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Taisyklingosios trikampės piramidės ABCS tūris lygus 8, piramidės aukštinė SO yra $$2\cdot \sqrt {3}$$ ilgio. Apskaičiuokite piramidės pagrindo ABC aukštinės ilgį.
Sprendimas.
Piramidės tūrio formulė $$V = \frac{1}{3}\cdot S_{pagr}\cdot h$$.
Aukštinė h (SO) lygi $$2\cdot \sqrt {3}$$, tūris lygus 8.
Rasime pagrindo plotą:
V =
* S_pagr* h
V = * S_pagr* h$$V$$ = $$\frac{1}{3}\cdot S_{pagr}\cdot h$$ 8 = 2* S_pagr* saknis(3) |
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$$8$$ = $$\frac{2\cdot S_{pagr}\cdot \sqrt {3}}{3}$$ 8* 3 = 2* S_pagr* saknis(3)$$8\cdot 3$$ = $$2\cdot S_{pagr}\cdot \sqrt {3}$$ = S_pagr* saknis(3)$$\frac{8\cdot 3}{2}$$ = $$S_{pagr}\cdot \sqrt {3}$$ 8* 3 |
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/ 2/ saknis(3) |
= S_pagr$$\frac{8\cdot 3}{2\cdot \sqrt {3}}$$ = $$S_{pagr}$$ S_pagr = 8* 3 |
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/ 2/ saknis(3) |
$$S_{pagr}$$ = $$\frac{8\cdot 3}{2\cdot \sqrt {3}}$$
$$V$$ = $$\frac{1}{3}\cdot S_{pagr}\cdot h$$
$$8$$ = $$\frac{1}{3}\cdot S_{pagr}\cdot 2\cdot \sqrt {3}$$
$$8$$ = $$\frac{2\cdot S_{pagr}\cdot \sqrt {3}}{3}$$
$$\frac{8\cdot 3}{2\cdot \sqrt {3}}$$ = $$S_{pagr}$$
$$S_{pagr}$$ = $$\frac{8\cdot 3}{2\cdot \sqrt {3}}$$
$$S_{pagr}$$ = $$\frac{24}{2\cdot \sqrt {3}}$$
$$S_{pagr}$$ = $$\frac{12}{\sqrt {3}}$$
8 = 1/3*S_pagr*2*saknis(3)
Pagrindo plotas: $$\frac{12}{\sqrt {3}}$$.
Kadangi piramidė taisyklinga, jos pagrindas lygiakraštis trikampis.
Lygiakraščio trikampio ploto formulė $$S = \frac{a^{2}\cdot \sqrt {3}}{4}$$
Rasime kraštinę a:
S =
a^2* saknis(3) |
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/ 4 |
S = a^2* saknis(3) |
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/ 4 |
$$S$$ = $$\frac{a^{2}\cdot \sqrt {3}}{4}$$ 12 |
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/ saknis(3)/ saknis(3) |
= $$\frac{12}{\sqrt {3}\cdot \sqrt {3}}$$ = $$\frac{a^{2}}{4}$$ 12* 4 |
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/ saknis(3)/ saknis(3) |
= a^2$$\frac{12\cdot 4}{\sqrt {3}\cdot \sqrt {3}}$$ = $$a^{2}$$ a^2 = 12* 4 |
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/ saknis(3)/ saknis(3) |
$$a^{2}$$ = $$\frac{12\cdot 4}{\sqrt {3}\cdot \sqrt {3}}$$ a^2 = $$a^{2}$$ = $$\frac{48}{3}$$
$$S$$ = $$\frac{a^{2}\cdot \sqrt {3}}{4}$$
$$\frac{12}{\sqrt {3}}$$ = $$\frac{a^{2}\cdot \sqrt {3}}{4}$$
$$\frac{12\cdot 4}{\sqrt {3}\cdot \sqrt {3}}$$ = $$a^{2}$$
$$a^{2}$$ = $$\frac{12\cdot 4}{\sqrt {3}\cdot \sqrt {3}}$$
$$a^{2}$$ = $$\frac{48}{3}$$
$$\sqrt {a^{2}}$$ = $$\sqrt {16}$$
12/saknis(3) = a^2*saknis(3)/4
12*4/saknis(3)/saknis(3) = a^2
a^2 = 12*4/saknis(3)/saknis(3)
Kraštinė lygi 4.
Kita trikampio ploto formulė $$S = \frac{1}{2}\cdot a\cdot h$$
Rasime aukštinę h (CD):
S =
* a* h
S = * a* h$$S$$ = $$\frac{1}{2}\cdot a\cdot h$$ 12 |
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/ 2/ saknis(3) |
= h$$\frac{12}{2\cdot \sqrt {3}}$$ = $$h$$ h = 12 |
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/ 2/ saknis(3) |
$$h$$ = $$\frac{12}{2\cdot \sqrt {3}}$$ h = $$h$$ = $$\frac{6\cdot \sqrt {3}}{\sqrt {3}\cdot \sqrt {3}}$$ h = 6* saknis(3) |
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/ 3 |
$$h$$ = $$\frac{6\cdot \sqrt {3}}{3}$$
$$S$$ = $$\frac{1}{2}\cdot a\cdot h$$
$$\frac{12}{\sqrt {3}}$$ = $$\frac{1}{2}\cdot 4\cdot h$$
$$\frac{12}{\sqrt {3}}$$ = $$2\cdot h$$
$$\frac{12}{2\cdot \sqrt {3}}$$ = $$h$$
$$h$$ = $$\frac{12}{2\cdot \sqrt {3}}$$
$$h$$ = $$\frac{6}{\sqrt {3}}$$
$$h$$ = $$\frac{6\cdot \sqrt {3}}{\sqrt {3}\cdot \sqrt {3}}$$
$$h$$ = $$\frac{6\cdot \sqrt {3}}{3}$$
$$h$$ = $$2\cdot \sqrt {3}$$
h = 6*saknis(3)/saknis(3)/saknis(3)
Atsakymas: $$2\cdot \sqrt {3}$$
26 uždavinys28 uždavinys