有關pyramids既問題

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：
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