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Question Number 170202 by otchereabdullai@gmail.com last updated on 18/May/22

$$\mathrm{A}\mathrm{point}\mathrm{is}3\mathrm{cm},4\mathrm{cm}\mathrm{and}5\mathrm{cm}\mathrm{away}$$$$\mathrm{from}\mathrm{three}\mathrm{vertices}\mathrm{of}\mathrm{a}\mathrm{rectangle}.$$$$\mathrm{How}\mathrm{far}\mathrm{can}\mathrm{it}\mathrm{be}\mathrm{from}\mathrm{the}4\mathrm{th}\mathrm{vertex}.$$$$$$$$\mathrm{Find}\mathrm{all}\mathrm{solutions}$$

Commented by otchereabdullai@gmail.com last updated on 18/May/22

$$\mathrm{much}\mathrm{much}\mathrm{grateful}\mathrm{thanks}$$

Commented by mr W last updated on 18/May/22

$$thereareinfinitelymanysolutions.$$

Commented by MJS_new last updated on 18/May/22

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{either}0\vee 3\sqrt{2}\vee 4\sqrt{2}$$

Commented by otchereabdullai@gmail.com last updated on 18/May/22

$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{some}\mathrm{of}\mathrm{the}$$$$\mathrm{solutions}\mathrm{prof}\mathrm{W}$$

Commented by otchereabdullai@gmail.com last updated on 18/May/22

$$\mathrm{prof}\mathrm{mjs}\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{the}$$$$0\vee 3\sqrt{2}\vee 4\sqrt{2}$$

Commented by MJS_new last updated on 18/May/22

$$‘‘{\mathrm{fix}}^{″}\mathrm{the}\mathrm{vertices}$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}a\\ 0\end{array}\right)C=\left(\begin{array}{c}a\\ b\end{array}\right)D=\left(\begin{array}{c}0\\ b\end{array}\right)$$$$\mathrm{let}P=\left(\begin{array}{c}p\\ q\end{array}\right)$$$$\mathrm{equations}\mathrm{with}\alpha ,\beta ,\gamma \mathrm{unknown}\mathrm{at}\mathrm{first}$$$$\left(1\right)\mid AP{\mid }^2={\alpha }^2$$$$\left(2\right)\mid BP{\mid }^2={\beta }^2$$$$\left(3\right)\mid CP{\mid }^2={\gamma }^2$$$$\left(4\right)\mid DP{\mid }^2=x^2$$$$\mathrm{you}\mathrm{can}\mathrm{easily}\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{for}p,q\mathrm{and}$$$$\mathrm{get}\mathrm{a}\mathrm{formula}\mathrm{for}{x}^2.\mathrm{now}\mathrm{insert}\mathrm{all}\mathrm{possible}$$$$\mathrm{permutations}\mathrm{of}3,4,5\mathrm{for}\alpha ,\beta ,\gamma$$

Commented by mr W last updated on 18/May/22

$$youarerightsir!$$$$thoughthereareinfinitelymanysuch$$$$rectangles,butthedistancetothe$$$$fourthvertexisconstant.$$$$x=\sqrt{{\alpha }^2+{\beta }^2-{\gamma }^2}$$$$x=\sqrt{3^2+4^2-5^2}=0$$$$x=\sqrt{3^2+5^2-4^2}=3\sqrt{2}$$$$x=\sqrt{4^2+5^2-3^2}=4\sqrt{2}$$

Commented by mr W last updated on 18/May/22

Commented by mr W last updated on 18/May/22

$${\alpha }^2=a_1^2+b_2^2$$$${\beta }^2=a_2^2+b_1^2$$$$\Rightarrow {\alpha }^2+{\beta }^2=a_1^2+a_2^2+b_1^2+b_2^2$$$$x^2=a_2^2+b_2^2$$$${\gamma }^2=a_1^2+b_1^2$$$$\Rightarrow x^2+{\gamma }^2=a_1^2+a_2^2+b_1^2+b_2^2$$$$\Rightarrow {\mathit{x}}^2+{\mathit{\gamma }}^2={\mathit{\alpha }}^2+{\mathit{\beta }}^2$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

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