Perpendiculars in an isosceles triangle

[Radit2125]

I do not know how to solve this problem:
In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Perpendiculars are drawn from it, which divide the triangle into 3 similar triangles. Find the lengths of the perpendiculars; the distance from Q to the sides

 

[lev888]

I do not know how to solve this problem:
In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Perpendiculars are drawn from it, which divide the triangle into 3 similar triangles. Find the lengths of the perpendiculars; the distance from Q to the sides

Please post a diagram.

 

[Radit2125]

Please post a diagram.

I don't know how a diagram would look like for this problem, but I suppose that it is something close to this:

 

[lev888]

I don't know how a diagram would look like for this problem, but I suppose that it is something close to this:

The problem is that the 3 shapes are not triangles, no matter where the point is. Can you post a photo of the problem?

 

[Radit2125]

The problem is that the 3 shapes are not triangles, no matter where the point is. Can you post a photo of the problem?

No, I don't have one. My teacher gave me this problem without a graph. Is there a way to solve it without a graph?

 

[lev888]

No, I don't have one. My teacher gave me this problem without a graph. Is there a way to solve it without a graph?

If we can't draw a valid diagram it means there is something wrong with the problem. You can't just ignore it and look for some other way to solve it.

 

[Radit2125]

Can't we try to connect point Q to every angle of the triangle? Wouldn't we then get 3 similar triangles? I found a similar graph, however with an
equilateral triangle:

 

[lev888]

Can't we try to connect point Q to every angle of the triangle? Wouldn't we then get 3 similar triangles? I found a similar graph, however with an
equilateral triangle:

You get triangles, but 6, not 3. And I don't see in the problem anything about connecting Q with the vertices.

 

[Radit2125]

I would ask my teacher to give me the graph and then I would ask for help

 

[Radit2125]

My teacher told me that in the problem she missed adding to connect Q with the vertices. Saying that I found out that maybe we would become triangles with similar areas: triangle ABQ, triangle ACQ, triangle BCQ. Could You tell me what to do after this?

 

[lev888]

My teacher told me that in the problem she missed adding to connect Q with the vertices. Saying that I found out that maybe we would become triangles with similar areas: triangle ABQ, triangle ACQ, triangle BCQ. Could You tell me what to do after this?

In math it's very important to use precise terms. As you can see, perpendiculars are completely different from segments connecting Q with vertices. What's also different is "similar triangles" and "triangles with similar areas". And "similar areas" is not something we can use in calculations. Maybe it should be "equal areas"?
So, is this the correct problem statement?
"In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Q is connected with the 3 vertices, which divides the triangle into 3 triangles of equal areas. Find the lengths of the perpendiculars from Q to the sides of the triangle"

Hint: find the area of the original triangle. Since 3 smaller triangles have equal areas you can now find their areas. Next, use the triangle area formula that involves base and height to set up an equation and solve it for height for each of the small triangles.

 

[Radit2125]

In math it's very important to use precise terms. As you can see, perpendiculars are completely different from segments connecting Q with vertices. What's also different is "similar triangles" and "triangles with similar areas". And "similar areas" is not something we can use in calculations. Maybe it should be "equal areas"?
So, is this the correct problem statement?
"In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Q is connected with the 3 vertices, which divides the triangle into 3 triangles of equal areas. Find the lengths of the perpendiculars from Q to the sides of the triangle"

Hint: find the area of the original triangle. Since 3 smaller triangles have equal areas you can now find their areas. Next, use the triangle area formula that involves base and height to set up an equation and solve it for height for each of the small triangles.

I’m sorry that I made a mistake when writing the problem, but I’m still not used to asking for help for my math problems in English.? Thank You for the hint that You gave me. I would try to see how far I could go with the solution and I would text whether I get something.

 

[Steven G]

I would ask my teacher to give me the graph and then I would ask for help

There is no valid graph for your teacher to give. Just because a teacher gives a problem does not always mean that the problem is valid. Just consider every point inside the triangle, draw lines perpendicular to the sides of the triangles and if you do not get 3 triangles out of it then the problem is not valid.

 

[Dr.Peterson]

My teacher told me that in the problem she missed adding to connect Q with the vertices. Saying that I found out that maybe we would become triangles with similar areas: triangle ABQ, triangle ACQ, triangle BCQ. Could You tell me what to do after this?

As you've been told, "similar areas" is meaningless. And the problem now is entirely different from what you said at first, with nothing about perpendiculars; you have not just added something to it.

Please quote the entire problem exactly as given to you. If it's not in English, give the original and a translation, as accurately as you can.

 

[Radit2125]

Hi again,
could You tell me how to continue(my try to solve the problem):
Let's look at triangle ABC:
Its area= 1/2* a *h;
To find this area we need to build the height of the triangle ABC;
Now when it is built, we can look at triangle ACQ or QCB:
Because the height CQ divides AB into 2 equal parts this means that AQ= BQ= 4cm,
We can find the length of the height using the Pythagorean theorem:
AQ^2+CQ^2= AC^2
16+ +CQ^2= 25 => CQ= 3 cm
Because QM is perpendicular to AC, KL is perpendicular to AB and KN is perpendicular to BC=> QM, QL and QN are heights in the 3 triangles with similar areas.
The area of triangle ABC= (AB*QL)/2+ (BC*QN)/2+(QM*AC)/2
S△ABC=S△AQB+S△BQC+S△AQC
BC = AC = 5 cm and AB=8 cm; QM= QN=x and QL=y
Now the area is:
S△ABC=(AC+BC)*x/2+ (AB*y)/2
The area of △ABC= 1/2* a *h= 1/2* 8*3= 12cm^2
12= 5x+ 4y

From now on, I don't know how to proceed.

 

[Dr.Peterson]

The original statement of the problem was:

In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Perpendiculars are drawn from it, which divide the triangle into 3 similar triangles. Find the lengths of the perpendiculars; the distance from Q to the sides

You later changed it:

My teacher told me that in the problem she missed adding to connect Q with the vertices. Saying that I found out that maybe we would become triangles with similar areas: triangle ABQ, triangle ACQ, triangle BCQ. Could You tell me what to do after this?

But you haven't yet stated what the real problem is. I would have guessed something like this:

In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Segments are drawn from it to each of the vertices A, B, and C, which divide the triangle into 3 triangles with the same area. Find the lengths of the lines drawn; the distance from Q to the vertices.​

That's a huge change, so we really need to see if it is correct.

Let's look at triangle ABC:
Its area= 1/2* a *h;
To find this area we need to build the height of the triangle ABC;
Now when it is built, we can look at triangle ACQ or QCB:
Because the height CQ divides AB into 2 equal parts this means that AQ= BQ= 4cm,
We can find the length of the height using the Pythagorean theorem:
AQ^2+CQ^2= AC^2
16+ +CQ^2= 25 => CQ= 3 cm
Because QM is perpendicular to AC, KL is perpendicular to AB and KN is perpendicular to BC=> QM, QL and QN are heights in the 3 triangles with similar areas.
The area of triangle ABC= (AB*QL)/2+ (BC*QN)/2+(QM*AC)/2
S△ABC=S△AQB+S△BQC+S△AQC
BC = AC = 5 cm and AB=8 cm; QM= QN=x and QL=y
Now the area is:
S△ABC=(AC+BC)*x/2+ (AB*y)/2
The area of △ABC= 1/2* a *h= 1/2* 8*3= 12cm^2
12= 5x+ 4y

From now on, I don't know how to proceed.

But your work is entirely wrong for either version! Your Q is not inside the triangle, but K is. Your drawing doesn't show an isosceles triangle.

If Q is the foot of the perpendicular, then in order to say it divide AB into equal parts, you are assuming that AC = BC, which is good; and you have the right altitude, CQ = 3. Then you seem to suppose that QM has been drawn as a perpendicular, so you have somehow retained that idea even though your Q is in the wrong place.

Please tell us the actual problem. I'm not going any further without knowing what it is.

 

[Radit2125]

Here is the whole problem:
In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Segments are drawn from it not only to each of the vertices A, B, and C, which divide the triangle into 3 triangles with the same area. Also from the same point Q are drawn perpendiculars to the sides of the triangle. Find the lengths of the perpendiculars drawn; the distance from Q to the sides of the triangle.

 

[lev888]

Here is the whole problem:
In an isosceles triangle with sides 5 cm, 5 cm, 8 cm, a point Q inside the triangle is chosen. Segments are drawn from it not only to each of the vertices A, B, and C, which divide the triangle into 3 triangles with the same area. Also from the same point Q are drawn perpendiculars to the sides of the triangle. Find the lengths of the perpendiculars drawn; the distance from Q to the sides of the triangle.

Please post updated diagram and solution.

 

[Radit2125]

Here is my corrected solution and diagram to the problem:
Let's look at triangle ABC:
Its area= 1/2* a *h;
To find this area we need to build the height of the triangle ABC;
Now when it is built, we can look at triangle ACT or TCB:
Because the height CT divides AB into 2 equal parts this means that AT= BT= 4cm,
We can find the length of the height using the Pythagorean theorem:
AT^2+CT^2= AC^2
16+ +CT^2= 25 => CT= 3 cm
Because QE is perpendicular to AC, QG is perpendicular to AB and QF is perpendicular to BC=> QE, QG and QF are heights in the 3 triangles with similar areas(△ABQ; △BQC; △AQC)
The area of triangle ABC= (AB*QG)/2+ (BC*QF)/2+(QE*AC)/2
S△ABC=S△AQB+S△BQC+S△AQC
BC = AC = 5 cm and AB=8 cm;Here I'm not sure but I think that QE= QF=x and QG=y
Now the area is:
S△ABC=(AC+BC)*x/2+ (AB*y)/2
The area of △ABC= 1/2* a *h= 1/2* 8*3= 12cm^2
12= 5x+ 4y

 

[lev888]

Here is my corrected solution and diagram to the problem:
Let's look at triangle ABC:
Its area= 1/2* a *h;
To find this area we need to build the height of the triangle ABC;
Now when it is built, we can look at triangle ACT or TCB:
Because the height CT divides AB into 2 equal parts this means that AT= BT= 4cm,
We can find the length of the height using the Pythagorean theorem:
AT^2+CT^2= AC^2
16+ +CT^2= 25 => CT= 3 cm
Because QE is perpendicular to AC, QG is perpendicular to AB and QF is perpendicular to BC=> QE, QG and QF are heights in the 3 triangles with similar areas(△ABQ; △BQC; △AQC)
The area of triangle ABC= (AB*QG)/2+ (BC*QF)/2+(QE*AC)/2
S△ABC=S△AQB+S△BQC+S△AQC
BC = AC = 5 cm and AB=8 cm;Here I'm not sure but I think that QE= QF=x and QG=y
Now the area is:
S△ABC=(AC+BC)*x/2+ (AB*y)/2
The area of △ABC= 1/2* a *h= 1/2* 8*3= 12cm^2
12= 5x+ 4y

Do you understand the approach I suggested in post 11?