Question Number 200377 by mr W last updated on 18/Nov/23
Commented by mr W last updated on 18/Nov/23
$${the}\:{side}\:{lengths}\:{of}\:{the}\:{hexagon}\:{are} \\ $$$${given}.\:{find}\:{x}+{y}+{z}=? \\ $$
Answered by mr W last updated on 18/Nov/23
Commented by mr W last updated on 18/Nov/23
Commented by mr W last updated on 18/Nov/23
$${red}\:{diagonal}\:={z} \\ $$$${blue}\:{diagonal}\:={y} \\ $$$${green}\:{diagonal}\:={x} \\ $$
Commented by mr W last updated on 18/Nov/23
$${applying}\:{Ptolemy}'{s}\:{theorem} \\ $$$${to}\:\Box{ADEF}: \\ $$$${z}^{\mathrm{2}} =\mathrm{81}^{\mathrm{2}} +\mathrm{81}{y}\:\:\:…\left({i}\right) \\ $$$${to}\:\Box{ACDF}: \\ $$$${y}^{\mathrm{2}} =\mathrm{81}^{\mathrm{2}} +{xz}\:\:\:…\left({ii}\right) \\ $$$${to}\:\Box{ABCD}: \\ $$$${xz}=\mathrm{31}×\mathrm{81}+\mathrm{81}{y}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${from}\:\left({ii}\right)\:\&\:\left({iii}\right): \\ $$$${y}^{\mathrm{2}} −\mathrm{81}{y}−\mathrm{81}×\mathrm{112}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\mathrm{81}+\mathrm{9}×\mathrm{23}}{\mathrm{2}}=\mathrm{144}\:\:\:\checkmark \\ $$$${from}\:\left({i}\right): \\ $$$${z}=\sqrt{\mathrm{81}^{\mathrm{2}} +\mathrm{81}×\mathrm{144}}=\mathrm{135}\:\checkmark \\ $$$${from}\:\left({ii}\right): \\ $$$${x}=\frac{{y}^{\mathrm{2}} −\mathrm{81}^{\mathrm{2}} }{{z}}=\frac{\mathrm{144}^{\mathrm{2}} −\mathrm{81}^{\mathrm{2}} }{\mathrm{135}}=\mathrm{105}\:\checkmark \\ $$$$\Rightarrow{x}+{y}+{z}=\mathrm{105}+\mathrm{144}+\mathrm{135}=\mathrm{384}\:\checkmark \\ $$
Commented by ajfour last updated on 18/Nov/23
$${i}\:{see},\:{thank}\:{you}\:{sir}. \\ $$