Question

przekatna przekroju osiowego walca ma dlugosc 40 cm i tworzy z jego podstawa kat a. oblicz pole pow. calkowitej tego walca, jesli:

a. sin a= pierwiastek3 / 2

b. cos a= 0.8

Answer 1

h - wysokość tego walca

r = promień walca

a)

Mmay

h/ 40 = sin alfa = p(3)/2

zatem 2*h = 40 *p(3)  / : 2

h = 20 p(3)

h = 20 p(3) cm

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Z tw/ Pitagorasa mamy

( 2 r)^2 + h^2 = 40^2

4 r^2 = 1 600 - [ 20 p(3)]^2 = 1 600 - 1200 = 400

r^2 = 100

r = 10

r = 10 cm

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Pc = 2 Pp + Pb = 2 pi*r^2 + 2 pi*r*h

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Pc = 2 * pi* 10^2 + 2 pI*10*20 p(3) = 200 pi + 400 p(3)* pi

Odp.

Pc = 200 pi*[ 1 + 2 p(3)]  cm^2

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b)

) 2 r)/ 40 = cos alfa = 0,8

2 r = 0,8* 40 = 32

r = 16

r = 16 cm

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Z tw. Pitagorasa mamy

h^2 + ( 2 r)^2 = 40^2

h^2 = 1 600 - 4 r^2 = 1 600 - 4*16^2 = 1 600 - 4*256 = 1 600 - 1024 = 576

czyli

h = 24

h = 24 cm

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Pc = 2 Pp + Pb = 2 pi*r^2 + 2 pi*r*h

Pc = 2 pi*16^2 + 2 pi*16*24 = 512 pi + 768 pi = 1280 pi

Odp. Pc = 1 280 pi cm62

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