Answer to Question #144913 in Geometry for ihatemath

Question

A concrete monument is a regular pentagonal pyramid with a base edge of
40 cm and one of its lateral edges measure 50 cm.

8. What is the monument’s altitude?
9. If the lateral surface of the monument is to be covered with glass, how much glass is needed?

Answer 1

$SABCDE$ is a regular pentagonal pyramid.

$ABCDE$ is regular pentagon. $O$ is the centre of $ABCDE$

Triangle $BOA$

$AB=40\ cm, \angle AOB=\dfrac{360\degree}{5}=72\degree$

$OK\perp AB, AK=\dfrac{1}{2}AB=20\ cm$

$$OA=R=\dfrac{AK}{\sin(\dfrac{\angle AOB}{2})}=\dfrac{20}{\sin(\dfrac{72\degree}{2})}cm$$

$\sin36\degree=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

$$OA=\dfrac{80}{\sqrt{10-2\sqrt{5}}}\ cm$$

8. Right triangle $SAO$

$$H=\sqrt{SA^2-OA^2}$$

Given $SA=50\ cm$

$$H=\sqrt{(50)^2-(\dfrac{80}{\sqrt{10-2\sqrt{5}}})^2}\ cm=$$

$$=\dfrac{10\sqrt{186-50\sqrt{5}}}{\sqrt{10-2\sqrt{5}}}=10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}}\approx$$

$$\approx36.636\ cm$$

9. Right triangle $OAK$

$$OK=r=AK\tan36\degree=20\cdot\dfrac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}\ cm$$

Right triangle $SKA$

$$SK=\sqrt{SA^2-AK^2}=\sqrt{(50\ cm)^2-(20\ cm)^2}=$$

$$=10\sqrt{21}\ cm$$

$$S_{lateral}=5\cdot S_{ABS}=5\cdot(\dfrac{1}{2})AB\cdot SK=$$

$$=\dfrac{5}{2}\cdot 40\cdot 10\sqrt{21}\ cm^2=1000\sqrt{21}\ cm^2\approx$$

$$\approx4583\ cm^2\approx0.46 \ m^2$$

$0.46\ m^2$ of glass.