# Triangle A has an area of 12  and two sides of lengths 8  and 7 . Triangle B is similar to triangle A and has a side of length 5 . What are the maximum and minimum possible areas of triangle B?

Jan 23, 2018

Case - Minimum Area :

$D 1 = \textcolor{red}{{D}_{\min}} = \textcolor{red}{1.3513}$

Case - Maximum Area :

$D 1 = \textcolor{g r e e n}{{D}_{\max}} = \textcolor{g r e e n}{370.3704}$

#### Explanation:

Let the two similar triangles be ABC & DEF.

Three sides of the two triangles be a,b,c & d,e,f and the areas A1 & D1.

Since the triangles are similar,

$\frac{a}{d} = \frac{b}{e} = \frac{c}{f}$

Also $\frac{A 1}{D 1} = {a}^{2} / {d}^{2} = {b}^{2} / {e}^{2} = {c}^{2} / {f}^{2}$

Property of a triangle is sum of any two sides must be greater than the third side.

Using this property, we can arrive at the minimum and maximum value of the third side of triangle ABC.

Maximum length of third side $c < 8 + 7$ , say 14.9 (corrected upto one decimal.

When proportional to maximum length, we get minimum area.

Case - Minimum Area :

$D 1 = \textcolor{red}{{D}_{\min}} = A 1 \cdot {\left(\frac{f}{c}\right)}^{2} = 12 \cdot {\left(\frac{5}{14.9}\right)}^{2} = \textcolor{red}{1.3513}$

Minimum length of third side $c > 8 - 7$ , say 0.9 (corrected upto one decimal.

When proportional to minimum length, we get maximum area.

Case - Maximum Area :

$D 1 = \textcolor{g r e e n}{{D}_{\max}} = A 1 \cdot {\left(\frac{f}{c}\right)}^{2} = 12 \cdot {\left(\frac{5}{0.9}\right)}^{2} = \textcolor{g r e e n}{370.3704}$