Help with surface area, volumes and measures

• Mar 16th 2008, 07:12 AM
sammyfox07
Help with surface area, volumes and measures
Hello is there anyone who can help me out with these questions, i would really appreciate it. Thanks

Q1. Work out the surface area of this triangular prism which has a cross-section in the shape of an equilateral triangle of side 6cm and a length of 12cm.

Q2. 8400mm(squared) to cm(squared)
Q3. 2.5cm(cubed) to mm(cubed)

Q4 If a,b,c,r and h represent lengths, use dimensions theory to deduce whether each of these expressions is length,area,volume or none of these.

a) abc
b) 2thirdsbc
c) 3bh + πr
d) 5h(squared)a
e) a + πb
f) ab(squared) + r(squared)
g) a(b+2r)
h) halfπr(squared)h
• Mar 16th 2008, 07:58 AM
earboth
Quote:

Originally Posted by sammyfox07
Hello is there anyone who can help me out with these questions, i would really appreciate it. Thanks

Q1. Work out the surface area of this triangular prism which has a cross-section in the shape of an equilateral triangle of side 6cm and a length of 12cm.

Q2. 8400mm(squared) to cm(squared)
Q3. 2.5cm(cubed) to mm(cubed)

Q4 If a,b,c,r and h represent lengths, use dimensions theory to deduce whether each of these expressions is length,area,volume or none of these.

a) abc
b) 2thirdsbc
c) 3bh + πr
d) 5h(squared)a
e) a + πb
f) ab(squared) + r(squared)
g) a(b+2r)
h) halfπr(squared)h

To Q1:

Let s denote the side of an equilateral triangle then the area of the triangle is calculated by:

$a_{eT} = \frac14\cdot s^2 \cdot \sqrt{3}$

The surface area consists of 2 triangles (base and top) and 3 rectangles whose length is the height of the prism and whose width is the side of the triangle.

I've got $a_{p} \approx 247.18\ cm^2$

to Q2:
Use the ratio: $1\ cm^2 = 100\ mm^2$

to Q3:
Use the ratio: $1\ cm^3 = 1000\ mm^3$

to Q4:

You know that the volume $V=lll$ and the area $a = ll$

a) abc ....... $lll~\rightarrow~V$
b) 2thirdsbc ....... $ll~\rightarrow~a$ ....... $\frac23$ is a constant without a dimension!
c) 3bh + πr ....... This one is really funny because you add a length ( $\pi r$) to an area... So I assum that there is a typo and you probably meant $3bh + \pi r^2$
d) 5h(squared)a ....... $lll~\rightarrow~V$ ...... (Do you see why?)

I'll leave the rest for you:

e) a + πb
f) ab(squared) + r(squared)
g) a(b+2r)
h) halfπr(squared)h