Question

HELP ME GUYS PLEASE, I'M BEGGING YOU HUHU :(
1. A right circular cylinder has 4in as its diameter and 6in as its height.
a. Find the lateral area
b. Find the volume of the regular hexagonal prism inscribed inside the cylinder
2. A rectangular solid is 4ft long and 3ft wide
a. Find the volume if its diagonal is square root of 41

Answer 1

lateral surface area= 2 x 3.14 x 4/2 x6

Answer 2

how about letter b in number 1?

Answer 3

do you know what a "regular hexagonal prism " is?

Answer 4

i have very cloudy information about it. my teacher is not that good. :(

Answer 5

also, what's the answer for number 2?

Answer 6

hmm do you know what a right circular cylinder is?

Answer 7

oh i remember, hexagonal prism has a hexagon as its base. the bases are perpendicular to each other right?

Answer 8

we are not yet discussing it :(

Answer 9

seems to me you'd need to cover all that first, it doesn't help trying to solve what you haven't dunno what it's

Answer 10

its an assignment, can you help me answer it please? :(

Answer 11

also for number 2 please?

Answer 12

a) http://www.mathwords.com/r/right_circular_cylinder.htm

Answer 13

how about b?

Answer 14

b) http://www.mathwords.com/s/segment_of_a_circle.htm
you get the volume of the cylinder then get the 6 segments NOT OCCUPIED by the hexagonal prism and subtract it from the cylinder's volume

Answer 15

how about number 2?

Answer 16

2 a) http://www.ajdesigner.com/phpgeometricformulas/rectangular_solid_diagonal_d.php

Answer 17

using the diagonal equation, you can find "h" height
you have the length and the width, once you have the height

volume = length * width * height

Answer 18

how do you find the height? what's the formula?

Answer 19

is it d= square root of(l/^2 +w^2 + h^2? am I right?

Answer 20

A rectangular solid is 4ft long and 3ft wide
a. Find the volume if its diagonal is square root of 41

\(\bf diagonal = \sqrt{41}\qquad length=4\qquad width=3\qquad height={\color{red}{ h}}\\ \quad \\
d=\sqrt{l^2+w^2+{\color{red}{ h}}^2}\implies \sqrt{41}=\sqrt{l^2+w^2+{\color{red}{ h}}^2}\implies 41=l^2+w^2+{\color{red}{ h}}^2\)

Answer 21

\(\bf diagonal = \sqrt{41}\qquad length=4\qquad width=3\qquad height={\color{red}{ h}}\\ \quad \\
d=\sqrt{l^2+w^2+{\color{red}{ h}}^2}\implies \sqrt{41}=\sqrt{l^2+w^2+{\color{red}{ h}}^2}\\ \quad \\\implies 41=l^2+w^2+{\color{red}{ h}}^2\)

Answer 22

oh i see? but i still cannot understand letter b. can you please explain it to me again?

Answer 23

what's the volume of the cylinder?

Answer 24

keeping in mind that cylinder's volume = \(\Large \bf \pi r^2 h\)

Answer 25

75.40 in^3

Answer 26

ok... .so we know the volume of the cylinder... so let's see an incribed hexagonal

a REGULAR hexagonal has 6 EQUAL sides
it's a closed polygon
so if we divide the circle in 6 EQUAL pieces, each piece will have an angle of 360/6

so we know that the angle for a regular hexagonal is 60 degrees for each side
we also know the radius is "2" so so we want to know the SEGMENT's area that's not occupied by the hexagonal, \(\bf \textit{segment of a circle}=\cfrac{\theta r^2 \pi}{360}\implies \cfrac{{\color{blue}{ 60}} \cdot {\color{red}{ 2}}^2 \pi}{360}\)

Answer 27

the volume of that segment will be the height times the Area, so 6 * Area
you have 6 segments, so sum them up, then subtract their volume from the 75.40 from the cylinder, and what's left is the hexagonal's volume

Answer 28

hmm wait a sec... I used .. .dohh the wrong ... equation for the segment, I got the SECTOR instead, lemme rewrite that

Answer 29

so 60 degrees and radius =2 \(\bf \cfrac{r^2}{3}\left(\cfrac{\pi \theta}{180}-sin(\theta)\right)\implies \cfrac{{\color{blue}{ 2}}^2}{3}\left(\cfrac{\pi \cdot {\color{red}{ 60}}}{180}-sin({\color{red}{ 60^o}})\right)\)

Answer 30

so what's your asnwer in lettter b?

Answer 31

the volume of that segment will be the height times the Area, so 6 * Area
you have 6 segments, so sum them up, then subtract their volume from the 75.40 from the cylinder, and what's left is the hexagonal's volume

Answer 32

is your asnwer 66.7?

Answer 33

hmmm

Answer 34

I just notice a typo ark

Answer 35

my bad

Answer 36

\(\bf \cfrac{r^2}{2}\left(\cfrac{\pi \theta}{180}-sin(\theta)\right)\implies \cfrac{{\color{blue}{ 2}}^2}{2}\left(\cfrac{\pi \cdot {\color{red}{ 60}}}{180}-sin({\color{red}{ 60^o}})\right)\)

Answer 37

\(\bf \cfrac{r^2}{2}\left(\cfrac{\pi \theta}{180}-sin(\theta)\right)\implies \cfrac{{\color{blue}{ 2}}^2}{2}\left(\cfrac{\pi \cdot {\color{red}{ 60}}}{180}-sin({\color{red}{ 60^o}})\right)\implies 2\left(\cfrac{\pi}{3}-sin(60^o)\right)\\ \quad \\

2\left(1.047-\frac{\sqrt{3}}{2}\right)\approx 0.3623\\ \quad \\
\textit{volume of the segment will be } 0.3623 \cdot 6\implies 2.1741\\ \quad \\
\textit{6 segments, so }2.1741\cdot 6 = 13.0444\\ \quad \\
{\color{red}{ 75.4}}-13.044 \approx 62.3556\)

Answer 38

THANK YOU SO MUCH! :D

Answer 39

yw