Question

Proszę o pomoc! zadania w załączniku

Answer 1

z.1

h = 6 cm,   l = 10 cm

Mamy

r^2 + h^2 = l^2   =>   r^2 =  l^2 - h^2 = 10^2 - 6^2 = 100  - 36 = 64

Pp = pi*r^2 = pi* 64 = 64 pi

Odp. Pp = 64 pi cm^2

===================

z.2

a = 3 cm

h = 7 cm

 

więc

V = (1/3) Pp*h = (1/3) a^2 *h = (1/3) *3^2 *7 = (1/3)*9*7 = 3*7 = 21

Odp. V = 21 cm^3

================

z.3

a = 6

b = 12

h1 - wysokośc trójkąta równobocznego

więc

h1 = a*p(3) / 2 = 6 *p(3) / 2 = 3 p(3)

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x = (2/3) *h1 = (2/3)* 3 p(3) = 2 p(3)

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Z tw. Pitagorasa mamy

x^2 + h^2 = b^2   =>  h^2 =  b^2 - x^2 = 12^2 - ( 2 p(3))^2 = 144 -  4*3 = 144 - 12 = 132

h^2 = 132 = 4*33

więc

h = p( 4*33) = 2 p(33)   - wysokość ostrosłupa

------------------------------

V = (1/3) Pp*h = ( 1/3)* [ a^2 p(3) / 4] *h = (1/12) a^2 p(3)*h

V = (1/12) *6^2 *p(3) * 2 p(33) = (72 / 12)*p( 3*33) =  6* p(99) = 6* p( 9*11) =

= 6*  3*p(11) = 18 p(11)

Odp. V = 18 p(11)

=================

z.4

 a^2 = 64   => a = p(64) = 8

zatem

a = 8 cm

oraz

2 r = a = 8 cm   =>  r =  4 cm

h = a = 8 cm

więc

V = Pp*h = pi*r^2 *h = pi*4^2* 8 = pi*16*8 = 128 pi

Odp. V = 128 pi  cm^3

====================

z.5

b = 8 cm

alfa = 30 st

h - wysokość ostrosłupa

Mamy

h /b = sin 30 st = 1/2

h = (1/2) *b = (1/2)* 8 cm = 4 cm

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d = 2*x    - długość przekatnej kwadratu

 x/ b = cos 30 st

x = b* cos 30 st = 8 cm* p(3)/ 2 = 4 p(3) cm

więc

d = 2* 4 p(3) cm = 8 p(3) cm

----------------------------------------

zatem

V = (1/3) Pp *h = (1/3) *(1/2) d^2 *h = (1/6) d^2 *h = (1/6)* [ 8 p(3)]^2 * 4 =

= (1/6) * 64*3*4 = 128

V = 128 cm^3

===============

z.6

Walec

h = 4

Pb = 32 pi

więc

 Pb = 2 pi*r*h = 32 pi  

2 pi* r * 4 = 32 pi  / : pi

8 r = 32  / : 8

r = 4

------

V = Pp*h = pi *r^2 *h = pi* 4^2 * 4 = pi*16*4 = 64 pi

Odp. V = 64 pi

================

z.7

Ostrosłup prawidłowy sześciokątny

 

a = 10

b = 13

Mamy

r = a = 10

h  - wysokość ostrosłupa

Z tw. Pitagorasa mamy

h^2 + r^2 = b^2

h^2 = b^2 - r^2 = 13^2 - 10^2 = 169 - 100 = 69

więc

h = p(69)

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Pole podstawy składa sie  z pól 6 trójkątów równobocznych o a = 10,więc

Pp = 6* [ a^2 p(3) / 4] = 1,5 a^2 *p(3)

Pp = 1,5 *10^2 *p(3) = 1,5 *100 *p(3) = 150 p(3)

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h1 - wysokość ściany bocznej  czyli trójkąta równoramiennego o podstawie

a = 10  i ramieniu  b = 13

Z tw. Pitagorasa mamy

(h1)^2 + ( a/2)^2 = b^2

(h1)^2 = 13^2 - 5^2 = 169 - 25 = 144

więc

h1 = p(144) = 12

------------------------

Pole powierzchni bocznej składa sie z 6 pól  trójkątów przystających, więc

Pb = 6* (1/2) a*h1 = 3 a *h1 = 3*10* 12 = 360

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Pole powierzchni całkowitej

Pc = Pp + Pb =  150 p(3) + 360

==========================

p(3)  - pierwiastek kwadratowy z 3

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z.8

Stożek

 a = 6

więc

2 r = 6    , to  r = 3

l = a = 6

Pc = Pp + Pb = pi*r^2 = pi*r*l = pi*3^2 + pi*3*6 = 9 pi + 18 pi = 27 pi

========================================================

z.9

 

P = 100 pi  cm^2

więc

P = 4 pi*r^2 = 100 pi  / : 4 pi

r^2 = 25

r = 5

------

Objętośc kuli

V = (4/3) pi*r^3 = (4/3) pi*5^3 = (4/3)* 125* pi = ( 500 / 3) *pi

Odp. V = ( 500 / 3) * pi  cm^3

===========================

 z.10

Stożek

 l = 15

h - wysokośc stożka

r - promień podstawy stożka

alfa = 60 st

Mamy

h / l = sin 60 st = p(3) / 2

więc

h = l * p(3)/2 = 15 * p(3) / 2 = 7,5 p(3)

--------------------------------------------------

r / l = cos 60 st = 1/2

r / 15 = 1/2

więc

r = (1/2)*15 = 7,5

------------------------

V = (1/3) Pp*h = (1/3) pi*r^2 *h = (1/3) pi* ( 7,5)^2 * 7,5 p(3) =

= 2,5 pi* 56,25 *p(3) = 140, 625 *p(3) * pi

===================================

z.11

Ostrosłupa

Pp = 21 cm^2

V = 21 cm^3

Mamy

V = ( 1/3) Pp*h     / * 3

3 V = Pp*h

więc

h = ( 3 V ) / Pp = ( 3* 21 ) / 21 = 3

h = 3  cm

==============

z.12

Walec

V = 36 m^2

r = 2 m

 

 Mamy

V = Pp*h = pi*r^2 *h

czyli

pi* 2^2 * h = 36

pi*4* h = 36

h =  36 / ( 4 pi) = 9 / pi

-------------------------------

Pc =  2 Pp + Pb = 2 pi*r^2 + 2 pi*r*h

Pc = 2 pi* 2^2 + 2 pi*2* [ 9 / pi ] = 8 pi + 36

Pc = 36 + 8 pi

================

z.13

 Czworościan foremny

Pc = 25 p(3) cm^2

 

Pc = a^2 p(3)

więc

a^2 p(3) = 25 p(3) cm^2

a^2 = 25 cm^2

a = 5 cm

---------------

V = [ a^3  p(2) ] /12 =  [ 5^3 * p(2) ] / 12 = [ 125 p(2) ] / 12 = ( 125 / 12) * p(2)

 

V = ( 125 / 12 ) p(2) cm^3

=======================

 

 z.14

Kula

 V = 27 pi

 

Mamy

V = ( 4/3) pi*r^3

czyli

( 4/3) pi*r^3 = 27 pi  / : pi

(4/3) r^3 = 27   / * ( 3/4)

r^3 = 81/4

r =  3/ p3st( 4)

-------------------

zatem

Pc = 4 pi*r^2 = 4 pi* [  3/ p 3st (4) ]^2 = 4 pi* [ 9/ p3st ( 16)] = 36 pi / p3st( 16) =

= 36 pi / p 3st ( 8*2) = 36 pi / ( 2 p3st(2)) =  18 pi / p3st (2)

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p 3st ( 2 )  - pierwiastek 3 stopnia  z  2

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