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Show that the whole perimeter of the cardioide $r=a(1+cos\theta)$ is $8a$. Also prove that the arc of the upper half of the cardioide is bisected at $\theta = \frac{\pi}{3}$
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$ \text {Length of Cardiode =} 2 \int_0^{\pi} \sqrt {r^2 +(\frac{dr}{d\theta})^2 } d\theta \\ $

$ = 2\int_0^{\pi} \sqrt{ a^2(1+cos\theta)^2 + a^2 sin^2\theta} d\theta \\ $

$ = 2a\int_0^{\pi} \sqrt{ 1 + cos^2\theta +2 cos\theta + sin^2\theta} d\theta \\ $

$ = 2a\int_0^{\pi} \sqrt{ 2 + 2 cos\theta } d\theta \\ $

$ = 2\sqrt2 a \int_0^{\pi} \sqrt 2 cos\frac{\theta}{2} d\theta \\ $

$ = 4a \int_0^{\pi} cos\frac{\theta}{2} d\theta \\ $

$ = 4a \left [\frac{ \frac{sin\theta}{2} }{\frac{1}{2}} \right]_0^{\pi} \\ $

$ = 8a [ sin{\pi}{2} - sin 0 ] \\ $

$ = 8a \\ $

$ \text{ Now the arc where the line } \theta = \frac{\pi}{3} \text{ divides the cardiode is given by } \\ $

$ \text{arc }OA = \int_0^\frac{\pi}{3} 2acos\frac{\theta}{2} d\theta = 2a[ 2 sin\frac{\theta}{2} ]_0^\frac{\pi}{3} \\ $

$ = 4a sin\frac{\pi}{6} = 4a * \frac{1}{2} = 2a \\ $

$ \text{Arc OA =} \frac{1}{2} \text{Arc OB } \\ $

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