Is this question doable? Finding triangle side length

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Homework Help Overview

The discussion revolves around a triangle ABC with a given area of 5.5 cm² and one side length, AB = 2√3 + 1 cm. Participants are exploring whether it is possible to determine the length of side AC with the provided information, considering the triangle does not have any special properties.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of having only one side length and the area, questioning if a unique solution exists. They consider various approaches, including using the area formula and trigonometric relationships, while expressing concerns about missing information or constraints.

Discussion Status

The conversation is ongoing, with participants sharing different interpretations and approaches. Some suggest drawing diagrams to visualize the problem, while others express frustration over the complexity of the equations derived from their attempts. There is no explicit consensus on a method, but several lines of reasoning are being explored.

Contextual Notes

Participants note that the triangle is a generic normal triangle without special properties, which raises questions about the uniqueness of the solution. There is also mention of potential complexities arising from the calculations, such as encountering complex numbers.

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Given a triangle ABC, whose area is 5.5 cm square, and length of AB = 2√3 + 1 cm, find the value of length AC.

Given triangle has no special properties like isoceles etc.

Only 1 length and area of the triangle are given. Is it possible to solve such a question? thanks.
 
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Welcome!
Are there any triangles that have two equal sides?
 
Consider the side AB to be the base of the triangle, then you can calculate the height of the triangle from the area. The question then is whether there are triangles with that height but with different length non-base sides. What would you say?
PS. If this is a homework problem, you need to use the homework format.
 
What do you mean by "doable"? A unique solution, perhaps?
 
Lnewqban said:
Welcome!
Are there any triangles that have two equal sides?
No there isn't, it is just a normal triangle.
 
shirozack said:
No there isn't, it is just a normal triangle.
Do you know the equation for the area of any triangle?
 
Lnewqban said:
Do you know the equation for the area of any triangle?
yea, but it feels like something is missing? given only 1 length and the area, is it possible to find the other length without any special properties of the triangle like isoceles, equilateral etc?

i tried 1/2 base height, 1/2 ab sin(c), sin rule, cos rule
 
FactChecker said:
Consider the side AB to be the base of the triangle, then you can calculate the height of the triangle from the area. The question then is whether there are triangles with that height but with different length non-base sides. What would you say?
PS. If this is a homework problem, you need to use the homework format.
it feels like something is missing? given only 1 length and the area, is it possible to find the other length without any special properties of the triangle like isoceles, equilateral etc?

i tried 1/2 base height, 1/2 ab sin(c), sin rule, cos rule

it is just a generic normal triangle.
 
DaveE said:
What do you mean by "doable"? A unique solution, perhaps?
I mean is it solvable?

it feels like something is missing? given only 1 length and the area, is it possible to find the other length without any special properties of the triangle like isoceles, equilateral etc?

i tried 1/2 base height, 1/2 ab sin(c), sin rule, cos rule

it is just a generic normal triangle.
 
  • #10
shirozack said:
I mean is it solvable?
Yes, there are solutions.
 
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  • #11
DaveE said:
Yes, there are solutions.
How should i approach?

I tried to replace sin(a) in 1/2absinc with sin rule, trigo, area half base and they all just cancel each other out.

Area formula 1/2 ab sin(c) => 5.5 = ½ (2√3 + 1)(AC) sin(a)

Area formula half base height => ½ (2√3 + 1) h = 5.5

h = 11 / (2√3 + 1)

sin(a) = h / AC

Sin rule => h/sin(a) = AC/sin(90)

h/AC = sin(a)
 
  • #12
shirozack said:
yea, but it feels like something is missing? given only 1 length and the area, is it possible to find the other length without any special properties of the triangle like isoceles, equilateral etc?

i tried 1/2 base height, 1/2 ab sin(c), sin rule, cos rule
Did the problem give you a specific type of triangle?
 
  • #13
Lnewqban said:
Did the problem give you a specific type of triangle?
no it is just a generic normal triangle. just draw any generic triangle free hand. it looks like that.
 
  • #14
shirozack said:
Area formula half base height => ½ (2√3 + 1) h = 5.5

h = 11 / (2√3 + 1)
OK, so now you know the length AB and the height of point C above the line through A, B. Try drawing some pictures of different triangles (different locations of point C). Which, if any, of those locations of point C meet your constraints?
 
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  • #15
shirozack said:
no it is just a generic normal triangle. just draw any generic triangle free hand. it looks like that.
Therefore, are you free to make it an isosceles triangle, or a right-angle one?
If not, you know that the three angles must add up to 180 degress.

angles-classifiesd-by-angles-and-by-side-1080x1080.jpg
 
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  • #16
shirozack said:
it feels like something is missing? given only 1 length and the area, is it possible to find the other length without any special properties of the triangle like isoceles, equilateral etc?

i tried 1/2 base height, 1/2 ab sin(c), sin rule, cos rule

it is just a generic normal triangle.
Without doing any calculations, draw a triangle with a horizontal base. Now draw a horizontal height line above the triangle. How many triangles can you draw with the top vertex on the height line? Is their area different? Is the length of the sides up to the height line different?

EDIT: Or just look at the triangles in @Lnewqban's post above. The base lengths and heights are identical, so what can you say about the areas of the triangles? What about the lengths of the sides going up to the point?
 
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  • #17
FactChecker said:
Without doing any calculations, draw a triangle with a horizontal base. Now draw a horizontal height line above the triangle. How many triangles can you draw with the top vertex on the height line? Is their area different? Is the length of the sides up to the height line different?

EDIT: Or just look at the triangles in @Lnewqban's post above. The base lengths and heights are identical, so what can you say about the areas of the triangles? What about the lengths of the sides going up to the point?
ok let's assume the triangle looks like this and i simplified the values. Given AB = 4 and we found the height through area formula to be 3. I am trying to get length of DB i.e. (L) through similar triangles ratio and then solve for x through pythagoras theorem AD,CD,x. Problem is, when I tried to solve for L, the quadratic equation, it starts to hit complex numbers, which can't be right? where am i wrong? are my values not set well or is my approach wrong?

1660885629184.png
 

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  • #18
Lnewqban said:
Therefore, are you free to make it an isosceles triangle, or a right-angle one?
If not, you know that the three angles must add up to 180 degress.

View attachment 312959
ok let's assume the triangle looks like this and i simplified the values. Given AB = 4 and we found the height through area formula to be 3. I am trying to get length of DB i.e. (L) through similar triangles ratio and then solve for x through pythagoras theorem AD,CD,x. Problem is, when I tried to solve for L, the quadratic equation, it starts to hit complex numbers, which can't be right? where am i wrong? are my values not set well or is my approach wrong?
1660886247968.png
 
  • #19
shirozack said:
ok let's assume the triangle looks like this and i simplified the values. Given AB = 4 and we found the height through area formula to be 3. I am trying to get length of DB i.e. (L) through similar triangles ratio and then solve for x through pythagoras theorem AD,CD,x. Problem is, when I tried to solve for L, the quadratic equation, it starts to hit complex numbers, which can't be right? where am i wrong? are my values not set well or is my approach wrong?
View attachment 312973
Those aren't similar triangles unless the angle at vertex C is a right angle.

You need to involve ##x## in your calculations. Otherwise, ##L## can take any value less than ##4##.
 
  • #20
PeroK said:
Those aren't similar triangles unless the angle at vertex C is a right angle.

You need to involve ##x## in your calculations. Otherwise, ##L## can take any value less than ##4##.
How should i approach it?

I have tried area formulas 1/2baseheight, 1/2absinc, sin law, cos law, and they all seem to just cancel each other out when i try to substitute them into solve the equations.
 
  • #21
Heron's formula?
 
  • #22
shirozack said:
How should i approach it?

I have tried area formulas 1/2baseheight, 1/2absinc, sin law, cos law, and they all seem to just cancel each other out when i try to substitute them into solve the equations.
##L## can be anything less than ##4##. But, if you specify ##x## then you can use Pythagoras:
$$x^2 = 3^2 + (4-L)^2$$
 
  • #23
PeroK said:
##L## can be anything less than ##4##. But, if you specify ##x## then you can use Pythagoras:
$$x^2 = 3^2 + (4-L)^2$$
yea i thought about pythagoras too, but how do i find L? the question only states the length of AB and the area. then it asks to find length of AC which is x in this case. i need to get rid of L somehow.
 
  • #24
shirozack said:
yea i thought about pythagoras too, but how do i find L? the question only states the length of AB and the area. then it asks to find length of AC which is x in this case. i need to get rid of L somehow.
You can't find ##L##. ##L## can be anything less than ##4##. There is more than one triangle with a base of ##4## and a height of ##3##.
 
  • #25
PeroK said:
You can't find ##L##. ##L## can be anything less than ##4##. There is more than one triangle with a base of ##4## and a height of ##3##.
does it then mean that this question is not doable?
the original question is they gave the length AB as 2sqrt(3) + 1
and the area as 5.5 cm square, and was asked to find AC, expressing it in some integer b sqrt(3).
So since L can be anything , i assume this question cannot be done?
 
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  • #26
shirozack said:
So since L can be anything , i assume this question cannot be done?
There is no unique solution for the length AC. The area and a single side length does fully specify a triangle. There must be another piece of data missing.
 
  • #27
PeroK said:
There is no unique solution for the length AC. The area and a single side length does fully specify a triangle. There must be another piece of data missing.
i see thanks everyone
 
  • #28
$$(a+b)(a-b)=a^2-b^2$$
 

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