## Basic College Mathematics (9th Edition)

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.
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$^{2}$