Answer to Question #144914 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.

10. How much concrete is needed to build the monument?

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 $SA=50\ cm$

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

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

$$\approx36.636\ cm$$

$$S_{ABCDE}=5(\dfrac{1}{2})(OK)(AB)=\dfrac{5}{2}(AK\tan36\degree)(AB)=$$

$$=\dfrac{5}{4}(AB)^2\tan36\degree$$

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

$\cos^2 36\degree=1-\sin^236\degree=1-\dfrac{10-2\sqrt{5}}{16}=\dfrac{6+2\sqrt{5}}{16}$

$\tan^236\degree=\dfrac{10-2\sqrt{5}}{6+2\sqrt{5}}$

$$\tan36\degree=\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}$$

$$S_{ABCDE}=\dfrac{5}{4}(40cm)^2\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}$$

$$V=\dfrac{1}{3}S_{ABCDE}H=$$

$$=\dfrac{1}{3}(2000)\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}(10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}})cm^3=$$

$$=\dfrac{20000}{3}\sqrt{\dfrac{93-25\sqrt{5}}{3+\sqrt{5}}}cm^3\approx$$

$$\approx17745.29cm^3\approx1.8m^3$$