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

1 Answer

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 through midpoint M of AB, is the set of all points that have equal distances to A and B.

Related questions

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

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 : 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 : 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.

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 : 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.

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

Last Answer : Solution :-

Description : Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. -Maths 9th

Last Answer : Solution :-

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

Last Answer : Solution :-

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : First observe that the altitudes from any vertex to the opposite sides of an equilateral triangle are all of equal length. Hence we can define the height of an equilateral triangle as this common value of three ... ∠MAB=30∘ and ∠MAC=30∘, with B and C on XY. Then ABC is the required triangle.

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : s= 2 a+b+c​ = 2 35+54+61​ =75 Area, A= s(s−a)(s−b)(s−c)​ = 75(75−35)(75−54)(75−61)​ =420 5​ cm 2 Now, Area of the triangle is also given as A= 2 1​ ×a×h Where, h is the longest altitude. Therefore, 2 1​ ×a×h=420 5​ Hence, h=24 5​ cm

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Description : An equilateral triangle, if its altitude is 3.2 cm. -Maths 9th

Last Answer : We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60°. Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC ... ∠DBA = 60° Similary, ∠DCA = 60° Thus, ∠A = ∠B=∠C = 60° Hence, ΔABC is an equilateral triangle.

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : The length of its longest altitude

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

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

Description : Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line XY. (ii) Construct perpendicular PD at any point D on the line XY. (iii) From point D, cut-off line segment AD = 6 cm. (iv) Construct ∠BAD = ∠CAD ... 30 °+ 30° = 60° and AD perpendicular BC therefore, △ABC is an equilateral triangle with altitude AD = 6 cm.

Description : Find the area of an equilateral triangle having altitude h cm -Maths 9th

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

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

Last Answer : answer:

Description : A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Last Answer : answer:

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.

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

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

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

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........

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

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

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

Last Answer : According to question prove that P, Q, R and D are concyclic.

Description : 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

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

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.

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

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

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

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........

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

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

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

Last Answer : According to question prove that P, Q, R and D are concyclic.

Description : 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

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

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

Last Answer : NEED ANSWER

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

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

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

Last Answer : Solution :-

Description : Prove that each angle of an Euilateral triangle is 60 degree. -Maths 9th

Last Answer : Solution :-

Description : If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -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 : Prove that in a triangle,other than an euilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle. -Maths 9th

Last Answer : Solution :-

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.

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

Last Answer : We have ∠BED = ∠BAD (Angles in the same segment) ⇒ ∠BED = 1/2∠A ...(i) Also, ∠BEF = ∠BCF (Angles in the same segment) ⇒ ∠BEF = 1/2∠C ...(ii) From (i) and (ii) ∠BED + ∠BEF = 1/2∠A + 1/2∠C ∠DEF ... ∠A + ∠C) ⇒ ∠DEF = 1/2(180° - ∠B) (Since, ∠A + ∠B + ∠C = 180°) ⇒ ∠DEF = 90° - 1/2∠B

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

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

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

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

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...

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

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