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

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According to question the area of ΔGBC = area of the quadrilateral AFGE.

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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 : 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]

Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

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

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

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

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Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

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

Last Answer : It is isosceles.

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

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

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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.

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 .

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

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

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Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

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Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

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Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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Description : Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

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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

Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

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Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

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Description : In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Last Answer : Area of ΔABC = \(rac{1}{2}\) x 3 x 4 cm2 = 6 cm2. Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.∴ Area of ΔABC = \(rac{1}{2}\) x BD x AC ⇒ 6 = \(rac{1}{2}\) BD x 5 ⇒ BD = \(rac{12}{5}\) cm.Now in ΔABD, AD = \(\ ... \(rac{1}{2}\)x AD x BD = \(rac{1}{2}\) x \(rac{9}{5}\) x \(rac{12}{5}\) = \(rac{54}{25}\) cm2.

Description : In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral -Maths 9th

Last Answer : (b) 7√3 cm2.AB = 6 cm, ∠C = 60º (∴ ∠A = ∠B = 60º) ∴ ΔABC is an equilateral triangle Area of ΔABC = \(rac{\sqrt3}{4}\) × (6)2 = 9√3 Area of (ΔADE + ΔBFG) = 2 x \(\bigg(rac{\sqrt3}{4} imes(2)^2\bigg)\) = 2√3 ∴ Area of pentagon = 9√3 - 2√3 = 7√3 cm2.

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

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Description : If a, b, c are the sides of a non-equilateral triangle, then the expression (b + c – a) (c + a – b) (a + b – c) – abc is -Maths 9th

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Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

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Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

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