# Problem of the week #26 - September 24th, 2012

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#### Jameson

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One side of a triangle has length 8 and a second side has length 5. Which of the following could be the area of the triangle?

a) 24
b) 20
c) 5
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#### Jameson

##### Administrator
Staff member
Congratulations to the following members for their correct solutions:

1) Sudharaka
2) soroban
3) Reckoner

Solution (from soroban):

One side of a triangle has length 8 and a second side has length 5.
Which of the following could be the area of the triangle?

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

Code:
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$$\text{If }\theta = 0^o\text{, the triangle has a minimum area of 0.}$$

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

$$\text{Therefore: }(b)\,20\text{ and }(c)\,5\text{ are possible areas.}$$

Two alternate solutions (from Reckoner):

Solution 1:

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.

Solution 2, with calculus:

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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