# Find the length of d

#### Akuseli

##### New member
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.
Thanks

#### Subhotosh Khan

##### Super Moderator
Staff member
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

And

What are the "similar" triangles in the given figure ?

Last edited:

#### lex

##### Full Member
Find the area of the rectangle (length $$\displaystyle \times$$ breadth).
Half that, and you have got the area of the triangle ACB.

You know the area of this triangle is: half base $$\displaystyle \times$$ height
(base is 275, height d).

Hopefully you can now find d.

#### nasi112

##### Full Member
Pythagoras theorem is applied here, and you are done.

lex

#### lex

##### Full Member
If I were to do this problem,
I would first determine -
AB^2 + BC^2 = AC^2
Pythagoras theorem is applied here, and you are done.
Yes, I suppose we should confirm that $$\displaystyle \angle ABC$$ is a right angle.

#### nasi112

##### Full Member
Yes, I suppose we should confirm that $$\displaystyle \angle ABC$$ 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

#### lex

##### Full Member
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
You could do it using Pythagoras' Theorem, but it doesn't appear the easiest method:
$$\displaystyle x^2+d^2=165^2$$
$$\displaystyle (275-x)^2+d^2=220^2$$

#### nasi112

##### Full Member
You could do it using Pythagoras' Theorem, but it doesn't appear the easiest method:
$$\displaystyle x^2+d^2=165^2$$
$$\displaystyle (275-x)^2+d^2=220^2$$
lol
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

#### Subhotosh Khan

##### Super Moderator
Staff member
lol
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
In similar triangles, use your strength in geometry and solve one linear equation

#### lex

##### Full Member
In similar triangles, use your strength in geometry and solve one linear equation
Likewise with the areas.

#### Subhotosh Khan

##### Super Moderator
Staff member
Likewise with the areas.
No no .... with "areas" you have to solve two equations.... I win.....

#### pka

##### Elite Member
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.
Add the label $$E$$ as the foot of the altitude from $$B$$ so $$BE=d$$. Now $$d$$ is a mean proportional between $$AE~\&~CE$$

#### lex

##### Full Member
No no .... with "areas" you have to solve two equations.... I win.....
No - there's only one equation to be solved! For the area of the triangle ABC:
$$\displaystyle \tfrac{1}{2}275d=\tfrac{1}{2}220\times165$$

Yours wins in beauty, mine I think in conceptual simplicity and simplicity of execution.

My reflection on the solutions!!
Conceptually I think area comes before similar triangles which comes before Pythagoras' Theorem.
In ease of execution I think the triangle area, similar triangles (have to be identified and correctly used), then the Pythagoras' Theorem solution.
In mathematical beauty similar triangles, then area/Pythagoras.

(Again special praise to you too, for pointing out that for our solutions the converse of Pythagoras' Theorem is needed to ensure $$\displaystyle \angle ABC$$ is a right-angle)!