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

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

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 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)

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

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 the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

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

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

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

Last Answer : Solution :-

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Description : In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Last Answer : AD = 24, BC = 12. In ΔCBE and ΔADE, ∠CBA = ∠CDA, ∠BCE = ∠DAE (Angles in the same segment are equal) ∠BEC = ∠DEA (vertical opposite angles are equal) ⇒ ΔBCE and ΔDEA are similar Δs with sides in the ratio 1 : 2. ∴ Ratio of areas = Ratio of square of sides = 12 : 22 = 1 : 4

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)

A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Description : A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Last Answer : When semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes slant height of the cup and the semi-circular part of the sheet becomes the circumference of the base of the cone. ∴ Slant height of the ... 3 x 22/7 x 7 x 7 x 7√3 = 1078/3.√3 = 359.3 x 1.732 = 622.31 cm3

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.

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.

In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Description : In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Last Answer : Given In a figure, two chords AB and CD intersecting each other at point E.

In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Description : In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Last Answer : Given In a figure, two chords AB and CD intersecting each other at point E.

In the following figure, if `/_ AOB =60^(@)` then `/_ ACB=30^(@)`. [True `//` False `//` Cannot say ]

Description : In the following figure, if `/_ AOB =60^(@)` then `/_ ACB=30^(@)`. [True `//` False `//` Cannot say ]

Last Answer : In the following figure, if `/_ AOB =60^(@)` then `/_ ACB=30^(@)`. [True `//` False `//` Cannot say ]

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC 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.

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

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.

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC 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.

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

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.

Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Description : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Last Answer : clear .

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

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 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

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

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 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

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:

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

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 .

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

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 .

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

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) = \(rac{\sqrt3a^2}{32}\)

Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3). -Maths 9th

Description : Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3). -Maths 9th

Last Answer : For three points to be collinear, the area of the triangle formed by the three points should be zero. ∴ Area of D formed by the given three points= \(rac{1}{2}\) [a((c + a) - (a + b)) + b((a + b) - (b + c)) + c((b ... ba - bc + cb - ca] = 0.Hence (a, b + c), (b, c + a) and (c, a + b) are collinear.

If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Description : If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Last Answer : Since A and B lie on the circle having centre O. Therefore, OA = OB root(4−2)2+(3−3)2​=(root(x−2)2+(5−3)2​2=root(x−2)2+4​(x−2)2+4=4 (x−2)2=0 x−2=0 x=2 hope it helps thank 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 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

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).