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

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

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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.

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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.

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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.

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Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

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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.

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

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.

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.

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.

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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 .

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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

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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.

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

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.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

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If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

Prove that altitude of a triangle are equal -Maths 9th

Description : Prove that altitude of a triangle are equal -Maths 9th

Last Answer : 1 Answer. Proof: To prove this, I must first prove that the three perpendicular bisectors of a triangle are concurrent. From Figure 1, consider triangle ABC, I know that the perpendicular bisector of AB, passing ... midpoint M of AB, is the set of all points that have equal distances to A and B.

Prove that altitude of a triangle are equal -Maths 9th

Description : Prove that altitude of a triangle are equal -Maths 9th

Last Answer : Proof: To prove this, I must first prove that the three perpendicular bisectors of a triangle are concurrent. From Figure 1, consider triangle ABC, I know that the perpendicular bisector of AB, passing ... on the perpendicular bisector. Because, PM is congruent to PM, MA is congruent to MB and

Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th

Description : Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th

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

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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.

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Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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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°

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

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°

Prove that a triangle must have atleast two acute angles. -Maths 9th

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

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If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

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

In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Description : In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Last Answer : The sum of angles of a triangle is180° If one aangke is of 90° then the sum of two angles is 90° It means that the angle forming 90° is biggest angle We know , Angle opposite to the longest side is largest. It means hypotenuse is the biggest side of right angled triangle

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........

state and prove angle sum property of triangle -Maths 9th

Description : state and prove angle sum property of triangle -Maths 9th

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If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

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Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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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.

In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

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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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state and prove angle sum property of triangle -Maths 9th

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If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

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Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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