體積和三角形計算複習問題

(ai) By Heron's formula, area of triangle of DEF
= sqrt [ 8(8 - 6)(8 - 5)(8 - 5)] = sqrt (8 x 2 x 3 x 3) = 3 sqrt 16 = 3 x 4 = 12 cm^2.
(aii) Let M and N be the point on AE and BF respectively such that CM is perpendicular to AE and CN is perpendicular to BF.
Volume of prism MCNEDF = base area x height = 12 x 25 = 300 cm^3.
By Pythagoras thm., height of triangle DEF = sqrt ( 5^2 - 3^2) = 4 cm.
So height of triangle MCN = 4 cm.
So volume of pyramid AMBNC = area of AMBN x height of pyramid x 1/3
= 6 x (30- 25) x 4 x 1/3 = 6 x 5 x 4/3 = 40 cm^3.
So volume of ABCDEF = 300 + 40 = 340 cm^3.
(bi) By Pythagoras thm.
AC = sqrt (5^2 + 5^2) = 5 sqrt 2 = 7.0711 = BC. AB = 6.
Again by Heron's formula, area of traingle ABC
= sqrt [ 10.0711(10.0711 - 7.0711)(10.0711 - 7.0711)(10.0711 - 6)]
= 19.21 cm^2.
(bii)
Total surface area of ABCDEF
= 19.21 + 12 + 6 x 30 + 5(25 + 30) = 19.21 + 12 + 180 + 275 = 486.21 cm^2.
(c)
Let radius of circle that can be put inside traingle ABC be r.
So area of triangle ABC = 7.0711r/2 + 7.0711r/2 + 6r/2 = 10.0711r
which is equal to 19.21 cm^2.
so r = 19.21/10.0711 = 1.9074 cm.
Therefore, a metallic circle of radius 2 cm. cannot be put inside triangle ABC.