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

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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 . 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)

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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 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

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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 : 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.

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Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

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Description : In the given figure, line DE is parallel to line AB. CD = 3 while DA = 6. Which of the following must be true? -Maths 9th

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Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

Description : In the figure if l parallel m, then find the value of x -Maths 9th

Last Answer : as L ll m Step-by-step explanation: :. 30 + 40 + y = 180 --------------------------------------(let's take the third angle as y) (because of angle sum property of triangle) 70 + y = 180 y = 110 ... + x = 180 ----------------------(co interior angles) :. x = 180 - 110 = 70 hence solved!!!!

Description : In the figure if l parallel m, then find the value of x -Maths 9th

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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

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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

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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

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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

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Description : Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

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Description : If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Last Answer : Given, (a−b)3=a3−3ab(a−b)−b3 (3x−2y)3=27x3−8y3−3(3x)(2y)(3x−2y) 113=27x3−8y3−18(12)(11) 27x3−8y3=3707

Description : If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

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Description : if 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2. -Maths 9th

Last Answer : Given 2x+3y=8 and xy=2, formula, (a+b)2=a2+b2+2ab ∴(2x+3y)2=4x2+9y2+2(2x)(3y) (2x+3y)2=4x2+9y2+12xy 82=4x2+9y2+12(2) ∴4x2+9y2=64−24=40

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Description : Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th

Last Answer : Consider the equation 3x + 2y = 12 Now, square both sides: (3x + 2y)2 = 122 => 9x2 + 12xy + 4y2 = 144 =>9x2 + 4y2 = 144 – 12xy From the questions, xy = 6 So, 9x2 + 4y2 = 144 – 72 Thus, the value of 9x2 + 4y2 = 72