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Question Number 79609 by mr W last updated on 26/Jan/20

Commented by mr W last updated on 26/Jan/20

$$ifthelargesquarehasthesidelength$$$$10,findtheradiusrofthesmallest$$$$circle.$$

Answered by ajfour last updated on 26/Jan/20

$$letbigsquareside=10=2a$$$$sideofsmallerone=2b$$$$circleradius=r$$$$4a^2=4b^2+{(a+b)}^2$$$$\Rightarrow 3a^2=5b^2+2ab$$$$\lambda =\frac{a}{b}\Rightarrow 3{\lambda }^2-2\lambda -5=0$$$$\lambda =\frac{1}{3}+\sqrt{\frac{1}{9}+\frac{5}{3}}=\frac{5}{3}$$$${(2a-r)}^2=a^2+{(2b+r)}^2$$$$\Rightarrow$$$$3a^2-4ar=4b^2+4br$$$$r=\frac{3a^2-4b^2}{4(a+b)}=\frac{a(3\times \frac{25}{9}-4)}{4(\frac{5}{3}+1)}$$$$=\frac{13a}{32}=\frac{65}{32}.$$$$$$

Commented by ajfour last updated on 26/Jan/20

Commented by mr W last updated on 26/Jan/20

$$thanksforthesolutionsir!$$

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