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Circles

Short Answer Type

261. A circle is touching the side BC of ΔABC at P and touching AB and AC produced at Q and R respectively. Prove that

AQ space equals space 1 half space left parenthesis Perimeter space of space increment space ABC right parenthesis

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262. In the given figure, the incircle of ΔABC, touches the sides BC, CA and AB at D, E and F respectively. Show that AF + BD + CE = AE + BF + CD

equals space 1 half left parenthesis perimeter space of space increment space ABC right parenthesis

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Long Answer Type

263. ABCD is a quadrilateral such that ∠D = 90°. A cirlce C (O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, Find r.

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Short Answer Type

264. In figure. XP and XQ are tangents from X to the circle with ccentre O. R is a point on the circle. Prove that XA + AR = XB + BR?

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265. In figure a circle touches the side BC of ΔABC at P and touches AB and AC produced at Q and R respectively. If AQ = 5 cm, find the perimeter of ΔABC.

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Long Answer Type

266. In the given figure, ABC is a right-angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle.

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Short Answer Type

267.

From a point P outside a circle, with centre O tangents PA and PB are drawn as shown in the figure, prove that

(i) ⇒AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB.

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Long Answer Type

268. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

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Short Answer Type

269. In two concentric circles, prove that all chords of the outer circle which touches the inner circle are of equal length.

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270. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.

Given : XY and X‘Y’ are two parallel tangents. AB is the intercept of tangents which intersects circle at point C.
To prove : ∠AOB    = 90°
Proof: ∠XAB + ∠X'BA = 180° (Consecutive interior angles)
⇒ 2∠OAB + 2∠OBA = 180°
[∵ If two tangents are drawn to a circle from an external point, then they are equally
inclined to the segment, joining the centre to that point]
⇒ ∠OAB + ∠OBA    = 90°...(i)
Now, in ΔAOB
∠OAB + ∠OBA + ∠AOB = 180°
⇒ 90° + ∠AOB    = 180° [Using (i)]
⇒ ∠AOB    = 90°.

625 Views