# Triangle A has an area of 12  and two sides of lengths 3  and 8 . Triangle B is similar to triangle A and has a side of length 15 . What are the maximum and minimum possible areas of triangle B?

Jun 30, 2018

Maximum possible area of triangle B is $300$ sq.unit
Minimum possible area of triangle B is
$36.99$ sq.unit

#### Explanation:

Area of triangle $A$ is ${a}_{A} = 12$

Included angle between sides $x = 8 \mathmr{and} z = 3$ is

$\frac{x \cdot z \cdot \sin Y}{2} = {a}_{A} \mathmr{and} \frac{8 \cdot 3 \cdot \sin Y}{2} = 12 \therefore \sin Y = 1$

$\therefore \angle Y = {\sin}^{-} 1 \left(1\right) = {90}^{0}$ Therefore, Included angle between

sides $x = 8 \mathmr{and} z = 3$ is ${90}^{0}$

Side $y = \sqrt{{8}^{2} + {3}^{2}} = \sqrt{73}$. For maximum area in triangle

$B$ Side ${z}_{1} = 15$ corresponds to lowest side $z = 3$

Then ${x}_{1} = \frac{15}{3} \cdot 8 = 40 \mathmr{and} {y}_{1} = \frac{15}{3} \cdot \sqrt{73} = 5 \sqrt{73}$

Maximum possible area will be $\frac{{x}_{1} \cdot {z}_{1}}{2} = \frac{40 \cdot 15}{2} = 300$

sq. unit. For minimum area in triangle $B$ Side ${y}_{1} = 15$

corresponds biggest side $y = \sqrt{73}$

Then ${x}_{1} = \frac{15}{\sqrt{73}} \cdot 8 = \frac{120}{\sqrt{73}}$ and

${z}_{1} = \frac{15}{\sqrt{73}} \cdot 3 = \frac{45}{\sqrt{73}}$. Minimum possible area will be

$\frac{{x}_{1} \cdot {z}_{1}}{2} = \frac{1}{2} \cdot \left(\frac{120}{\sqrt{73}} \cdot \frac{45}{\sqrt{73}}\right) = \frac{60 \cdot 45}{73}$

$\approx 36.99 \left(2 \mathrm{dp}\right)$ sq.unit [Ans]