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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

334 Views

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.

358 Views

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?

149 Views

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.

265 Views

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.

421 Views

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.

1204 Views

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.

Given : Two concentric circles C₁ and C₂of radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C₁) and BD is a tangent to the smaller circle (C₂).
∴ The tangent at any point of a circle is perpendicular to the radius through the point of contact.

$therefore space space OD space perpendicular space BD$

Now, in right triangle BOD, we have

OB² = OD² = BD² [ Using Pythagoras theorem]

$rightwards double arrow$ (13)²= (8)²+ BD²

$rightwards double arrow$ 169 = 64 + BD²
$rightwards double arrow$ BD² = 169 - 64
$rightwards double arrow space$ BD² = 105

$rightwards double arrow space space space BD space equals space square root of 105 space cm$
Since, perpendicular drawn from the centre to the chord bisects the chord.

$rightwards double arrow space space space space space BD space equals space DE space equals space square root of 105 space cm rightwards double arrow space space space space BE space equals space 2 BD rightwards double arrow space space space space space space space space space equals space 2 square root of 105 space cm Now comma space in space increment space BOD space and space increment space BAE comma angle BDo space equals space angle BEA space equals space 90 degree$

(∵ ∠BEA = 90°, angle in semicircle is right angle] and, ∠OBD = ∠ABE (common)

Therefore, using AAS similar condition ΔBOD ~ ΔBAE

$rightwards double arrow space space BD over BE equals OD over AE$

[Proportional sides of two similar triangles]

$rightwards double arrow space space fraction numerator square root of 105 over denominator 2 square root of 105 end fraction equals 8 over AE rightwards double arrow space space space 1 half equals 8 over AE rightwards double arrow space space space space AE space equals space 16 space cm$

Now, in right triangle ADE, we have

AD² = AE² + DE²

$rightwards double arrow space space AD squared space equals space 16 squared space plus space left parenthesis square root of 105 right parenthesis squared$
= 256 + 105
= 361

$rightwards double arrow$ AD = 19 cm.