# Triangle A has an area of 24  and two sides of lengths 8  and 12 . 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?

Apr 20, 2018

${A}_{s} = 4.16$
${A}_{L} = 14.36$

#### Explanation:

Using either given side as a base, the third side can be calculated.
$A = \frac{1}{2} \times b \times h$ ; $24 = \frac{1}{2} \times 8 \times h$

$h = 6$ We use this as part of a right triangle to find the third side.
$C = 6.46$

NOW we know that the SMALLEST similar triangle "B" must have the largest side as 5, and the LARGEST similar triangle will have the smallest side with a length of 5. The corresponding ratios are:
$12 : 8 : 6.46$ (original)
$5 : 3.33 : 2.69$ (smallest)
$9.29 : 6.19 : 5$ (largest)

Using Herron's Rule directly saves some further angle computation.
$s = \frac{a + b + c}{2}$
A = sqrt(((s(s-a)(s-b)(s-c))

$5 : 3.33 : 2.69$ (smallest)
$s = \frac{5 + 3.33 + 2.69}{2} = 5.51$
A = sqrt(((5.51(5.51-5)(5.51-3.33)(5.51-1.74))
${A}_{s} = 4.16$

$9.29 : 6.19 : 5$ (largest)
$s = \frac{9.29 + 6.19 + 5}{2} = 10.24$
A = sqrt(((10.24(10.24-9.29)(10.24-6.19)(10.24-5))
${A}_{L} = 14.36$