Totally pathetically easy maths question

quad_red

Lads, I know this is noob of the century question but it's been a long time since I looked at a maths book.

Can someone quickly give me the equation for solving this?

(4 x 6) + (0.5 x 6 x 2)

quad_red

Deleted User wrote: »

(4 x 6) + (0.5 x 6 x 2)

Sorry...

how did you come to that? Are you splitting it into triangles?

Yakuza

It was split into a 4*6 rectangle and a right-angled triangle with a base of 6 and a perpendicular height of 2 (area of such a triangle is 1/2 * base * perp. height).

You could also treat it as a 6 * 6 square minus a 6*2 triangle, you'll get the same answer

Ascii

Alternatively you could use 6*5......The give and take rule

Yakuza

The what now?

quad_red

quad_red wrote: »

Sorry...

how did you come to that? Are you splitting it into triangles?

Cut off the top triangle and took them both as seperate objects

rectangle = length by breadth
triangle = half the base by the perpendicular height

embeeheart

quad_red wrote: »

Lads, I know this is noob of the century question but it's been a long time since I looked at a maths book.

Can someone quickly give me the equation for solving this?

Area of rectangle first 4x6 = 24, then get area of triangle at top - half the base 3 x perpendicular height 2 = 6.
24 + 6 = 30

ahnowbrowncow

Yakuza wrote: »

The what now?

I'm guessing average of both lengths * width

Yakuza

Here's a funkier way of working it out

I give up, I can't post wolfram alpha links:

Paste this in and you'll see what I mean:

http://www.wolframalpha.com/input/?i=integrate+y%3Dx%2F3+%2B4+between+0+and+6

harry21

Yakuza wrote: »

Here's a funkier way of working it out

Given the OP's original question, I doubt they have ever done any kind of integration.

Yakuza

Fair enough, I'm not trying to blind the OP with science, just trying to show a practical application of calculus from schooldays.

Breaking such a shape into two smaller, regular ones is by far the most straightforward way to go.

Michael Collins

If you're interested this shape is known as a trapezium, and therefore the area formula is simply

[latex] \displaystyle \left(\frac{a+b}{2}\right) h = \left(\frac{4+6}{2}\right) 6 = 30. [/latex]

quad_red

Guys,

Thanks for the input. This was rather panicked post in response to trying to do sample psychometric tests (in advance of an actual one next week) and then not being sure what I was doing with that question was correct.

Have done a bit of warmup and I'm up to speed now (and cringing given I've done maths to a reasonable level. Just had a bit of blank and very rusty brain - need to put down the bloody calculator every now and again).

I actually got sort of a fright at how clueless I found myself after not having looked at any maths beyond arithmetic/percentages for a decade.

Anyway, I shall now reverse out of this forum now before I make an even bigger dunce out of myself.

Cheers, Quad

2+1 =?

764dak

Ascii wrote: »

Alternatively you could use 6*5......The give and take rule

Yakuza wrote: »

The what now?

ahnowbrowncow wrote: »

I'm guessing average of both lengths * width

Michael Collins wrote: »

If you're interested this shape is known as a trapezium, and therefore the area formula is simply

[latex] \displaystyle \left(\frac{a+b}{2}\right) h = \left(\frac{4+6}{2}\right) 6 = 30. [/latex]

Yup, the average of both lengths multiplied by the width.

764dak

Yakuza wrote: »

Fair enough, I'm not trying to blind the OP with science, just trying to show a practical application of calculus from schooldays.

Breaking such a shape into two smaller, regular ones is by far the most straightforward way to go.

This isn't science. Plus, averaging both lengths and multiplying it by the width is easier.

Yakuza

It's an idiom : http://idioms.thefreedictionary.com/blind+with+science.

harry21

Yakuza wrote: »

It's an idiom : http://idioms.thefreedictionary.com/blind+with+science.

Now thats some nice schooling!:D

trodsky

Is going to be a very small shed. Are you sure you units are correct?

userfriendly2

4 14 11 31
35 26 73 ?



7 8 20 1
5 6 2 ?



41 44 72 78
36 66 62 ?

These are a few questions from a basic numerical test that I can't answer. there are 3 questions there, but I cant figure out the sequence??? Its probably very basic and staring me in the face:-( cheers in advance.

basill

Did you get them from here? Answers are at the back.

http://www.psychometric-success.com/practice-papers/Psychometric%20Success%20Numerical%20Ability%20-%20Reasoning%20Practice%20Test%201.pdf

Razzen

4 14 11 31 (Seq = A*2+3)
35 26 73 ? = 55



7 8 20 1
5 6 2 ?



41 44 72 78 (Seq = A*2-10)
36 66 62 ? = 122

Razzen

I don't see the pattern for the second one, but looking at the format given in the pdf posted by Basil, I'm guessing its the total of each of the four numbers (7+8+5+6) = (20+1+2+?) ? = 3

Not sure of this one...Maybe someone else spots the sequence?

userfriendly2

Razzen wrote: »

I don't see the pattern for the second one, but looking at the format given in the pdf posted by Basil, I'm guessing its the total of each of the four numbers (7+8+5+6) = (20+1+2+?) ? = 3

Not sure of this one...Maybe someone else spots the sequence?

Cheers man thanks - when u say seq=(A*2+3) what is A? Prob anoda stupid question haha