FREE lectures this September + October | HSC Head Start + Exam Revision: book here | QCE Head Start: book here | VCE Exam Revision: book here

Welcome, Guest. Please login or register.

September 17, 2019, 09:11:20 pm

Author Topic: Where do I even start?  (Read 346 times)  Share 

0 Members and 1 Guest are viewing this topic.

Jsy443

  • Adventurer
  • *
  • Posts: 7
  • Respect: 0
Where do I even start?
« on: January 27, 2019, 04:22:50 pm »
0
"The length of a rectangle is 4cm more than the width. If the length were to be decreased by 5cm and the width decreased by 2cm, the perimeter would be 18cm. Calculate the dimensions of the rectangle."

I don't understand where to start with this question. I suck at problem solving questions, so a detailed explaintion for this question will be very much appreciated. Thank you.

mango8

  • Trendsetter
  • **
  • Posts: 122
  • Respect: +54
Re: Where do I even start?
« Reply #1 on: January 27, 2019, 04:53:51 pm »
+4
First thing to remember, is for questions like these, always draw a diagram!
Here is my working, let me know if anything doesn't make sense! Hope I'm right lol :)

working out!

Edit: Okay so I found the dimensions of the altered rectangle. For the original rectangle as darkz said, just sub in x=6 into (x+4)=10. So yes, original dimensions are 6cm x 10cm.



« Last Edit: January 27, 2019, 05:34:12 pm by mango8 »

darkz

  • MOTM: APR 18
  • Forum Obsessive
  • ***
  • Posts: 278
  • Respect: +94
Re: Where do I even start?
« Reply #2 on: January 27, 2019, 05:03:39 pm »
+3
"The length of a rectangle is 4cm more than the width. If the length were to be decreased by 5cm and the width decreased by 2cm, the perimeter would be 18cm. Calculate the dimensions of the rectangle."

I don't understand where to start with this question. I suck at problem solving questions, so a detailed explaintion for this question will be very much appreciated. Thank you.
\[
\text{Let the width of the rectangle be }x\\
\therefore \text{the width}=x\text{ and the length}=x+4\\
\text{If we decrease the length by 5 and decrease the width by 2, then}\\
\text{width}=x-2,\,\text{length}=x+4-5=x-1\\
\begin{aligned}\\
\text{Perimeter}=2(x-2)+2(x-1)&=18\\
4x-6&=18\\
x&=6\\
\end{aligned}\\
\therefore \text{ the dimensions of the original rectangle are: } 6cm\times10cm\\
\]
Edit: fixing my latex
2018: Biology [50 + Prems], Mathematical Methods
2019: English, Latin, Chemistry, Specialist Mathematics
Free Biology Unit 3/4 Notes

Jimmmy

  • Trendsetter
  • **
  • Posts: 144
  • Respect: +50
Re: Where do I even start?
« Reply #3 on: January 27, 2019, 05:34:42 pm »
0
First thing to remember, is for questions like these, always draw a diagram!
Here is my working, let me know if anything doesn't make sense! Hope I'm right lol :)

working out!
Spot on working out Mango, just forgot to sub x into the original rectangle to find its dimensions.  :) Darkz got it.
2018: Business Management

2019: English Language, Maths Methods, Legal Studies, Global Politics, Philosophy