In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th
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Answer :

In △OBC, OB = BC ⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal] ∠OBA is the exterior angle of △BOC So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles] Similarly, ∠AOD is the exterior angle of △AOC  ∴ x = 2y + y = 3y
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Related questions

In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Description : In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th

Last Answer : In △OBC, OB = BC ⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal] ∠OBA is the exterior angle of △BOC So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles] Similarly, ∠AOD is the exterior angle of △AOC ∴ x = 2y + y = 3y

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

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

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Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

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Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

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Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

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In the figure below, CD is a chord of the semi circle with centre O. OA is the radius of the circle. If `CD=10` cm, `AB=2` cm and `bar(OA)_|_bar(CD)`

Description : In the figure below, CD is a chord of the semi circle with centre O. OA is the radius of the circle. If `CD=10` cm, `AB=2` cm and `bar(OA)_|_bar(CD)`

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Description : A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then, show that OA^2 + OC^2 = OB^2 + OD^2. -Maths 9th

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Description : Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

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

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

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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

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Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

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l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

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If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

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If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Description : ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Last Answer : answer:

A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Description : A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Last Answer : answer:

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th

Description : What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th

Last Answer : Solution :- Perpendicular line from the centre bisect the chord.

Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.

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Description : The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th

Last Answer : Let PQ be a chord of a circle with centre O and radius 13cm such that PQ = 24cm. From O, draw OM perpendicular PQ and join OP. As, the perpendicular from the centre of a circle to a chord bisects the chord. ∴ PM ... Hence, the distance of the chord from the centre is 5cm.

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Description : The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Last Answer : There are two parallel chords of length 8 cm and 6 cm. The distance between center and shortest chord (6cm chord) is 4cm. So, the perpendicular distance from the center to the shortest chord is 4cm. The ... 32+42 =5. Since the radius is 5. The distance from center to largest chord is 52−42 =3.

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Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

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The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

Description : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

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a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Description : In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Last Answer : AB is the longest chord because it is passing through the Centre hence it is a diameter ThereforeAB>CD

In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Description : In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Last Answer : AB is the longest chord because it is passing through the Centre hence it is a diameter ThereforeAB>CD

In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Last Answer : answer:

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Description : A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Last Answer : According to question prove that the perpendicular bisectors of AB, BC and CA are concurrent.

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Description : A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Last Answer : According to question prove that the perpendicular bisectors of AB, BC and CA are concurrent.

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)

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.

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.