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Question-223935




Question Number 223935 by fantastic last updated on 10/Aug/25
Commented by fantastic last updated on 10/Aug/25
PERIMETER
$${PERIMETER} \\ $$
Answered by dionigi last updated on 11/Aug/25
P = AB^(⌢)  + DE^(⌢)  +AD+EB  P = πr + ((πr)/2) + d + (r−d) = ((3πr)/2) + r
$${P}\:=\:\overset{\frown} {{AB}}\:+\:\overset{\frown} {{DE}}\:+{AD}+{EB} \\ $$$${P}\:=\:\pi{r}\:+\:\frac{\pi{r}}{\mathrm{2}}\:+\:{d}\:+\:\left({r}−{d}\right)\:=\:\frac{\mathrm{3}\pi{r}}{\mathrm{2}}\:+\:{r} \\ $$
Answered by mehdee7396 last updated on 12/Aug/25
DE=AC=r  ⇒P=d+r−d+(3/2)πr=r+((3π)/2)r
$${DE}={AC}={r}\:\:\Rightarrow{P}={d}+{r}−{d}+\frac{\mathrm{3}}{\mathrm{2}}\pi{r}={r}+\frac{\mathrm{3}\pi}{\mathrm{2}}{r} \\ $$
Answered by MirHasibulHossain last updated on 12/Aug/25
arc(AB)=(1/2)×2π×CB=πr     [∵ AC=CB=r]  CB=AC=r  ⇒AC+CE=r+d  ⇒AE=r+d  ⇒AE−AD=r+d−d  ⇒DE=r  arc(DE)=(1/2)×2π×((DE)/2)=π×(r/2)=(1/2)πr  EB=CB−CE=r−d  Perimeter=AD+arc(DE)+EB+arc(AB)                         =d+(1/2)πr+r−d+πr                         =(3/2)πr+r                         =r((3/2)π+1)          [ANSWER]
$$\mathrm{arc}\left(\mathrm{AB}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\pi×\mathrm{CB}=\pi\mathrm{r}\:\:\:\:\:\left[\because\:\mathrm{AC}=\mathrm{CB}=\mathrm{r}\right] \\ $$$$\mathrm{CB}=\mathrm{AC}=\mathrm{r} \\ $$$$\Rightarrow\mathrm{AC}+\mathrm{CE}=\mathrm{r}+\mathrm{d} \\ $$$$\Rightarrow\mathrm{AE}=\mathrm{r}+\mathrm{d} \\ $$$$\Rightarrow\mathrm{AE}−\mathrm{AD}=\mathrm{r}+\mathrm{d}−\mathrm{d} \\ $$$$\Rightarrow\mathrm{DE}=\mathrm{r} \\ $$$$\mathrm{arc}\left(\mathrm{DE}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\pi×\frac{\mathrm{DE}}{\mathrm{2}}=\pi×\frac{\mathrm{r}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{r} \\ $$$$\mathrm{EB}=\mathrm{CB}−\mathrm{CE}=\mathrm{r}−\mathrm{d} \\ $$$$\mathrm{Perimeter}=\mathrm{AD}+\mathrm{arc}\left(\mathrm{DE}\right)+\mathrm{EB}+\mathrm{arc}\left(\mathrm{AB}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{d}+\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{r}+\mathrm{r}−\mathrm{d}+\pi\mathrm{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\pi\mathrm{r}+\mathrm{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{r}\left(\frac{\mathrm{3}}{\mathrm{2}}\pi+\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\left[\mathrm{ANSWER}\right] \\ $$

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