# Triangle A has an area of 9  and two sides of lengths 3  and 9 . Triangle B is similar to triangle A and has a side with a length of 7 . What are the maximum and minimum possible areas of triangle B?

Jul 19, 2016

Maximum possible Area of B: $10 \frac{8}{9}$ sq.units

Minimum possible Area of B: $0.7524$ sq.units (approximately)

#### Explanation:

If we use the side of A with length $9$ as the base
then the height of A relative to this base is $2$
(since the area of A is given as $9$ and $\text{Area"_triangle=1/2xx"base"xx"height}$)

Note that there are two possibilities for $\triangle A$: The longest "unknown" side of $\triangle A$ is obviously given by Case 2 where this length is the longest side possible.

In Case 2
$\textcolor{w h i t e}{\text{XXX}}$the length of the "extension" of the side with length $9$ is
$\textcolor{w h i t e}{\text{XXXXXX}} \sqrt{{3}^{2} - {2}^{2}} = \sqrt{5}$
$\textcolor{w h i t e}{\text{XXX}}$and the "extended length" of the base is
$\textcolor{w h i t e}{\text{XXXXXX}} 9 + \sqrt{5}$
$\textcolor{w h i t e}{\text{XXX}}$So the length of the "unknown" side is
$\textcolor{w h i t e}{\text{XXXXXX}} \sqrt{{2}^{2} + {\left(9 + \sqrt{5}\right)}^{2}}$
$\textcolor{w h i t e}{\text{XXXXXXXX}} = \sqrt{90 + 18 \sqrt{5}}$
$\textcolor{w h i t e}{\text{XXXXXXXX}} = 3 \sqrt{10 + 2 \sqrt{5}}$

The Area of a geometric figure varies as the square of its linear dimensions.

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The maximum area of $\triangle B$ will occur when $B$'s side of length $7$ corresponds to the shortest side of $\triangle A$ (namely $3$)

$\left(\text{Area of "triangleB)/("Area of } \triangle A\right) = {7}^{2} / {3}^{2}$
and since $\text{Area of } \triangle A = 2$
$\Rightarrow \text{Area of } \triangle B = \frac{{7}^{2}}{{3}^{2}} \times 2 = \frac{98}{9} = 10 \frac{8}{9}$

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The minimum area of $\triangle b$ will occur when $B$'s side of length $7$ corresponds to the longest possible side of $\triangle A$ (namely $3 \sqrt{10 + 2 \sqrt{5}}$ as shown above).

$\left(\text{Area of "triangleB)/("Area of } \triangle A\right) = {7}^{2} / \left({\left(3 \sqrt{10 + 2 \sqrt{5}}\right)}^{2}\right)$

and since $\text{Area of } \triangle A = 2$

$\Rightarrow \text{Area of } \triangle B = \frac{{7}^{2}}{{\left(3 \sqrt{10 + 2 \sqrt{5}}\right)}^{2}} \times 2 = \frac{98}{90 + 19 \sqrt{5}} \approx 0.7524$