Question Number 94291 by I want to learn more last updated on 17/May/20
Commented by PRITHWISH SEN 2 last updated on 18/May/20
$$\mathrm{diagonal}=\sqrt{\left(\mathrm{14}+\mathrm{9}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }\:\:=\mathrm{17}\sqrt{\mathrm{2}} \\ $$$$\therefore\:\boldsymbol{\mathrm{x}}=\mathrm{17} \\ $$
Commented by I want to learn more last updated on 18/May/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by Ar Brandon last updated on 18/May/20
$$\mathrm{H}=\sqrt{\left(\mathrm{14}+\mathrm{9}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\mathrm{17}\sqrt{\mathrm{2}} \\ $$$$\mathrm{H}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \Rightarrow\mathrm{x}=\sqrt{\frac{\mathrm{H}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{17cm} \\ $$
Commented by I want to learn more last updated on 18/May/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$