Question Number 177253 by BaliramKumar last updated on 02/Oct/22
$$ \\ $$The base of a right pyramid is a hexagon of side 16 cm, and its lateral surface is 720 sq. cm. Is. the height of the pyramid will be
Answered by HeferH last updated on 02/Oct/22
$$\sqrt{\mathrm{33}\:}\:?\: \\ $$
Answered by a.lgnaoui last updated on 02/Oct/22
$${soit}\:\mathrm{AB}\:\mathrm{base}\:\mathrm{du}\:\mathrm{pyramide}\:\mathrm{hexagonsle} \\ $$$$\:\:\:\:\:\:\:\mathrm{hauteur}\:\mathrm{du}\:\mathrm{pyramide} \\ $$$$\mathrm{OA}\:\:\mathrm{rayon}\:\mathrm{du}\:\mathrm{ceecle}\:\mathrm{circonscrit} \\ $$$$\mathrm{syrfsce}\:\mathrm{de}\:\mathrm{la}\:\mathrm{base}\:\mathrm{est}\:\frac{\mathrm{AB}×\mathrm{CD}}{\mathrm{2}}\:\:\:\: \\ $$$$\mathrm{Surface}\:\mathrm{laterale}\:\:\mathrm{S}=\mathrm{6}×\frac{\mathrm{AB}×\mathrm{CD}}{\mathrm{2}}=\mathrm{3AB}×\mathrm{CD} \\ $$$$\mathrm{CD}=\frac{\mathrm{S}}{\mathrm{3AB}}=\frac{\mathrm{720}}{\mathrm{48}}=\mathrm{15} \\ $$$$\mathrm{CD}^{\mathrm{2}} =\mathrm{OC}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \:\:\:\:\left(\mathrm{x}=\mathrm{haureur}\right) \\ $$$$\mathrm{OC}=\mathrm{OAsin}\:\mathrm{60}=\mathrm{16}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{CD}^{\mathrm{2}} −\mathrm{OC}^{\mathrm{2}} =\mathrm{225}−\mathrm{192}=\mathrm{33} \\ $$$$\mathrm{x}=\sqrt{\mathrm{33}}\:=\mathrm{5},\mathrm{74}{cm} \\ $$
Commented by a.lgnaoui last updated on 02/Oct/22
Commented by a.lgnaoui last updated on 02/Oct/22
$${x}={h}={hauteur}\:{du}\:{Pyramide} \\ $$
Commented by BaliramKumar last updated on 03/Oct/22
$${thanks} \\ $$