# Thread: Area of rectangle problem.

1. ## Area of rectangle problem.

First off, hey guys. I've never been a fan of the introduce yourself threads, "who am I introducing myself to"? Everybody? I don't walk into a room and shout, "hey, everybody, I'm here!". I strike up a conversation with someone and introduce myself during such conversation, so, that's what this is. How are you doing?

Well, my head is frazzled today. I'm doing an intense IT course in which there is a very hefty maths module. I have average junior cert maths at best (we do junior cert maths at around 14/15 in Ireland) so I'm more or less a newb to this and we are now covering Calculus and only in our 4th week. I have an exam on Thursday of this week and that is the end of the module, so you can image it's pretty intense stuff.

Anyhow, that's a little bit of my personality and a little bit of my situation out of the way. Here's the problem:

A rectangle has length 2root(x)cm and width root(x)cm. The length of a diagonal of the rectangle is root(45)cm.

i) Find the area of the rectangle.

I thought about this for a few hours to consider what I had when it occurred to me, that as I have the diagonal and the 2 sides in terms of x, I could treat it as 2 right angles triangles (solve one triangle and double it if you follow me. I've taken the liberty of drawing you a picture of the triangle that ensued. It's poor drawing but Da Vinci wasn't taking requests at the time so I had to do it myself.

<img src="http://postimg.org/image/8149f5q5n/">

I pondered for another while when something came to me like a flash. Would I be right in saying that Pythagoras' theorem might be used to solve this? What I mean is, "x^2+2x^2=45". Does that statement describe my triangle?

I thank you very much for your help and sorry to rush straight into it but time is of the essence. The course is very tough, and while some might buckle, others have the opportunity to step out from the pack, and that's where I fit in. I know this is elementary to some of you math magicians out there but I'm amazed at how much I've learned in the last number of weeks and I believe I can keep learning. I love how I hear things now and I instantly think geometric series or arithmetic series, or that's a linear statement, linear movement even. In fact, even the word "linear" itself makes more sense. It's great.

So, yeah, a bit about me and there's my problem. Feel free excerpt some verbal diarrhea of your own when replying - well, the equation wouldn't balance if I was the only one that spewed from the gut. Ya gotta do the same on both sides so my tutor keeps telling me.

2. ## Re: Area of rectangle problem.

Actually, I've just realised another problem that I've given myself but I'm not sure of so I'll ask here.

It doesn't cause me any problems for this but I've assumed the angles 45 on both occasions in the picture I put up. 45 degree angles aren't a given in a triangle. I've answered my own question there so just ignore this drivel if you don't mind.

Still, I'm happy I've come this far with the problem.

3. ## Re: Area of rectangle problem.

Originally Posted by MathsRedemption
First off, hey guys. I've never been a fan of the introduce yourself threads, "who am I introducing myself to"? Everybody? I don't walk into a room and shout, "hey, everybody, I'm here!". I strike up a conversation with someone and introduce myself during such conversation, so, that's what this is. How are you doing?

Well, my head is frazzled today. I'm doing an intense IT course in which there is a very hefty maths module. I have average junior cert maths at best (we do junior cert maths at around 14/15 in Ireland) so I'm more or less a newb to this and we are now covering Calculus and only in our 4th week. I have an exam on Thursday of this week and that is the end of the module, so you can image it's pretty intense stuff.

Anyhow, that's a little bit of my personality and a little bit of my situation out of the way. Here's the problem:

A rectangle has length 2root(x)cm and width root(x)cm. The length of a diagonal of the rectangle is root(45)cm.

i) Find the area of the rectangle.

I thought about this for a few hours to consider what I had when it occurred to me, that as I have the diagonal and the 2 sides in terms of x, I could treat it as 2 right angles triangles (solve one triangle and double it if you follow me. I've taken the liberty of drawing you a picture of the triangle that ensued. It's poor drawing but Da Vinci wasn't taking requests at the time so I had to do it myself.

<img src="http://postimg.org/image/8149f5q5n/">

I pondered for another while when something came to me like a flash. Would I be right in saying that Pythagoras' theorem might be used to solve this? What I mean is, "x^2+2x^2=45". Does that statement describe my triangle?

Almost but not quite. The Pythagorean theorem says that $\displaystyle a^2+ b^2= c^2$. Here, $\displaystyle a= \sqrt{x}$ so that $\displaystyle a^2= x$ and $\displaystyle b= 2\sqrt{x}$ so that $\displaystyle b^2= 4x$. So $\displaystyle a^2+ b^2= x+ 4x= 5x= 45$. From that is it easy to see that x= 9 so that the two sides of the rectangle are 3 and 2(3)= 6. What is the area of a rectangle of side length 3 and 6?

I thank you very much for your help and sorry to rush straight into it but time is of the essence. The course is very tough, and while some might buckle, others have the opportunity to step out from the pack, and that's where I fit in. I know this is elementary to some of you math magicians out there but I'm amazed at how much I've learned in the last number of weeks and I believe I can keep learning. I love how I hear things now and I instantly think geometric series or arithmetic series, or that's a linear statement, linear movement even. In fact, even the word "linear" itself makes more sense. It's great.

So, yeah, a bit about me and there's my problem. Feel free excerpt some verbal diarrhea of your own when replying - well, the equation wouldn't balance if I was the only one that spewed from the gut. Ya gotta do the same on both sides so my tutor keeps telling me.