Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th
Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)
Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th
Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.
Description : In the given figure, YZ is parallel to MN, XY is parallel is LM and XZ is parallel to LN. Then MY is -Maths 9th
Last Answer : answer:
Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th
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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 : 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 : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th
Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0
Description : In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th
Last Answer : Given , LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinates of point P=(3,2) [since,its perpendicular distance from X-axis is 2] Coordinate of point ... L=3 , abscissa of point M=3 ∴∴ Difference between the abscissa of the points L and M =3-3=0
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 x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th
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Last Answer : According to question find the value of z
Description : If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th
Last Answer : if pqrs is a trapezium so pq and RS are parallel is you draw a diagonal you will divide the trapezium into two parts such that two equal triangle so
Description : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at origin. (ii) Find the points on the curve`4x^(2)+9 y^(2
Last Answer : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at ... ` at which the normal drawn is parallel to X-axis.
Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th
Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC
Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th
Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ 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 : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th
Description : Which one of the following is one of the factors of x^2 (y – z) + y^2 (z – x) – z (xy – yz – zx) ? -Maths 9th
Description : If a = (xy)/(x+y), b = (xz)/(x+z), and c = (yz)/(y+z), where a, b and c are non-zero, then what is x equal to ? -Maths 9th
Description : If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th
Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th
Last Answer : Solution :-
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 : MN and PS are two equal chords of a circle drawn on either side of centre O of the circle . Both the chords are produced to meet at point A. If the ra
Last Answer : MN and PS are two equal chords of a circle drawn on either side of centre O of the circle . Both the chords ... 12 `cm and OA `=17` cm , then find NA.
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 : 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 : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th
Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)
Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th
Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF=ABFB=1 from (1) ⇒LF=LD so, DF=2DL
Description : At what point does the graph of the linear equation 2x + 3y = 9 meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis? -Maths 9th
Description : Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th
Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -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
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
Description : If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th
Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3
Description : Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th
Last Answer : No, (xy) is necessarily an irrational only when x ≠0. Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since ... wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.
Description : Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th
Last Answer : x3 + y3 = x+y (x2 + y2 -xy) RHS = x+y (x2 + y2 -xy) On expanding, = x3 + xy2 - x2y + x2y + y3 - xy2 = x3 + y3 LHS = RHS
Description : If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th
Description : x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th
Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th
Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th
Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square
Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th