Simillar triangles

[Chikis]

How I wish I could show this problem in drawing. This problem has to deal with the concept of similar triangle.

 

In a diagram, PQ//YZ. PQ is parallel to YZ,

|XP| = 2cm, |PY| = 3cm, |PQ|= 6cm and the area of triangle [math]XPQ = 24cm^2[/math]

Calculate the area of trapezium PQZY.

 

Let me describe the shape. We have a large triangle XYZ. The large triange XYZ overllaps a small triangle XPQ

 

I was able to find YZ using PX/YX = PQ/YZ

i.e 2/5 =6/YZ

YZ = 30/2 = 15 cm

 

My greatest problem now is how to find the lenght XQ of the small triangle XPQ using the area [math]24cm^2[/math] given. I find it hard because I can't figure out the height. Though am taking PQ as the base. Could XQ be the height? I don't think so.

 

If you understand my problem could you please draw the shape so that I can confirm it. I have the diagram with me; is just that I don't have the means to show it.

[studiot]

1) Is this your diagram?

 

 

 

 

 

2) Why do you think you need the length of XQ?

 

3) What is the relationship between the areas of similar triangles? : If you don't know look it up.

 

4) What is your strategy for solving this?

[imatfaal]

A. take a photo of your diagram if you can - a smart phone will do. Save photo to drive. Click "more reply options". Click "Browse". Navigate to file. Select file. Click "Attach this file". Wait. Click "Add to Post".

 

B. Studiot's point three is crucial. You can surely work it out by thinking about the triangle formed by dividing a square along diagonal, or splitting an equilateral triangle at half way along each edge.

 

C. A free picture of the British Museum cos I can never remember the exact sequence

 

[Chikis]

1) Is this your diagram?

 

simtri1.jpg

Yes and yes. You made it. How did you do it? Edited July 29, 2014 by Chikis

[studiot]

OK it was a bit tricky with your description.

 

To help you in the future, here is a statement of you diagram.

 

Draw triangle XYZ with base YZ and vertex X

Position P on XY such that XP:PY = 2:3

Draw PQ through P parallel to YZ, to intersect XZ in Q

 

Now what about my questions designed to help?

Have you looked up the similar triangle areas theorem?

Edited July 29, 2014 by studiot

[Chikis]

I have looked it up. The scale factor for the areas of the triangle is 4: 25

Using the scale factor, the area of

[math]\triangleXYZ[/math]

= [math]\frac{25\times{24}}{4}[/math] = [math]150cm^2[/math]

 

To get the area of the trapezium PQZY, I subtract:

Area of PQZY = 150-24

= [math]126cm^2[/math]

 

Oh no! I still have problems of this nature that showing diagrams is making it impossible for me to bring in for help.

Edited July 29, 2014 by Chikis

[studiot]

You got it.

 

Can you write it out formally as an solution now?

 

Edit I would have thought you could have used Paint to draw something as simple as that triangle for upload, if you don't have a scanner/camera.

 

 

Edited July 29, 2014 by studiot

[Chikis]

A REPOST TO CORRECT WHAT IS NOT CORRECT

 

I have looked it up. The scale factor for the areas of the triangle is 4: 25

Using the scale factor, the area of

[math]\bigtriangleup XYZ[/math]

= [math]\frac{25\times{24}}{4}[/math] = [math]150cm^2[/math]

To get the area of the trapezium PQZY, I subtract:

Area of PQZY = 150-24

= [math]126cm^2[/math]

Oh no! I still have problems of this nature that showing diagrams is making it impossible for me to bring in for help.

Edited July 30, 2014 by Chikis

[imatfaal]

What sort of diagram are you trying to upload? I gave instructions for uploading a photo - it is a piece of cake once you get used to it.

[Chikis]

Is not easy for me. My phone can't do it.

[studiot]

Do you not use a computer?

[Chikis]

I don't. I use a phone.

[studiot]

 

I don't. I use a phone.

 

OK, I understand, I seem to remember you mentioned this before.

 

Then you will need to become slick and quick at writing out maths in words.

 

This is why I offered a description in my post#5 and also what I meant by my question in post#7.

 

These were meant for practice.

 

So in words

 

Since PQ is parallel to YZ

Angle XPQ is equal to angle XYZ and angle XQP is equal to angle XZY

 

Thus triangle XYZ is similar to triangle XPQ (Three angles the same)

 

Thus the area XYZ : area XPQ is in the ratio of the square of the sides ie XPQ / XYZ = (2/5)²

 

Thus XYZ = (25*24)/4 = 150 cm²

 

Area XYZ = Area XQP + Area PQZY

 

Thus Area PQZY = 150 - 24 = 126 cm²

 

If you can learn how to layout your working so the examiner can follow it you will get better marks.

You will also find it easier to check your own work and follow it at a later date.

[imatfaal]

Chikis

 

What sort of phone - I will have a dig and see if we can come up with a workaround that allows you to send diagrams from camera on phone to the board; that's presuming the camera on the phone works

 

Studiot above sounds disturbingly like my maths teacher when I was around 14-15 (as he was probably biggest positive influence at school that's a cmpliemtn of sorts) . But he is so correct - as was Mick Marmion - layout and an unbreakable line of logic that is impossible not to follow is crucial in maths; especially if you are still being examined.