# problems with quite a few questions

• Nov 25th 2005, 12:37 PM
rpatel
problems with quite a few questions
hello.

as part of my revision i carried many questions at home. these are questions which i couldn't answer and wasn't able to find the solutions from textbooks.

i have attached the questions.

thank you

regards

:D
• Nov 25th 2005, 07:56 PM
Jameson
I'll start with #8. Area is measured in two dimensions, so you know that it is going to reduce down to some form of l x w, or length times width. So for instance take the surface area of a sphere, $4\pi{r^2}$. You can see that the term r squared implies two demensions and the formula makes sense in that aspect. Two of your expressions are expressed in this basic form, with constants thrown in that won't affect them in making them an area.
• Nov 26th 2005, 01:15 AM
CaptainBlack
I'll continue with #18.

The two cylinders are similar, which means that they have the
same proportions but different sizes. The surface areas of similar
solids scale as the square of the ratio of corresponding linear
dimensions. Thus if our cylinders had heights of 6 and 12 cm the
surface area of the larger would be four times that of the smaller.

The volumes of similar solids scale as the cube of the ratio of
corresponding linear dimensions. Thus if our cylinders had heights
of 6 and 12 cm the volume of the larger would be eight times
that of the smaller.

So if $C1$ and $C2$ are our two cylinders we have:

$\mbox{SurfArea}(C1)/\mbox{SurfArea}(C2)\ =\ [\mbox{Height}(C1)/\mbox{Height}(C2)]^2$.
$\mbox{Volume}(C1)/\mbox{Volume}(C2)\ =\ [\mbox{Height}(C1)/\mbox{Height}(C2)]^3$.

With this the solution of #18 should be easy.

RonL
• Nov 26th 2005, 02:36 AM
CaptainBlack
Continuing with #11.

This is about the properties of similar triangles - Since QT parallel to
RS implies that triangle PQT is similar to triangle PRS.

a) Hence from the properties of ratios of corresponding sides of similar
triangles we may conclude:
$\frac{PQ+QR}{PQ}\ =\ \frac{PS}{PT}$,
which will allow you to find $QR$.

b) Similarly we have::
$\frac{QT}{SR}\ =\ \frac{PT}{PS}$,
which will allow you to find $QT$.

RonL
• Nov 26th 2005, 06:29 AM
CaptainBlack
Moving on to #22.

What you need to know to do this problem is that in any
polygon with vertices $V_1,\ V_2,\ ...\ V_n$, that:
$\vec{V_1V_2}+\vec{V_2V_3}+\ ...\ +\ \vec{V_{n-1}V_n}\ =\ 0$.

and that for two segments $AB$ and $CD$ to be
parallel means that for some real number k:
$\vec{AB}\ =\ k.\vec{CD}$.

RonL
• Nov 26th 2005, 12:18 PM
rpatel
thank you !!!! :D