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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

$$\mathrm{yes}.$$

Commented by Gulnoza last updated on 16/Jan/21

ok help me , then

Commented by MJS_new last updated on 16/Jan/21

$$\mathrm{sorry},\mathrm{I}\mathrm{love}\mathrm{to}\mathrm{exactly}\mathrm{answer}\mathrm{questions};-)$$

Answered by bemath last updated on 16/Jan/21

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

Answered by MJS_new last updated on 16/Jan/21

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

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