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Question Number 129449 by Gulnoza last updated on 15/Jan/21

Answered by MJS_new last updated on 16/Jan/21

yes.

$$\mathrm{yes}. \\ $$

Commented by Gulnoza last updated on 16/Jan/21

ok help me , then

Commented by MJS_new last updated on 16/Jan/21

sorry, I love to exactly answer questions ;−)

$$\left.\mathrm{sorry},\:\mathrm{I}\:\mathrm{love}\:\mathrm{to}\:\mathrm{exactly}\:\mathrm{answer}\:\mathrm{questions}\:;−\right) \\ $$

Answered by bemath last updated on 16/Jan/21

yes...too

$$\mathrm{yes}...\mathrm{too} \\ $$

Answered by MJS_new last updated on 16/Jan/21

rectangular triangle with the centers of the 3 circles as vertices has sides a=6, b=12−r, c=6+r ⇒ 6^2 +(12−r)^2 =(6+r)^2 ⇒ r=4 ⇒ blue area =(1/4)12^2 π−(1/2)6^2 π−(1/2)4^2 π=10π

$$\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{3}\:\mathrm{circles}\:\mathrm{as}\:\mathrm{vertices}\:\mathrm{has}\:\mathrm{sides} \\ $$$${a}=\mathrm{6},\:{b}=\mathrm{12}−{r},\:{c}=\mathrm{6}+{r} \\ $$$$\Rightarrow \\ $$$$\mathrm{6}^{\mathrm{2}} +\left(\mathrm{12}−{r}\right)^{\mathrm{2}} =\left(\mathrm{6}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{r}=\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{blue}\:\mathrm{area}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{12}^{\mathrm{2}} \pi−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{6}^{\mathrm{2}} \pi−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{4}^{\mathrm{2}} \pi=\mathrm{10}\pi \\ $$

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