# Triangle A has an area of 18  and two sides of lengths 8  and 12 . Triangle B is similar to triangle A and has a side with a length of 9 . What are the maximum and minimum possible areas of triangle B?

##### 2 Answers
Oct 10, 2017

Maximum area of $\Delta$ B 729/32 & Minimum area of $\Delta$ B 81/8

#### Explanation:

If sides are 9:12, areas will be in their square.
Area of B $= {\left(\frac{9}{12}\right)}^{2} \cdot 18 = \frac{81 \cdot 18}{144} =$ 81/8

If the sides are 9:8,
Area of B $= {\left(\frac{9}{8}\right)}^{2} \cdot 18 = \frac{81 \cdot 18}{64} =$ 729/32

Aliter :
For similar triangles, ratio of corresponding sides are equal.

Area of triangle A =18 and one base is 12.
Hence height of $\Delta$ A $= \frac{18}{\left(\frac{1}{2}\right) 12} = 3$
If $\Delta$ B side value 9 corresponds to $\Delta$ A side 12, then the height of $\Delta$ B will be $= \left(\frac{9}{12}\right) \cdot 3 = \frac{9}{4}$

Area of $\Delta$ B $= \frac{9 \cdot 9}{2 \cdot 4} =$ 81/8

Area of $\Delta$ A = 18 and base is 8.
Hence height of $\Delta$ A $= \frac{18}{\left(\frac{1}{2}\right) \left(8\right)} = \frac{9}{2}$
I$\Delta$ B side value 9 corresponds to $\Delta$ A side 8, then
the height of $\Delta$ B $= \left(\frac{9}{8}\right) \cdot \left(\frac{9}{2}\right) = \frac{81}{16}$

Area of $\Delta$ B $= \left(\frac{9 \cdot 81}{2 \cdot 16}\right) =$729/32

$\therefore$ Maximum area 729/32 & Minimum area 81/8

Oct 13, 2017

Minimum possible area 81/8
Maximum possible area 729/32

#### Explanation:

Alternate Method :

Sides ratio 9/12=3/4.Areas ratio will be ${\left(\frac{3}{4}\right)}^{2}$
$\therefore$ Min. possible area $= 18 \cdot \left({3}^{2} / {4}^{2}\right) = 18 \cdot \left(\frac{9}{16}\right) = \frac{81}{8}$

Sides ratio = 9/8.
$\therefore$ Max. possible area $= 18 \cdot \left({9}^{2} / {8}^{2}\right) = \frac{729}{32}$