# Similar Triangle

• January 20th 2009, 06:13 PM
Serialkisser
Similar Triangle
Given two similar triangles http://tdsb.elearningontario.ca/cont...s/triangle.gifABChttp://tdsb.elearningontario.ca/cont.../wigleqils.gifhttp://tdsb.elearningontario.ca/cont...s/triangle.gif and DEF, determine the measurement of the unknown sides.
AB = 8cm
DF = 10cm
http://tdsb.elearningontario.ca/cont...ages/angle.gifA = http://tdsb.elearningontario.ca/cont...ages/angle.gifD = 60°
AC = 12cm
EF = 9cm

How should I draw a triangle of ABC and DEF from these values ? If AB and DF, how can I get ABC ? And what unknown sides ? Im confused :S

Another question I have:

2 sides of one triangle are in proportion to 2 sides of a second triangle. One angle - not the contained angle - is equal to another angle - not the contained angle - in the second triangle.

Is this a similar triangle ? If not why.
• January 20th 2009, 06:35 PM
mollymcf2009
There are several ways to deal with this depending on what subject this is for. Is this for a trig class?
• January 20th 2009, 07:03 PM
Serialkisser
Grade 10 Math, the whole unit is named "similar triangles" the next unit will be trigonometry.
• January 20th 2009, 08:17 PM
pyrosilver
Have you covered the law of cosines yet?
• January 20th 2009, 08:36 PM
Serialkisser
Not in this Unit. Unknown values were solved by doing something with the corresponding sides. We also covered solving proportions.
• January 20th 2009, 10:14 PM
mollymcf2009
Given two similar triangles http://tdsb.elearningontario.ca/cont...s/triangle.gifABChttp://tdsb.elearningontario.ca/cont.../wigleqils.gifhttp://tdsb.elearningontario.ca/cont...s/triangle.gif and DEF, determine the measurement of the unknown sides.
AB = 8cm
DF = 10cm
http://tdsb.elearningontario.ca/cont...ages/angle.gifA = http://tdsb.elearningontario.ca/cont...ages/angle.gifD = 60°
AC = 12cm
EF = 9cm

You are drawing 2 separate triangles. Triangle ABC & Triangle DEF

I assume you have used the pathagorean theorum? a^2 + b^2 = c^2

Draw two triangles that are similar. Similar triangles have the exact same angle measurements on all three angles, but they can have different lengths of sides.
On your triangles, mark the three corners ABC on one and DEF on the other, make sure that you A on the first is in the same place as D on the other, since their angles are both 60 degrees. Then mark your other corners and write the measurements of the sides that are given. Once you have your triangles labeled, use the pathagorean theorum to solve for your unknown sides. Remember that c^2 is always your hypotenuse!! Hope that helps! Good luck!
• January 23rd 2009, 07:32 PM
Serialkisser
I have marked the triangles,and I also have A and D on the same place in 2 similar triangles with the same degree. On Triangle 1 I have:
A-B=8
A-C=12
B-C= ?

using a²+b²=c2
I get 64+144=208 --> sqr(208)= 14.4
So, is 14.4 correkt for B-C ?

I did the same thing on Triangle # 2.
• January 23rd 2009, 10:37 PM
mollymcf2009
I have marked the triangles,and I also have A and D on the same place in 2 similar triangles with the same degree. On Triangle 1 I have:
A-B=8
A-C=12
B-C= ?

using a²+b²=c2
I get 64+144=208 --> sqr(208)= 14.4
So, is 14.4 correkt for B-C ?

Right idea, but you need to subtract! 144 - 64 = 80
Attached a picture so you can see what all this looks like. It is kinda hard to explain without a picture!

$\sqrt{80}$ * I would just leave it as a square root. The answer as a decimal is really messy and long.

a^2 and b^2 are the two legs of the triangle and c^ is your hypotenuse. You are trying to find the measurement of the other leg of the triangle.
• January 23rd 2009, 11:32 PM
Similar Triangles
Hello
Quote:

Originally Posted by Serialkisser
Given two similar triangles http://tdsb.elearningontario.ca/cont...s/triangle.gifABChttp://tdsb.elearningontario.ca/cont.../wigleqils.gifhttp://tdsb.elearningontario.ca/cont...s/triangle.gif and DEF, determine the measurement of the unknown sides.
AB = 8cm
DF = 10cm
http://tdsb.elearningontario.ca/cont...ages/angle.gifA = http://tdsb.elearningontario.ca/cont...ages/angle.gifD = 60°
AC = 12cm
EF = 9cm

How should I draw a triangle of ABC and DEF from these values ? If AB and DF, how can I get ABC ? And what unknown sides ? Im confused :S

I'm sorry, but this question can't be correct. The information given is contradictory.

See the attached diagram.

Bear in mind that you can only use Pythagoras' Theorem when you know that the triangle is right-angled - and you don't know this here.

You could now find the length of the side BC using the Cosine Rule: $a^2 = b^2 + c^2 -2bc\cos A$. Then find another angle using the Sine Rule, and the third angle using angle sum = 180 degrees.

Looking at the triangle DEF, you can find the angle E using the Sine Rule, and hence the other angle and sides.

But, given those measurements, the triangles are not then similar - the angles of one triangle have to be the same as the angles of the other for them to be similar, and the sides in the same ratio.

Quote:

Originally Posted by Serialkisser
Another question I have:

2 sides of one triangle are in proportion to 2 sides of a second triangle. One angle - not the contained angle - is equal to another angle - not the contained angle - in the second triangle.

Is this a similar triangle ? If not why.

No, the equal angles must be opposite to the corresponding sides, and you cannot be certain that this is so.

For example, if the triangles are ABC and DEF, and we know that the side BC corresponds to side DF, and AC corresponds to DE - in other words the ratios are equal, BC : DF = AC : DE - then the angles opposite these pairs of corresponding sides must be equal: A = E and B = F. And you cannot be sure that this is so from the information given.