Triangle question
- Thread starter Lisa1
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If a triangle has side lengths \(a,b,c\) then the area \(A\) of the triangle is:
[MATH]A=\sqrt{s(s-a)(s-b)(s-c)}[/MATH]
Where \(s\) is the semi-perimeter:
[MATH]s=\frac{a+b+c}{2}[/MATH]
I would first compute the semi-perimeter, and then apply and simplify the result.
[MATH]A=\sqrt{s(s-a)(s-b)(s-c)}[/MATH]
Where \(s\) is the semi-perimeter:
[MATH]s=\frac{a+b+c}{2}[/MATH]
I would first compute the semi-perimeter, and then apply and simplify the result.
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Okay, what do you get for \(s\)?
Dr.Peterson
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Since the question as stated doesn't really make sense (as stated in posts #2 and #8), you have probably copied it incorrectly. (It is also grammatically incorrect.)
Please show us the problem exactly as given to you.
Dr.Peterson
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No, 4.5 is just s, not the area. But once that is calculated, it will be the answer to the problem as stated.
Dr.Peterson
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Did you forget to attach the problem?
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If I were going to try to find the area of the given triangle without using Heron's formula, I would use some coordinate geometry instead.
I would consider a circle of radius 3 centered at the origin, and a circle of radius 2 centered at (4,0). The quadrant 1 intersection of these circles would be the third vertex of the triangle, in addition to the centers of the two circles:
And so we know the base of the triangle is 4 units, and we need to find where the circles intersect in quadrant I, specifically the \(y\)-coordinate to get the triangle's altitude:
[MATH]x^2+y^2=3^2[/MATH]
[MATH](x-4)^2+y^2=2^2[/MATH]
Subtracting the latter from the former, we obtain:
[MATH]x^2-(x-4)^2=5[/MATH]
[MATH](x+x-4)(x-x+4)=5[/MATH]
[MATH]8(x-2)=5[/MATH]
[MATH]x-2=\frac{5}{8}[/MATH]
[MATH]x=\frac{21}{8}[/MATH]
[MATH]y=\sqrt{3^2-\left(\frac{21}{8}\right)^2}=\frac{3\sqrt{15}}{8}[/MATH]
And so the area of the triangle is:
[MATH]A=\frac{1}{2}(4)\left(\frac{3\sqrt{15}}{8}\right)=\frac{3\sqrt{15}}{4}[/MATH]
And this result agrees with that given by Heron's formula.
I would consider a circle of radius 3 centered at the origin, and a circle of radius 2 centered at (4,0). The quadrant 1 intersection of these circles would be the third vertex of the triangle, in addition to the centers of the two circles:
And so we know the base of the triangle is 4 units, and we need to find where the circles intersect in quadrant I, specifically the \(y\)-coordinate to get the triangle's altitude:
[MATH]x^2+y^2=3^2[/MATH]
[MATH](x-4)^2+y^2=2^2[/MATH]
Subtracting the latter from the former, we obtain:
[MATH]x^2-(x-4)^2=5[/MATH]
[MATH](x+x-4)(x-x+4)=5[/MATH]
[MATH]8(x-2)=5[/MATH]
[MATH]x-2=\frac{5}{8}[/MATH]
[MATH]x=\frac{21}{8}[/MATH]
[MATH]y=\sqrt{3^2-\left(\frac{21}{8}\right)^2}=\frac{3\sqrt{15}}{8}[/MATH]
And so the area of the triangle is:
[MATH]A=\frac{1}{2}(4)\left(\frac{3\sqrt{15}}{8}\right)=\frac{3\sqrt{15}}{4}[/MATH]
And this result agrees with that given by Heron's formula.
Dr.Peterson
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Thanks. Do you realize that this is an entirely different problem than you posted?
To solve this, you might start by thinking about what happens if you make two sides 2 and 3 (for example) and allow the third side to vary in length. What would be the largest area you could get? It will not necessarily be when the third side is as long as possible, so there may be a triangle fitting these conditions that is larger than the 2-3-4 triangle we've been discussing.
Once you've given that some thought, you'll have some other things to try before making a conclusion, and then proving it.
Let us know your thoughts.
A triangle has sides of length at most 2, 3, and 4 respectively. Determine, with proof, the maximum possible area of the triangle.
To solve this, you might start by thinking about what happens if you make two sides 2 and 3 (for example) and allow the third side to vary in length. What would be the largest area you could get? It will not necessarily be when the third side is as long as possible, so there may be a triangle fitting these conditions that is larger than the 2-3-4 triangle we've been discussing.
Once you've given that some thought, you'll have some other things to try before making a conclusion, and then proving it.
Let us know your thoughts.
Dr.Peterson
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Again, please show your work or thoughts if you want more help. My suggestion should take you far.
Steven G
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Yes, it was posted that s=4.5
Using the Doctor’s approach
“ make two sides 2 and 3 and allow the third side to vary in length “
you could let third side = x.
Do you think you could use Heron’s formula to first write semiperimeter(s) in terms of x and then following substitution express A² in terms of x?
After simplifying you should end up with
A² = B( )( )where B represents a fraction and each set of brackets holds a quadratic expression which is a difference of two squares.
The next part you have to think about carefully. If one set of brackets represents a length and the second set of brackets represents a width what shape would give the maximum value for A² and therefore maximum area?
How do the two quadratic expressions relate to each other?
Once you know the answer to this you’re almost there. See if you can finish off.
“ make two sides 2 and 3 and allow the third side to vary in length “
you could let third side = x.
Do you think you could use Heron’s formula to first write semiperimeter(s) in terms of x and then following substitution express A² in terms of x?
After simplifying you should end up with
A² = B( )( )where B represents a fraction and each set of brackets holds a quadratic expression which is a difference of two squares.
The next part you have to think about carefully. If one set of brackets represents a length and the second set of brackets represents a width what shape would give the maximum value for A² and therefore maximum area?
How do the two quadratic expressions relate to each other?
Once you know the answer to this you’re almost there. See if you can finish off.