## Basic College Mathematics (10th Edition)

Perimeters: $P1 = 36cm$ $P2 = 24cm$ Areas: $A1 = 62.4cm^2$ $A2 = 27.6cm^2$
As the first triangle has all the sides of $12cm$ then all the sides of the second triangle are of $8cm$. The rate is $\frac{12}{8} = \frac{3}{2}$, then the height of the second triangle is $\frac{10.4*2}{3} = 6.9cm$. Perimeters: $P1 = 12 + 12 + 12 = 36cm$ $P2 = 8 + 8 + 8 = 24cm$ Areas: $A1 = \frac{12*10.4}{2} = 62.4cm^2$ $A2 = \frac{8*6.9}{2} = 27.6cm^2$