What are the angle measures of triangle ABC?

What are the angle measures of triangle ABC?
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The 3 interior angles of any triangle add up to 180 degrees
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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.

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........

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........

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

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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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

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

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

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Description : What Two angle measures of this triangle are 60 and deg. Which type of triangle is it?

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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

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.

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.

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.

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.

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

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

Last Answer : NEED ANSWER

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

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...

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]

D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : Solution :-

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

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

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

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

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...

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

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...

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...

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

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...

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

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.

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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.

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.

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