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
Last Answer : This is the sketch of the question but its hard to answer.
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.
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 :-
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
Last Answer : answer:
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
Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th
Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3
Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3
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 : 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 .
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.
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 : 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 : NEED ANSWER
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 : 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 : 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 : 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 : 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
Description : in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th
Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -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.
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
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.
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
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
Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th
Description : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th
Description : In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th
Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -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
Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -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
Description : Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th
Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th
Description : Find the area of triangle ABC whose vertices are A (-5, 7), B (-4, -5) and C (4, 5). -Maths 9th
Last Answer : the answer is 56.55u because height is 8.7 base is 13 cm
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 : In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th
Last Answer : First Find the points D and E by midpoint formula. (x₂+x₁/2 , y₂+y₁/2) For DE=1/2AB In ΔsCED and CAB ∠ECD=∠ACB and the ratio of the side containing the angle is same i.e, CD=1/2BC ⇒CD/BC=1/2 EC=1/2AC ⇒EC/AC=1/2 ∴,ΔCED~ΔCAB hence the ratio of their corresponding sides will be equal, DE=1/2AB
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