Triangular prism - regular

The regular triangular prism is 7 cm high. Its base is an equilateral triangle whose height is 3 cm. Calculate the surface and volume of this prism.

Correct result:

V =  36.373 cm3
S =  83.138 cm2

Solution:

v=3 cm h=7 cm  a2=v2+(a/2)2 4a2=4v2+a2 3a2=4v2  a=v 2/3=3 2/32 3 cm3.4641 cm  S1=34 a2=34 3.464123 3 cm25.1962 cm2  V=S1 h=5.1962 7=21 3=36.373 cm3v=3 \ \text{cm} \ \\ h=7 \ \text{cm} \ \\ \ \\ a^2=v^2 + (a/2)^2 \ \\ 4a^2=4v^2 + a^2 \ \\ 3a^2=4v^2 \ \\ \ \\ a=v \cdot \ 2/\sqrt{ 3 }=3 \cdot \ 2/\sqrt{ 3 } \doteq 2 \ \sqrt{ 3 } \ \text{cm} \doteq 3.4641 \ \text{cm} \ \\ \ \\ S_{1}=\dfrac{ \sqrt{ 3 } }{ 4 } \cdot \ a^2=\dfrac{ \sqrt{ 3 } }{ 4 } \cdot \ 3.4641^2 \doteq 3 \ \sqrt{ 3 } \ \text{cm}^2 \doteq 5.1962 \ \text{cm}^2 \ \\ \ \\ V=S_{1} \cdot \ h=5.1962 \cdot \ 7=21 \ \sqrt{ 3 }=36.373 \ \text{cm}^3
o=3 a=3 3.46416 3 cm10.3923 cm  S=2 S1+o h=2 5.1962+10.3923 7=48 3=83.138 cm2o=3 \cdot \ a=3 \cdot \ 3.4641 \doteq 6 \ \sqrt{ 3 } \ \text{cm} \doteq 10.3923 \ \text{cm} \ \\ \ \\ S=2 \cdot \ S_{1} + o \cdot \ h=2 \cdot \ 5.1962 + 10.3923 \cdot \ 7=48 \ \sqrt{ 3 }=83.138 \ \text{cm}^2

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Hola Qomo Estas
hola qomo estas ya se que esta mal escrito

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