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

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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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
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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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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

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Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

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

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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

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 .

The radius of a circle is 10cm. The length of a chord is 12 cm. Then the distance of the chord from the centre is `"__________________"`.

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A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

Description : A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

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A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

Description : A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th

Last Answer : Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e., AB = BO …(i) Join OA, AC and BC. Since, OA = OB= Radius of circle OA = AS = BO

If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th

Description : If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th

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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

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ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -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