X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th
by

1 Answer

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. Then, ar (ΔXMZ) = ar (ΔXLZ) …(i) On adding ar (ΔXYZ) both sides of Eq. (i), we get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.
by

Related questions

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

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)

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

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.

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

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)

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

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:

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 : 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

Last Answer : answer:

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

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

Last Answer : NEED ANSWER

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

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

Last Answer : According to question find the value of z

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 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

Last Answer : answer:

If x + y = 5 and xy = 4 , find x - y , using identities. -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

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

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.

If x + y = 5 and xy = 4 , find x - y , using identities. -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

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

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.

Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Description : Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Last Answer : NEED ANSWER

Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

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

If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

Description : If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

Last Answer : Solution :-

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 : 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

Last Answer : answer:

x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Description : x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Last Answer : answer:

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 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

Last Answer : answer:

If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -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

Last Answer : answer:

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 : 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

Last Answer : answer:

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

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

Last Answer : answer:

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

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

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

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

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Description : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.

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 : 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

Last Answer : answer:

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 : 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

Last Answer : answer:

Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x​ 2 ⇒h= 4 4y 2 −x 2 ​ ​ cm ∴Area= 2 1​ ×base×h = 2 x​ 4 4y 2 −x 2 ​ ​ cm 2

Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th

Description : Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th

Last Answer : hope this helps you

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

If the side of an equilateral triangle is x unit, then find the area of the triangle. -Maths 9th

Description : If the side of an equilateral triangle is x unit, then find the area of the triangle. -Maths 9th

Last Answer : Solution :- √3/4.x2 sq. unit

In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Description : In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : answer:

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

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

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

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

If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

Description : If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

Last Answer : answer:

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

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

If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

Description : If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

Last Answer : answer:

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

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

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 .

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

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 .

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Description : Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Last Answer : Solution :- The obtained triangle is an isosceles triangle.

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 :-

Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Description : Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Last Answer : (d) \(rac{3\sqrt3a^2}{32}\)Let AB = a be the side of the outermost square.Then AG = AH = \(rac{a}{2}\)⇒ GH = \(\sqrt{rac{a^2}{4}+rac{a^2}{4}}\) = \(rac{a}{\sqrt2}\)∴ Diameter of circle = \(rac{a} ... rac{\sqrt3}{2}\) = \(rac{\sqrt3a^2}{32}\)∴ Area of ΔPQR = 3 (Area of ΔPOQ) = \(rac{\sqrt3a^2}{32}\)