Finding area of a triangle using only Pythagorean theorem

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sel147

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So, the area of this triangle is to be calculated using only simple methods. I've calculated partial sides with the information I had, but found myself unable to proceed without the use of trigonometry. Any ideas on this one?
 
sel147 said:
View attachment 10954
So, the area of this triangle is to be calculated using only simple methods. I've calculated partial sides with the information I had, but found myself unable to proceed without the use of trigonometry. Any ideas on this one?
Click to expand...

1) I dare you to define "simple"?
2) What did you manage from the Pythagorean Theorem?
3) Did you notice that AC(extended)B is a Right Triangle? How about BC(extended)A
 
tkhunny said:
1) I dare you to define "simple"?
2) What did you manage from the Pythagorean Theorem?
3) Did you notice that AC(extended)B is a Right Triangle? How about BC(extended)A
Click to expand...

1) no use of trigonometry. Only Pythagorean theorem and perhaps transforming the triangle into parallelogram with matching sides?
2) Using info that hypothenuse and one side are given ( 15 and 5 cm) i found the third side, and also found that small extended piece using 15 and 8 as a data.
3) Yes and I've tried using that but no success
 
tkhunny said:
1) I dare you to define "simple"?
2) What did you manage from the Pythagorean Theorem?
3) Did you notice that AC(extended)B is a Right Triangle? How about BC(extended)A
Click to expand...

1) like first grade of highschool level, no trigonometry, only Pythagorean theoreme
2) I found the part of the side between the height of a triangle and B, and that short extended side on the right from C
3) yes, tried using that but no success
 
sel147 said:
View attachment 10954
So, the area of this triangle is to be calculated using only simple methods. I've calculated partial sides with the information I had, but found myself unable to proceed without the use of trigonometry. Any ideas on this one?
Click to expand...
Do you realize that the post does not say what is to be done?
I don't see what the Pythagorean theorem has to do with area.
Using only one half base times height the area of triangle ABC is \(\displaystyle 0.5(15)(6)=45cm^2\)
Now using the Pythagorean theorem, you can find the lengths of \(\displaystyle \overline{AB}~\&~\overline{AC}\)
 
sel147 said:
View attachment 10954
So, the area of this triangle is to be calculated using only simple methods. I've calculated partial sides with the information I had, but found myself unable to proceed without the use of trigonometry. Any ideas on this one?
Click to expand...
Hi, I am a curious type of person and I am just wondering which of the seven triangles you are trying to find the area of?
 
I think the OP is trying to find the area of the shaded, labeled triangle (ABC). :cool:
 
sel147 said:
1) no use of trigonometry. Only Pythagorean theorem and perhaps transforming the triangle into parallelogram with matching sides?
2) Using info that hypothenuse and one side are given ( 15 and 5 cm) i found the third side, and also found that small extended piece using 15 and 8 as a data.
3) Yes and I've tried using that but no success
Click to expand...

Perhaps you could show the work that you believe did not lead to success?
 
Denis said:
Seems to be an erroneous diagram:
you have (top right) 2 lines length 8 and 15 that
appear to be impossible...

EDIT: I'll be blunt, your diagram makes no sense!!
Click to expand...
Seriously, can you be more clear? Why not 15, 8 and sqrt(181)?
 
Jomo said:
Seriously, can you be more clear? Why not 15, 8 and sqrt(181)?
Click to expand...
Well, all I'll say is "try to draw a representative diagram"; impossible!
But might work if sqrt(181) switched with the 8; haven't tried that...
But if it does, then diagram is faulty...

If you're able to draw that mess to scale as presented,
you get "geometric immortality" :p
 
Last edited:
Jomo said:
Seriously, can you be more clear? Why not 15, 8 and sqrt(181)?
Click to expand...
First it is \(\displaystyle \sqrt{161}\approx 12.7\) not \(\displaystyle \sqrt{181}\) see here
Denis said:
Well, all I'll say is "try to draw a representative diagram"; impossible!
But might work if sqrt(181) switched with the 8; haven't tried that...But if it does, then diagram is faulty...If you're able to draw that mess to scale as presented.
Click to expand...
Yes the diagram is faulty. Have you ever seen a wedge with a dot for a right angle?
But allowing for its faults the problem makes some sense.
 
pka said:
Do you realize that the post does not say what is to be done?
I don't see what the Pythagorean theorem has to do with area.
Using only one half base times height the area of triangle ABC is \(\displaystyle 0.5(15)(6)=45cm^2\)
Now using the Pythagorean theorem, you can find the lengths of \(\displaystyle \overline{AB}~\&~\overline{AC}\)
Click to expand...

My idea was finding the length of the side AB and calculating the area with the simple (side*height)/2. But the height does not divide the triangle's base into two equal parts?
 
Denis said:
Seems to be an erroneous diagram:
you have (top right) 2 lines length 8 and 15 that
appear to be impossible...

EDIT: I'll be blunt, your diagram makes no sense!!
Click to expand...
I don't think there are two of the same length?:confused:
 
pka said:
First it is \(\displaystyle \sqrt{161}\approx 12.7\) not \(\displaystyle \sqrt{181}\) see here

Yes the diagram is faulty. Have you ever seen a wedge with a dot for a right angle?
But allowing for its faults the problem makes some sense.
Click to expand...
(· and Г· are both legit ways to mark a right angle in EU.
 
tkhunny said:
Perhaps you could show the work that you believe did not lead to success?
Click to expand...
I found the sqrt(161) side and 10*sqrt(2) side.
Then tried to do something about the fact that the extended ones are also right triangles...
 
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