S A B C D E SABCDE S A BC D E is a regular pentagonal pyramid.
A B C D E ABCDE A BC D E is regular pentagon. O O O is the centre of A B C D E ABCDE A BC D E
Triangle B O A BOA BO A
A B = 40 c m , ∠ A O B = 360 ° 5 = 72 ° AB=40\ cm, \angle AOB=\dfrac{360\degree}{5}=72\degree A B = 40 c m , ∠ A OB = 5 360° = 72°
O K ⊥ A B , A K = 1 2 A B = 20 c m OK\perp AB, AK=\dfrac{1}{2}AB=20\ cm O K ⊥ A B , A K = 2 1 A B = 20 c m
O A = R = A K sin ( ∠ A O B 2 ) = 20 sin ( 72 ° 2 ) c m OA=R=\dfrac{AK}{\sin(\dfrac{\angle AOB}{2})}=\dfrac{20}{\sin(\dfrac{72\degree}{2})}cm O A = R = sin ( 2 ∠ A OB ) A K = sin ( 2 72° ) 20 c m sin 36 ° = 10 − 2 5 4 \sin36\degree=\dfrac{\sqrt{10-2\sqrt{5}}}{4} sin 36° = 4 10 − 2 5
O A = 80 10 − 2 5 c m OA=\dfrac{80}{\sqrt{10-2\sqrt{5}}}\ cm O A = 10 − 2 5 80 c m 8. Right triangle S A O SAO S A O
H = S A 2 − O A 2 H=\sqrt{SA^2-OA^2} H = S A 2 − O A 2 Given S A = 50 c m SA=50\ cm S A = 50 c m
H = ( 50 ) 2 − ( 80 10 − 2 5 ) 2 c m = H=\sqrt{(50)^2-(\dfrac{80}{\sqrt{10-2\sqrt{5}}})^2}\ cm= H = ( 50 ) 2 − ( 10 − 2 5 80 ) 2 c m =
= 10 186 − 50 5 10 − 2 5 = 10 93 − 25 5 5 − 5 ≈ =\dfrac{10\sqrt{186-50\sqrt{5}}}{\sqrt{10-2\sqrt{5}}}=10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}}\approx = 10 − 2 5 10 186 − 50 5 = 10 5 − 5 93 − 25 5 ≈
≈ 36.636 c m \approx36.636\ cm ≈ 36.636 c m
S A B C D E = 5 ( 1 2 ) ( O K ) ( A B ) = 5 2 ( A K tan 36 ° ) ( A B ) = S_{ABCDE}=5(\dfrac{1}{2})(OK)(AB)=\dfrac{5}{2}(AK\tan36\degree)(AB)= S A BC D E = 5 ( 2 1 ) ( O K ) ( A B ) = 2 5 ( A K tan 36° ) ( A B ) =
= 5 4 ( A B ) 2 tan 36 ° =\dfrac{5}{4}(AB)^2\tan36\degree = 4 5 ( A B ) 2 tan 36° sin 36 ° = 10 − 2 5 4 \sin36\degree=\dfrac{\sqrt{10-2\sqrt{5}}}{4} sin 36° = 4 10 − 2 5
cos 2 36 ° = 1 − sin 2 36 ° = 1 − 10 − 2 5 16 = 6 + 2 5 16 \cos^2 36\degree=1-\sin^236\degree=1-\dfrac{10-2\sqrt{5}}{16}=\dfrac{6+2\sqrt{5}}{16} cos 2 36° = 1 − sin 2 36° = 1 − 16 10 − 2 5 = 16 6 + 2 5
tan 2 36 ° = 10 − 2 5 6 + 2 5 \tan^236\degree=\dfrac{10-2\sqrt{5}}{6+2\sqrt{5}} tan 2 36° = 6 + 2 5 10 − 2 5
tan 36 ° = 5 − 5 3 + 5 \tan36\degree=\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}} tan 36° = 3 + 5 5 − 5
S A B C D E = 5 4 ( 40 c m ) 2 5 − 5 3 + 5 S_{ABCDE}=\dfrac{5}{4}(40cm)^2\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}} S A BC D E = 4 5 ( 40 c m ) 2 3 + 5 5 − 5
V = 1 3 S A B C D E H = V=\dfrac{1}{3}S_{ABCDE}H= V = 3 1 S A BC D E H =
= 1 3 ( 2000 ) 5 − 5 3 + 5 ( 10 93 − 25 5 5 − 5 ) c m 3 = =\dfrac{1}{3}(2000)\sqrt{\dfrac{5-\sqrt{5}}{3+\sqrt{5}}}(10\sqrt{\dfrac{93-25\sqrt{5}}{5-\sqrt{5}}})cm^3= = 3 1 ( 2000 ) 3 + 5 5 − 5 ( 10 5 − 5 93 − 25 5 ) c m 3 =
= 20000 3 93 − 25 5 3 + 5 c m 3 ≈ =\dfrac{20000}{3}\sqrt{\dfrac{93-25\sqrt{5}}{3+\sqrt{5}}}cm^3\approx = 3 20000 3 + 5 93 − 25 5 c m 3 ≈
≈ 17745.29 c m 3 ≈ 1.8 m 3 \approx17745.29cm^3\approx1.8m^3 ≈ 17745.29 c m 3 ≈ 1.8 m 3
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