In figure, shown a sector OAP of a circle with centre O, containing $∠ θ,$ AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is $r [tan θ + sec θ + \frac{π θ}{180^{0}} - 1] .$

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Solution

From the given figure we have, in $Δ$ OAB,

$tan θ = \frac{A B}{O A} \Rightarrow A B = r tan θ$

[Since $O A = r$ ]....(1)

And $cos θ = \frac{O A}{O B} \Rightarrow O B = r sec θ$ .....(2).

Now $P B = O B - O P = r sec θ - r$ ......(3).[Using (2)].

Again $\frac{π θ}{180^{o}} = \frac{ˆ A P}{O A}$

or, $ˆ A P = r \frac{π θ}{180^{o}}$ ........(4).

Now perimeter of the shaded region is

$A B + P B + ˆ A P = r [sec θ + tan θ + \frac{π θ}{180^{o}} - 1]$ . [Using (1),(3) and (4)].

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∠ θ

r [t a n θ + s e c θ + \frac{π θ}{180} - 1]

θ^{o}

r (tan θ + sec θ + \frac{π θ}{180} - 1)

Figure shows a sector of a circle, centre O, containing an angle $θ^{\circ}$ . Prove that:

(i) Perimeter of the shaded region is $r (t a n θ + s e c θ + \frac{π θ}{180} - 1)$

(ii) Area of the shaded region is $\frac{r^{2}}{2} (t a n θ - \frac{π θ}{180})$

(\tan θ + s e c θ + \frac{πθ}{180} - 1)

\frac{r^{2}}{2} (\tan θ - \frac{πθ}{180})

θ^{\circ}

\frac{r^{2}}{2} (t a n θ - \frac{π θ}{180})

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