In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th
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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

Description : 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

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

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

Description : 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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Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

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

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

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

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Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

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Last Answer : Solution :- Let OP be perpendicular from O on line l. Since the perpendicular from the centre of a circle to a chord,bisects the chord.Therefore, AP = DP ...(i) BP = CP ...(ii) Subtracting (ii) from (i), we get AP - BP = DP - CP ⇒ AB = CD

For what value of x+y in fig.6.4 will ABC be a line? -Maths 9th

Description : For what value of x+y in fig.6.4 will ABC be a line? -Maths 9th

Last Answer : Solution :- For ABC to be a line, the sum of two adjacent angles must be 180°, i.e., x + y must be equal to 180°.

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

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

Last Answer : Solution :-

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Description : In Fig. 8.28, ABCD is a parallelogram. Find the value of x, y and z. -Maths 9th

Last Answer : Solution :-

In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Description : In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

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Description : In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Last Answer : hope its clearhope its clear

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Last Answer : Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral. ∠ACD+ ∠AED = 180° …(i) [sum of opposite angles in a cyclic quadrilateral is 180°]

In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Description : In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Last Answer : Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral. ∠ACD+ ∠AED = 180° …(i) [sum of opposite angles in a cyclic quadrilateral is 180°]

If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Description : If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Last Answer : (b) 16π cm2.AB = OA ⇒ AB = OA = OB (radii of circle are equal) ⇒ ΔAOB is equilateral. ∴ If ‘r’ is the radius of the circle,then area of ΔAOB = \(rac{\sqrt3}{4}\)side2⇒ \(rac{\sqrt3}{4}\)(r)2 = 4√3 (given)⇒ r2 = 16 ⇒ r = 4∴ Area of circle = πr2 = 16π cm2.

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

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

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Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

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Description : In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

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

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

Last Answer : Solution :- y = 2∠ ACB ⇒ y = 2 65° ⇒ y = 130° OA = ... Or 2x = 50° Or x = 25°

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Description : In Fig. 4.6, if ABC and ABD are equilateral triangles then find the coordinates of C and D. -Maths 9th

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If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

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Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

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In Fig. 6.9, find the value of x. -Maths 9th

Description : In Fig. 6.9, find the value of x. -Maths 9th

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In Fig. 6.17, AB| |CD. Find the value of x. -Maths 9th

Description : In Fig. 6.17, AB| |CD. Find the value of x. -Maths 9th

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

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

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In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

Description : In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th

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What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Description : What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Last Answer : (a) 4x + 3y = 24 Let the x-intercept = a. Then, y-intercept = 14 - a ∴ Eqn of the straight line is \(rac{x}{a}\) + \(rac{y}{14-a}\) = 1Since it passes through (3, 4), so\(rac{3}{a}\) + \(rac{4}{14-a}\) = 1⇒ 3(14 - ... = 1 ⇒ x + y = 7or \(rac{x}{6}\) + \(rac{y}{8}\) = 1 ⇒ 8x + 6y = 48 ⇒ 4x + 3y = 24.

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Last Answer : Solution :-

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

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 sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1. -Maths 9th

Description : If x sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1. -Maths 9th

Last Answer : xsin3θ+ycos3θ=sinθcosθ (xsinθ)sin2θ+(ycosθ)cos2θ=sinθcosθ (xsinθ)sin2θ+(xsinθ)cos2θ=sinθcosθ xsinθ=sinθcosθ x=cosθ Again, ycosθ=xsinθ ycosθ=cosθsinθ y=sinθ Therefore, x2+y2=sin2θ+cos2θ=1.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

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

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)

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

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

Last Answer : Solution :-

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 . 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)

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

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Description : In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : Solution :-

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

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

Last Answer : Solution :-

In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Description : In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Last Answer : Solution :-

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

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°