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- #1

- Jan 26, 2012

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a) 24

b) 20

c) 5

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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,055

a) 24

b) 20

c) 5

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

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1) Sudharaka

2) soroban

3) Reckoner

Solution (from soroban):

Which of the following could be the area of the triangle?

. . (a) 24 . . (b) 20 . . (c) 5

Code:

```
*
5 * .
* .
* θ .
* * * * * *
8
```

[tex]\text{If }\theta = 90^o\text{, the triangle has a maximum area of 20.}[/tex]

[tex]\text{Therefore: }(b)\,20\text{ and }(c)\,5\text{ are possible areas.}[/tex]

Two alternate solutions (from Reckoner):

Consider the side of length 8 to be the base of the triangle. Since the side of length 5 will be adjacent to this, the height of the triangle must necessarily be less than or equal to 5. So, using the area formula \(A = bh/2\) with \(b = 8\), we have \[\begin{eqnarray} 0 < h \leq 5 &\Rightarrow& 0 < \frac12b\cdot h \leq \frac12b\cdot5\\ &\Rightarrow& 0 < A \leq \frac12\cdot8\cdot5 \\ &\Rightarrow& 0 < A \leq 20.\end{eqnarray}\] Therefore, the triangle must have an area greater than 0 but less than or equal to 20. Thus 24 is not a possible area.

We shall call the length of the unknown side \(c\) and use Heron's formula to find the area of the triangle.

The semiperimeter of the triangle is \[s = \frac{5 + 8 + c}2 = \frac{13+c}2,\] so the area of the triangle is \[\begin{eqnarray}A &=& \sqrt{s(s - 5)(s - 8)(s - c)}\\&=& \sqrt{\left(\frac{13+c}2\right)\left(\frac{13+c}2 - 5\right)\left(\frac{13+c}2 - 8\right)\left(\frac{13+c}2 - c\right)}\\&=& \sqrt{\left(\frac{c + 13}2\right)\left(\frac{c + 3}2\right)\left(\frac{c - 3}2\right)\left(\frac{13 - c}2\right)}\\ &=&\frac14\sqrt{-c^4 + 178c^2 - 1521}.\end{eqnarray}\]

This expression is defined for \(c\in[3, 13]\). We do not need to consider negative values of \(c\) since \(c\) represents a length.

We then have \[\frac{dA}{dc} = \frac{-c^3 + 89c}{2\sqrt{-c^4 + 178c^2 - 1521}}\] so that, for \(c\in[3, 13]\), \[\begin{eqnarray}\frac{dA}{dc} = 0 &\Rightarrow& -c^3 + 89c = 0\\&\Rightarrow& c=\sqrt{89}.\end{eqnarray}\]

Testing the critical value \(c = \sqrt{89}\) with the endpoints of the interval \(c = 3\) and \(c = 13\), we see that \(c = \sqrt{89}\) produces a maximum area of 20, and that \(c = 3\) and \(c = 13\) produce a minimum area of 0 (these last two lengths would not form a valid triangle, but by taking values of \(c\) close to these we can create a triangle with area as close to zero as we wish).

Therefore, the area of the triangle could be any value \(A \in (0, 20]\). We cannot have \(A = 24\), as this exceeds the maximum area.

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