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Question Number 77988 by ajfour last updated on 12/Jan/20

Commented by ajfour last updated on 12/Jan/20

$$Ifthecentralrectanglehas$$$$hassideshalfofthatofouter$$$$one,thenfindAB.$$

Answered by mr W last updated on 12/Jan/20

$$C(0,c)$$$$A(a,0)$$$$B(b,0)$$$$Eqn.ofCA:$$$$\frac{y-1}{c-1}=\frac{x-\frac{d}{2}}{0-\frac{d}{2}}$$$$pointA:$$$$\frac{0-1}{c-1}=\frac{a-\frac{d}{2}}{0-\frac{d}{2}}$$$$\Rightarrow a=\frac{d}{2}(1+\frac{1}{c-1})$$$$$$$$Eqn.ofCB:$$$$\frac{y-2}{c-2}=\frac{x-d}{0-d}$$$$pointB:$$$$\frac{0-2}{c-2}=\frac{b-d}{0-d}$$$$\Rightarrow b=d(1+\frac{2}{c-2})$$$$AB=b-a=\frac{d}{2}(1+\frac{4}{c-2}-\frac{1}{c-1})$$$$AC=\sqrt{a^2+c^2}$$$$AC=AB$$$$\sqrt{a^2+c^2}=b-a$$$$c^2=b(b-2a)$$$$c^2=d(1+\frac{2}{c-2})[d(1+\frac{2}{c-2})-2\times \frac{d}{2}(1+\frac{1}{c-1})]$$$$c^2=d^2(1+\frac{2}{c-2})(\frac{2}{c-2}-\frac{1}{c-1})$$$$c^2=\frac{d^2{c}^2}{{(c-2)}^2(c-1)}$$$${(c-2)}^2(c-1)-d^2=0$$$$lett=c-2$$$$\Rightarrow t^3+t^2-d^2=0$$$$lett=s-\frac{1}{3}$$$${(s-\frac{1}{3})}^3+{(s-\frac{1}{3})}^2-d^2=0$$$$s^3-\frac{s}{3}+(\frac{2}{27}-d^2)=0$$$$s=\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}-\frac{1}{27}+\frac{d^2}{2}}-\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}+\frac{1}{27}-\frac{d^2}{2}}$$$$t=c-2=\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}-\frac{1}{27}+\frac{d^2}{2}}-\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}+\frac{1}{27}-\frac{d^2}{2}}-\frac{1}{3}$$$$\Rightarrow c=\frac{5}{3}+\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}-\frac{1}{27}+\frac{d^2}{2}}-\sqrt[3]{\sqrt{{(\frac{1}{27}-\frac{d^2}{2})}^2-\frac{1}{729}}+\frac{1}{27}-\frac{d^2}{2}}-\frac{1}{3}$$$$AB=\frac{d}{2}(1+\frac{4}{c-2}-\frac{1}{c-1})$$

Commented by ajfour last updated on 13/Jan/20

$$Awesomesolutionsir,thanks!$$

Commented by ajfour last updated on 13/Jan/20

$$itsalrightsir,ishallconsult$$$$youagain..$$

Commented by mr W last updated on 13/Jan/20

$$sir,pleaserecheck:$$$${(c-2)}^2(c-1)=d^{2}iscorrect.witht=c-2$$$$weget{t}^3+t=d^2.not{t}^3-t=2d!$$

Commented by MJS last updated on 13/Jan/20

$$\mathrm{to}\mathrm{mrW}:\mathrm{our}{\mathrm{friend}}^{\prime }\mathrm{s}\mathrm{new}\mathrm{ID}\mathrm{is}\mathrm{Pratah}$$

Commented by mr W last updated on 13/Jan/20

$$yes.iknewthisassoonthefirst$$$$questionwiththenewidwasposted.$$

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