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

Find the area of triangle ABC whose vertices are A (-5, 7), B (-4, -5) and C (4, 5). -Maths 9th
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the answer is 56.55u because height is 8.7 base is 13 cm
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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 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.

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).

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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

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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.

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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

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

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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

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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

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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 .

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

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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

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%

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

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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).

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)

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

Description : in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

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

In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

Description : In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

Last Answer : (a) (873 - 504√3) cm2.Since ∠CPO = ∠COP = 60º, therefore, PCO is also an equilateral triangle. Let each side of the square MNOP be x cm. Then PC = CO = PO = x cm Then in ΔPAM,\(rac{PM}{PA}\) = sin 60º⇒ \(rac{x ... most square = y2= \(\big(3(7-4\sqrt2)\big)^2\)= 9(49 + 48 - 56√3) = (873 - 504√3) cm2.

There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Description : There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

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The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

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

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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

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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).

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

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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

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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 = \(rac{-f-h}{b+k}\) = \(-\bigg(rac{f+h}{b+k}\bigg)\)\(\bigg[Slope = rac{y_2-y_1}{x_2-x_1}\bigg]\) ... \(-rac{1}{m_1}\)= \(rac{-1}{-\big(rac{f+h}{b+k}\big)}\) = \(\bigg(rac{b+k}{f+h}\bigg)\)

A right triangle ABC with sides 5 cm, -Maths 9th

Description : A right triangle ABC with sides 5 cm, -Maths 9th

Last Answer : Let ABC be a right triangle with AB = 12 cm, BC = 5 cm and AC = 13 cm. When △ABC is revolved about AB, it forms a right circular cone of radius BC = 5 cm and height AB = 12 cm. Thus, volume of cone formed = 1/3 πr2h = 1/3 x π x 52 x 12 = 100π cm3

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).

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

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Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

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

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.

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

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.

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

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

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.