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

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
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1 Answer

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
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Related questions

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

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

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.

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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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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

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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

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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

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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.

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

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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, 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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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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If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by (A) a = S.R/L (B) a = L.S/R (C) a = L.R/S (D) None of these

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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 .

The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Description : The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Last Answer : answer:

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

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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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:

PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to chord PS at M. If `bar(PS) =30 cm` and `bar(OM) =

Description : PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to chord PS at M. If `bar(PS) =30 cm` and `bar(OM) =

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The distances of two chords AB and CD from the centre of a circle are 6 cm and 8 cm respectively. Then, which chord is longer?

Description : The distances of two chords AB and CD from the centre of a circle are 6 cm and 8 cm respectively. Then, which chord is longer?

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If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -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 two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Description : If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Last Answer : Explanation of this question

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 two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Description : If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Last Answer : Explanation of this question

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

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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 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 .

What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Description : What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Last Answer : (d) 6 cmLet a = 35 cm, b = 44 cm, c = 75 cm. Thens = \(rac{a+b+c}{2}\) = \(rac{34+44+75}{2}\) cm = 77 cm∴ Area if triangle = \(\sqrt{77(77-35)(77-44)(77-75)}\) cm2= \(\sqrt{77 imes42 ... ) cm2 = 462 cm2∴ Radius of incircle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \(rac{462}{77}\) cm = 6 cm.

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

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 is the length of a sub-chord and is the radius of simple curve, the angle of deflection between its tangent and sub-chord, in minutes, is equal to (A) 573 S/R (B) 573 R/S (C) 1718.9 R/S (D) 1718.9 S/R

Description : If is the length of a sub-chord and is the radius of simple curve, the angle of deflection  between its tangent and sub-chord, in minutes, is equal to  (A) 573 S/R (B) 573 R/S (C) 1718.9 R/S (D) 1718.9 S/R

Last Answer : (D) 1718.9 S/R

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.

A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their -Maths 9th

Description : The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their -Maths 9th

Last Answer : answer:

In a circle , chord AB subtends an angle of `60^(@)` at the centre and chord CD subtends `120^(@)`, at it. Then which chord is longer ?

Description : In a circle , chord AB subtends an angle of `60^(@)` at the centre and chord CD subtends `120^(@)`, at it. Then which chord is longer ?

Last Answer : In a circle , chord AB subtends an angle of `60^(@)` at the centre and chord CD subtends `120^(@)`, at it. Then which chord is longer ?