ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: samiira on July 12, 2010, 02:55:40 pm

Title: please help.. urgentt!!!
Post by: samiira on July 12, 2010, 02:55:40 pm: heplp with this question..
Title: Re: please help.. urgentt!!!
Post by: brightsky on July 12, 2010, 03:23:15 pm: 1. a) $V = \mbox {Area of triangular cross section} \cdot \mbox{Length}$

$\therefore V = \frac{1}{2} \cdot 2x \cdot (3 - x) \cdot (x + 16)$

$V = x(3 - x)(x + 16)$

b) From the given dimensions, $2x > 0$ , $3-x>0$ and $x + 16 > 0$ (Because neither of the dimensions can be 0 if the triangular prism was to exist.)

So $x > 0$ , $x < 3$ , $x > -16$

Solving these inequalities simultaneously, we get: $0 < x < 3$ .
Title: Re: please help.. urgentt!!!
Post by: Martoman on July 12, 2010, 03:24:25 pm: I don't know whats so hard about this question but hey, we all have our days/questions.

It tells you $V = A_{base}\times h$

$V = \frac{1}{2}\times 2x\times(3-x)(x+16)$ = $x(3-x)(x+16)$

b) $x > 0$ obviously because you can't have a negative area..... on earth. And $x < 3$ as if it were say 4 then it will make the 3-x term 3-4 = -1 thus the whole thing negative and you can't have a negative area.

Soooo that leaves $0 < x < 3$

2) You know the triangle is $x(3-x)$ which is the same as the other triangle side.
There are three rectangles and we need to do Pythagoras to work out the width.

You get $x^2 +(3-x)^2 = width^2$
$width = \sqrt{2x^2 -6x+9}$

So area of rectangle is $A = \sqrt{2x^2 -6x+9} \times (x+16)$

TSA = $2triangles + 3 rectangles$

I'm sure you can now shove in the stuff and then for part C do some algebra :smitten: