# I need help with similar triangles.

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Posted (edited)

Hello. This is not even a problem but an example my book gives, but I want to understand it in order to do the exercises. It is as follows:

A light L is being raised up a pole. The light shines on the object Q, casting a shadow on the ground. At a certain moment the light is 40 meters off the ground, rising at 5 meters per minute. How fast is the shadow shrinking at that instant?

Then the book gives the answer that I don't understand. It reads: Let the height be y at time t and the lenght of the shadow be x. By similiar triangles, x/10=(x+20)/y; i.e., xy=10(x+20).The explanation goes on but I only want to know about the similar triangles part. I don't know where the equation and the 10 and 20 come from. Just that, I have no problem with the rest. Thank you in advance.

Edited by KFS

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You know why they are similar yes?  The angles are all the same....  this means that the ratio of the sides will be the same....

So- the triangle with base x has the same ratio of x/10  as the larger triangle of base 20+x....   the ration of it's sides being (20+X) / y.

Therefore we can say x/10 = (20+X)/y  and solve from there.  Any clearer?

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Posted (edited)

Thanks for the answer DrP. But please explain it more clear. I know nothing about similar triangles. Thank you.

Edited by KFS

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Posted (edited)

OK  -  If all the angles in a triangle are the same as the angles in a different triangle with longer or shorter sides...  the RATIO of the lengths of the sides will be the same.

Look at these 2 triangles:-

Triangle ABC is 'similar' to triangle ADE yes?   Angle BCA is the same as angle DEA right?   AB is parallel to AD and AC is parallel to AE.  The 2 triangles are said to be similar. We can make some statements...    Angel ABC is equivalent to ADE. Angle DAE is equivalent to BAC etc...  we can also say that the ratio of the lengths of the sides are the same...

So - Length AB divided by Length AC is the same as length AD divided by length AE.   AB/AC = AD/AE.   This goes for all the ratios of all the other side lengths as well, so,

In your diagram you are taking the length of the side of the small triangle 'x' and dividing it by the height 10m.  This is then equal to the same ratio of side in the larger triangle..  the length x+20 meters along the bottom divided by the height of the larger triangle 'y'.

You can then write an equation that shows that the ratio of the sides is equal,   x/10 = (x+20)/y.

In my triangle, comparing the same lengths as the ones in your example I can say that AC/BC = AE/DE

The rest is algebra/math and is about rearranging the equation.

Edited by DrP

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Posted (edited)

Ok, thank you very much for the explanation. But 10 and 20 were not in the problem' statement, so where do they come from?

Edited by KFS

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19 hours ago, KFS said:

Ok, thank you very much for the explanation. But 10 and 20 were not in the problem' statement, so where do they come from?

Yes they were  -  10m was the height of the shadow..... thus the height of the small triangle (from the diagram in the question - figure 2.4.3.)   20m is the distance from the shadow to the point on the ground perpendicular to the light....  so the length of the base of the larger triangle is this 20m plus the length of the shadow, x  =  (x+ 20)m

Look -

See?  It is in the diagram in the figure...   Light source at D (y distance off the ground), shadow goes from C to A.  Object Q at height 10m.

It can be thought of as 2 triangles that are similar. The big one from ADE and the smaller one from ABC.   The rest is just maths/algebra/calculus.

Hope that helps. Bye.