#### Answer

The area of the small triangle is $139.5 $m$^{2}$ and the perimeter is $54.6$ m
The area of the large triangle is $248$ m$^{2}$ and the perimeter is $72.8$ m.

#### Work Step by Step

1. Solve for $x$
$\frac{x}{18.6} = \frac{28}{21}$
$x = \frac{28 \times 18.6}{21}$
$x = 24.8$ m
2. Solve for $y$
$\frac{y}{20} = \frac{21}{28}$
$y = 15$ m
3. Find the perimeter and area of the small triangle
Let $P = $ perimeter of the small triangle
Let $A = $ area of the small triangle
$P = 21 + 18.6 + y$
$P = 21 + 18.6 + 15$
$P = 54.6$ m
$A = (18.6 \times 15) \div 2$
$A = \frac{279}{2}$
$A = 139.5$ m$^{2}$
4. Find the perimeter and area of the large triangle
Let $R = $perimeter of the large triangle
Let $T =$ area of the large triangle
$R = 20 + 28 + x$
$R = 20 + 28 + 24.8$
$R = 72.8$ m
$T = (20 \times 24.8) \div 2$
$T = \frac{496}{2}$
$T = 248$ m$^{2}$