Find the unknown in each of the following pairs of similar figures.

Attachment 24550

My solution:=

Area of DEF4^{2}

Area of ABC = 6

6x4cm^{2/}Area of ABC^{ =4/6x4/6}

Is this right? What is the correct answer?

Please help, thanks.

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- Aug 19th 2012, 06:00 PMwilliam24Similar figure problem
Find the unknown in each of the following pairs of similar figures.

Attachment 24550

My solution:=

Area of DEF__4__^{2}

Area of ABC = 6

6x4cm^{2/}Area of ABC^{ =4/6x4/6}

Is this right? What is the correct answer?

Please help, thanks. - Aug 19th 2012, 09:13 PMpickslidesRe: Similar figure problem
How is the area of DEF = 16 $\displaystyle cm^2$ ?

- Aug 20th 2012, 01:52 AMwilliam24Re: Similar figure problem
Triangle area: base x height/2

base=4 height=6

so, 4x6=24

But, I don't know that is it correct. - Aug 20th 2012, 02:36 AMcybertutorRe: Similar figure problem
Firstly, you need to consider the similar shape problem, before you start to think of the Triangle area.

Similar shapes:

Similar shapes have exactly the same angles, as a shape increases in size, the angles stay the same, but the side lengths increase in the same proportion.

the triangles are similar shapes in this question, so as triangle DEF has 2 lengths shown (4cm and 6cm), and triangle ABC has only one length shown, we can use this to help us find out more information. BC and EF in these triangles represent the same part of each shape, but BC is bigger than EF.

How much bigger though?

Well we need to find out the proportion.... BC/EF= the proportional increase....that's 6cm/4cm = 1.5 times bigger.

Therefore all sides in the ABC triangle will be 1.5 times bigger than the corresponding sides in the DEF triangle.

That means the the 6cm side in the DEF triangle x 1.5 = 9cm.....so this is the length of the vertical dotted line on the ABC triangle.

So now we have a little bit more information...which is actually key to finding the total solution to the triangle ABC's Area.

Your initial area of a triangle calculation wasn't a bad start, but you need to realise that 0.5 x Base x Height is for a RIGHT ANGLED triangle only. ABC is a scalene triangle.

To do this part, you need to consider 2 separate right angled triangles.

I have added another couple of points to the triangle:

"x" is the total length of the base of the triangle

"o" is the corner of the 2 dotted lines

Attachment 24556

Now we have 2 Right Angled triangles to consider:

ABO

ACO

To find ABC ( the scalene triangle), we can subtract the area of ACO from the area of ABO

Ok so here are the individual areas by using 1/2 x Base x Height:

Area(ABO) = 9 times X divided by 2 = 4.5X

Area(ACO) = 9 times (X-6) divided by 2 = 4.5X - 27

Area(ABO) - Area(ACO) = 4.5X - (4.5X - 27)

= 4.5X -4.5X + 27

=27 square cm

Hope this helps

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