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Angle Subtended by Diameter on the Circle

For a PQR w...

Question

For a $△ P Q R$ with points $P, Q$ and $R$ lying on a circle, if $R P$ is the diameter of the circle and $P Q = 5 c m$ and $Q R = 12 c m$ . Find radius of the circle.

6.5 c m

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6 c m

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8 c m

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5.5 c m

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Solution

The correct option is A $6.5 c m$
Because angle in a semicircle is always $90^{\circ},$ so
So $∠ R Q P = 90^{\circ}$ and $R P$ is the diameter
Thus by pythagorean theorem we have

$R P^{2} = R Q^{2} + Q P^{2}$
$= 12^{2} + 5^{2} = 169$

$R P = 13 = d = 2 r$

$r = 6.5 c m$

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Similar questions

Q. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS = 12 cm, find the perimeter and area of the shaded region.

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For a △PQR with points P,Q and R lying on a circle, if RP is the diameter of the circle and PQ=5 cm and QR=12 cm. Find radius of the circle.

The correct option is A 6.5 cmBecause angle in a semicircle is always 90∘, soSo ∠RQP=90∘ and RP is the diameterThus by pythagorean theorem we have RP2=RQ2+QP2 =122+52=169 RP=13=d=2r r=6.5 cm

For a $△ P Q R$ with points $P, Q$ and $R$ lying on a circle, if $R P$ is the diameter of the circle and $P Q = 5 c m$ and $Q R = 12 c m$ . Find radius of the circle.

The correct option is A $6.5 c m$
Because angle in a semicircle is always $90^{\circ},$ so
So $∠ R Q P = 90^{\circ}$ and $R P$ is the diameter
Thus by pythagorean theorem we have

$R P^{2} = R Q^{2} + Q P^{2}$
$= 12^{2} + 5^{2} = 169$

$R P = 13 = d = 2 r$

$r = 6.5 c m$