If I were to do this problem,This was some advanced math showed to our class by our teacher and I would like to get it solved if possible.
So the task is to find the length of d and all the picture has all the information that was given to us. There's no hurry.
ThanksView attachment 27270
If I were to do this problem,
I would first determine -
AB^2 + BC^2 = AC^2
Yes, I suppose we should confirm that [MATH]\angle ABC[/MATH] is a right angle.Pythagoras theorem is applied here, and you are done.
as along as we are given the right angle of the length d, this can be solve in many waysYes, I suppose we should confirm that [MATH]\angle ABC[/MATH] is a right angle.
as along as we are given the right angle of the length d, this can be solve in many ways
it can be solved as Khan said, similar triangles
It can be solved by your way lex, area and height
and finally and the easiest one is just Pythagorean
lolYou could do it using Pythagoras' Theorem, but it doesn't appear the easiest method:
[MATH]x^2+d^2=165^2[/MATH][MATH](275-x)^2+d^2=220^2[/MATH]
In similar triangles, use your strength in geometry and solve one linear equationlol
By easiest I meant you don't need to think too much, just use your strength in algebra by simplifying and solving the two equations
Likewise with the areas.In similar triangles, use your strength in geometry and solve one linear equation
Likewise with the areas.
Add the label \(E\) as the foot of the altitude from \(B\) so \(BE=d\). Now \(d\) is a mean proportional between \(AE~\&~CE\)This was some advanced math showed to our class by our teacher and I would like to get it solved if possible.
So the task is to find the length of d and all the picture has all the information that was given to us. There's no hurry.
No - there's only one equation to be solved! For the area of the triangle ABC:No no .... with "areas" you have to solve two equations.... I win.....