : The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th

Last Answer : answer:

Description : Find the coordinates of the point which divides the join of the points (8, 9) and (–7, 4) internally in the ratio 2 : 3. -Maths 9th

Last Answer : The circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(3, 0), B(-1, -6), and C(4, -1) be the vertices of ΔABC and P(x, y) be the circumcentre of this triangle. Then, PA = PB = PC ⇒ PA2 = PB2 = ... PC = \(\sqrt{(1-3)^2+(-3-0)^2}\) = \(\sqrt{4+9}\) = \(\sqrt{13}\) units.

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).

Description : Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).

Description : Find the coordinates of the point which divides externally the join of the points (3, 4) and (– 6, 2) in the ratio 3 : 2. -Maths 9th

Last Answer : (d) D lies on the boundary of ΔABC∵ Mid-point of BC = \(\bigg(rac{7+3}{2},rac{7+5}{2}\bigg)\), i.e, (5, 6). we can easily show that D lies on the boundary of ΔABC.

Description : The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Last Answer : (a) (–24, –2)Co-ordinates of the point of external division are\(\bigg(rac{m_1\,x_2-m_2\,x_1}{m_1-m_2},rac{m_1y_2-m_2y_1}{m_1-m_2}\bigg)\), i.e.,∴ Required point = \(\bigg(rac{3 imes-6-2 imes4}{3-2},rac{3 imes2-2 imes4}{3-2}\bigg)\)= \(\big(rac{-24}{1},rac{-2}{1}\big)\), i.e., (–24, –2).

Description : Determine the ratio in which the point P(m, 6) divides the join of A(– 4, 3) and B(2, 8). Also find the value of m. -Maths 9th

Last Answer : (b) 1 : 2Any point on the x-axis is (a, 0).Let the point (a, 0) divide the join of A(2, -3) and B(5, 6) in the ratio k : 1. Then the co-ordinates of the point of division are \(\bigg(rac{5k+2}{k+1},rac{6k-3}{k+1}\ ... 6k - 3 = 0 ⇒ k = \(rac{1}{2}\)Required ratio is k : 1 ⇒ \(rac{1}{2}\) : 1 = 1 : 2.

Description : If (–5, 4) divides the line segment between the co-ordinate axes in the ratio 1 : 2, then what is its equation ? -Maths 9th

Last Answer : (d) x = yThe equations of the given lines are: 4x + 3y = 12 ...(i) 3x + 4y = 12 ...(ii) Solving the simultaneous equations (i) and (ii), we get\(x\) = \(rac{12}{7}\), y = \(rac{12}{7}\)∴ Point of the ... )isy - 0 = \(\bigg(rac{rac{12}{7}-0}{rac{12}{7}-0}\bigg)\) (x - 0), i.e., y = x.

Description : The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Last Answer : (d) 2 : 1 externallyThe mid-point of the line joining the points (-10, 8) and (- 6, 12) is\(\bigg(rac{-10+(-6)}{2},rac{8+12}{2}\bigg)\), i.e., (-8, 10).Let (-8, 10) divide the join of (4 ... 6k = 12 ⇒ k = -2 Since the value of k is negative, it is a case of external division and the ratio is 2 : 1.

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : NEED ANSWER

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : NEED ANSWER

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : According to question find the median of the data

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : The child does not understand, that data has to be arranged in ascending or descending order before finding the median.

Description : A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th

Last Answer : In order to find the median, data has to be arranged in ascending or descending order before finding the median.

Description : Find the median of the numbers: 4, 4, 5, 7, 6, 7, 7, 12, 3. -Maths 9th

Last Answer : Arranging the data in ascending order, we get 3, 4, 4, 5, 6, 7, 7, 7, 12 Here, n = 9 Median = (9 +1)th/2 observation = 5th observation = 6.

Description : If AD is the median of DABC, then -Maths 9th

Last Answer : answer:

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:

Description : What median of an isosceles triangle is the same segment in the triangle as the leg bisector hypotenuse altitude?

Last Answer : It is the median which divides the side which is not one of theequal sides.

Description : If is both the altitude and median of triangle ABC then triangle ABC is?

Last Answer : It is isosceles.

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 cm²

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Last Answer : each interior opposite angles are 55

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Description : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

Description : Can a triangle have two obtuse angles? Give reason for your answer. -Maths 9th

Last Answer : No, because if the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.

Description : Prove that a triangle must have at least two acute angles. -Maths 9th

Last Answer : Given ΔABC is a triangle. To prove ΔABC must have two acute angles Proof Let us consider the following cases Case I When two angles are 90°. Suppose two angles are ∠B = 90° and ∠C = 90°

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Last Answer : each interior opposite angles are 55

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.