In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

In Fig. 6.7, if l||m, then find the value of x. -Maths 9th
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l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

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Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

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

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!!!!

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

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

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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

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Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved

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Description : In Fig. 10.11, find the value of x and y. -Maths 9th

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Description : In Fig. 10.17, O is the centre of a circle, find the value of x. -Maths 9th

Last Answer : Solution :- In △APB, ∠APB = 90° (Angle in the semi- circle) ∠APE + ∠ABP + ∠BAP = 180° 90° + 40° + ∠BAP = 180° ⇒ ∠BAP = 180° - 130° ⇒ ∠BAP = 50° Now, ∠BQP = ∠BAP (Angles in the same segment) ∴ x = 50°

In Fig. 6.19, AB| |CD. Determine x. -Maths 9th

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Find the value of x and y if l is parallel to m -Maths 9th

Description : Find the value of x and y if l is parallel to m -Maths 9th

Last Answer : the value should be zero...according to me as I think!!

If l parallel to m, find the value of x. -Maths 9th

Description : If l parallel to m, find the value of x. -Maths 9th

Last Answer : Pls provide diagram, how can we know where is x

Find the value of x and y if l is parallel to m -Maths 9th

Description : Find the value of x and y if l is parallel to m -Maths 9th

Last Answer : the value should be zero...according to me as I think!!

If l parallel to m, find the value of x. -Maths 9th

Description : If l parallel to m, find the value of x. -Maths 9th

Last Answer : Pls provide diagram, how can we know where is x

In the given figure, if l//m, tgen find the value of x -Maths 9th

Description : In the given figure, if l//m, tgen 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!!!!

In the given figure, if l//m, tgen find the value of x -Maths 9th

Description : In the given figure, if l//m, tgen 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!!!!

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Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

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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

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

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If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th

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

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

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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

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]

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

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In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

Description : In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

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In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Description : In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th

Last Answer : Solution :- As OD is perpendicular to AB ⇒ AC = AB (Perpendicular from the centre to the chord bisects the chord) ∴ AC = AB/2 = 8cm In right △OCA, OA2 = AC2 + OC2 (102) = 82 + OC2 OC2 = 100 - 64 OC2 = 36 ... = 6cm CD = OD - OC = 10 - 6 = 4cm [∴ OA = OD = 10cm (radii)]

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

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Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

In Fig.6.5,which of the two lines are parallel? -Maths 9th

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

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In Fig.8.6, BDEF and FDCE are parallelograms.Can you say that BD = CD? -Maths 9th

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Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.

In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

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In Fig. 8.12, ABCD is a square. Find x. -Maths 9th

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The sides BA and DC of quad. ABCD are produced as shown in Fig. 8.23. Prove that x + y = a + b -Maths 9th

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In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

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In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]