Question #144914
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?
1
Expert's answer
2020-11-19T16:54:34-0500

SABCDESABCDE is a regular pentagonal pyramid.

ABCDEABCDE is regular pentagon. OO is the centre of ABCDEABCDE

Triangle BOABOA

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

OKAB,AK=12AB=20 cmOK\perp AB, AK=\dfrac{1}{2}AB=20\ cm


OA=R=AKsin(AOB2)=20sin(72°2)cmOA=R=\dfrac{AK}{\sin(\dfrac{\angle AOB}{2})}=\dfrac{20}{\sin(\dfrac{72\degree}{2})}cm

sin36°=10254\sin36\degree=\dfrac{\sqrt{10-2\sqrt{5}}}{4}


OA=801025 cmOA=\dfrac{80}{\sqrt{10-2\sqrt{5}}}\ cm

8. Right triangle SAOSAO


H=SA2OA2H=\sqrt{SA^2-OA^2}

Given SA=50 cmSA=50\ cm


H=(50)2(801025)2 cm=H=\sqrt{(50)^2-(\dfrac{80}{\sqrt{10-2\sqrt{5}}})^2}\ cm=

=101865051025=109325555=\dfrac{10\sqrt{186-50\sqrt{5}}}{\sqrt{10-2\sqrt{5}}}=10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}}\approx

36.636 cm\approx36.636\ cm

SABCDE=5(12)(OK)(AB)=52(AKtan36°)(AB)=S_{ABCDE}=5(\dfrac{1}{2})(OK)(AB)=\dfrac{5}{2}(AK\tan36\degree)(AB)=

=54(AB)2tan36°=\dfrac{5}{4}(AB)^2\tan36\degree

sin36°=10254\sin36\degree=\dfrac{\sqrt{10-2\sqrt{5}}}{4}

cos236°=1sin236°=1102516=6+2516\cos^2 36\degree=1-\sin^236\degree=1-\dfrac{10-2\sqrt{5}}{16}=\dfrac{6+2\sqrt{5}}{16}

tan236°=10256+25\tan^236\degree=\dfrac{10-2\sqrt{5}}{6+2\sqrt{5}}


tan36°=553+5\tan36\degree=\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}

SABCDE=54(40cm)2553+5S_{ABCDE}=\dfrac{5}{4}(40cm)^2\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}

V=13SABCDEH=V=\dfrac{1}{3}S_{ABCDE}H=

=13(2000)553+5(109325555)cm3==\dfrac{1}{3}(2000)\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}(10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}})cm^3=

=200003932553+5cm3=\dfrac{20000}{3}\sqrt{\dfrac{93-25\sqrt{5}}{3+\sqrt{5}}}cm^3\approx

17745.29cm31.8m3\approx17745.29cm^3\approx1.8m^3

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