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 : If x(square) - 1 is a factor of ax(cube) + bx(square) + cx + d,show that a+c=0. -Maths 9th
Last Answer : Solution :-
Description : If (x^2 – 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, then : -Maths 9th
Last Answer : answer:
Description : If a, b, c are distinct positive integers, then ax^(b–c) + bx^(c–a) + cx^(a–b) is -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.
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 : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@)` and `/_ ABO=20^(@)`. If `/_ OCB=(1)/(2) /_ A
Last Answer : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@) ... OCB=(1)/(2) /_ ACO,` then find `/_ BOC.`
Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th
Description : If (x – 2) is a common factor of the expressions x^2 + ax + b and x^2 + cx + d, -Maths 9th
Description : If (x + k) is the HCF of ax^2 + ax + b and x^2 + cx + d, then what is the value of k ? -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 : If the roots of the equation x^3 – ax^2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b ? -Maths 9th
Last Answer : Let the roots of the equation x3 – ax2 + bx – c = 0 be (α – 1), α, (α + 1) ∴ S2 = (α – 1)α + α(α + 1) + (α + 1) ( ... ; 1 = b ⇒ 3α2 – 1 = b ∴ Minimum value of b = – 1, when α = 0.
Description : If the expression ax^2 + bx + c is equal to 4, when x = 0, leaves a remainder 4 when divided by x + 1 and leaves a -Maths 9th
Last Answer : Given exp. f(x) = ax2 + bx + c ∴ When x = 0, a.0 + b.0 + c = 4 ⇒ c = 4. The remainders when f(x) is divided by (x + 1) and (x + 2) respectively are f(–1) and f(–2). ∴ f( ... 2b = 2 ...(ii) Solving (i) and (ii) simultaneously we get, a = 1, b = 1.
Description : If (x – 1) is a factor of Ax^3 + Bx^2 – 36x + 22 and 2^B = 64^A, find A and B ? -Maths 9th
Last Answer : Solution:- x - 1 = 0 x = 1 Let p(x) = Ax³ + Bx² - 36x + 22 p(1) = A(1)³ + B(1)² - (36 × 1) + 22 ⇒ A + B - 36 + 22 =0 ⇒ A + B - 14 = 0 ⇒ A + B = 14 ....... ... 22 p(1) = 2(1)³ + 12(1)² - (36 × 1) + 22 ⇒ 2 + 12 - 36 + 22 ⇒ 36 - 36 = 0
Description : The values of a, b and c respectively for the expression f(x) = x^3 + ax^2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are : -Maths 9th
Last Answer : f′(x)=3x2+2ax+6 f(x)⇒f′(x)≥0 3x2+2ax+6≥0 ⇒D≤0 4[a2−3b]≤0 a2≤3b ∴P=6×6×6(16)×6=3616=94 Answer.
Description : If ax^3 + bx^2 + x – 6 has (x + 2) as a factor and leaves a remainder 4, when divided by (x – 2), the value of a and b respectively are : -Maths 9th
Last Answer : Let p(x) = ax³ + bx² + x - 6 A/C to question, (x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0 So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0 ⇒ ... --(2) solve equations (1) and (2), 4a = 0 ⇒a = 0 and b = 2 Then, equation will be 2x² + x - 6
Description : Arrange in descending order of magnitude : -Maths 9th
Last Answer : The given surds are of order 3,6 and 9 respectively . The L.C.M. of 3,6 and 9 is 18 . Reducing each surd to a surd of order 18.
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 : 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 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 : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th
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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 : 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 : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th
Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th
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
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 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 : 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 : 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 : Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th
Description : In quadratic equation ax^2 + bx + c = 0 , Identities the sum and product of Roots? -Maths 9th
Last Answer : If the two roots of the quadratic equation ax2 + bx + c = 0 obtained by the quadratic formula be denoted by a and b, then we have α = \(\frac{-b+\sqrt{b^2-4ac}}{2a},\) β = \(\frac{-b-\sqrt{b^2-4ac}}{2a}\) ∴ Sum of roots ... then α + β = - (- \(\frac{5}{6}\)) = \(\frac{5}{6}\), ab = \(\frac{7}{6}\).
Description : Which one of the following is the equation whose roots are respectively three times the roots of the equation ax^2 + bx + c = 0 ? -Maths 9th
Description : If the roots of the equation ax^2 + bx + c = 0 are equal in magnitude but opposite in sign, then which one of the following is correct ? -Maths 9th
Description : If the sum of the roots of the equation ax^2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct ? -Maths 9th
Last Answer : Given equation: ax2+bx+c=0 Let α and β be the roots of given quadratic equation Sum of the roots i.e. α+β=a−b Product of roots i.e. αβ=ac It is given that, Sum of the roots = Sum of squares of the roots i ... )2−2αβ i.e. a−b =(a−b )2−a2c i.e. −ab=b2−2ac i.e. ab+b2=2ac Hence, C is the correct option.
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 : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th