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有關pyramids既問題
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1.A regular tetrahedron is a pyramid formed by four congruent equilateral triangles.If the sides of a regular tetrahedron are 6cm long,find its total surface area. 我要列式+答案,plz help! 更新: 本書既答案係62.4cm^2
最佳解答:
1.A regular tetrahedron is a pyramid formed by four congruent equilateral triangles.If the sides of a regular tetrahedron are 6cm long,find its total surface area. Let the vertices of one equilateral triangle be A, B and C. Let D be the height from A to BC. So BD = DC = 6 cm/2 = 3 cm By Pyth. Throrem, (AD)2 = (AB)2 + (BD)2 (AD)2 = 62 + 32 (AD)2 = 36 + 9 (AD)2 = 45 AD = 3√5 Area of t△ABC = AD * BC = (3√5)(6) = 18√5 Total surface area of the regular tetrahedron = 4 * (Area of t△ABC) = 4 * (18√5) = 72√5 cm2 (or 160.997 cm2) 2006-12-08 23:42:52 補充： Sorry, correction:By Pyth. Throrem,(AD)2 = (AB)2 - (BD)2(AD)2 = 62 - 32(AD)2 = 36 - 9(AD)2 = 27AD = 3√3Area of t△ABC= AD * BC/2= (3√3)(6)/2= 18√3/2= 9√3Total surface area of the regular tetrahedron= 4 * (Area of t△ABC)= 4 * (9√3)= 36√3 cm2 (or 62.353 cm2)
Let see this much simplier version. Since the area of one triangle is (1/2) 6 * 6 sin 60 = 9 sqrt(3) Therefore the total surface area is 36 sqrt(3) ~= 62.4cm^2 done. 2006-12-30 10:40:54 補充： 第一次又答錯 個presentation又長....
有關pyramids既問題
發問:
1.A regular tetrahedron is a pyramid formed by four congruent equilateral triangles.If the sides of a regular tetrahedron are 6cm long,find its total surface area. 我要列式+答案,plz help! 更新: 本書既答案係62.4cm^2
最佳解答:
1.A regular tetrahedron is a pyramid formed by four congruent equilateral triangles.If the sides of a regular tetrahedron are 6cm long,find its total surface area. Let the vertices of one equilateral triangle be A, B and C. Let D be the height from A to BC. So BD = DC = 6 cm/2 = 3 cm By Pyth. Throrem, (AD)2 = (AB)2 + (BD)2 (AD)2 = 62 + 32 (AD)2 = 36 + 9 (AD)2 = 45 AD = 3√5 Area of t△ABC = AD * BC = (3√5)(6) = 18√5 Total surface area of the regular tetrahedron = 4 * (Area of t△ABC) = 4 * (18√5) = 72√5 cm2 (or 160.997 cm2) 2006-12-08 23:42:52 補充： Sorry, correction:By Pyth. Throrem,(AD)2 = (AB)2 - (BD)2(AD)2 = 62 - 32(AD)2 = 36 - 9(AD)2 = 27AD = 3√3Area of t△ABC= AD * BC/2= (3√3)(6)/2= 18√3/2= 9√3Total surface area of the regular tetrahedron= 4 * (Area of t△ABC)= 4 * (9√3)= 36√3 cm2 (or 62.353 cm2)
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其他解答:Let see this much simplier version. Since the area of one triangle is (1/2) 6 * 6 sin 60 = 9 sqrt(3) Therefore the total surface area is 36 sqrt(3) ~= 62.4cm^2 done. 2006-12-30 10:40:54 補充： 第一次又答錯 個presentation又長....
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