Q:

Area and perimeter of geometric functions

Accepted Solution

A:
Answer:C. 60Step-by-step explanation:From triangle ADB,[tex]|AD|^2+|BD|^2=|AB|^2[/tex][tex]|AD|^2+16^2=20^2[/tex][tex]|AD|^2=20^2-16^2[/tex][tex]|AD|^2=400-256[/tex][tex]|AD|^2=144[/tex][tex]|AD|=\sqrt{144}[/tex][tex]|AD|=12[/tex]From triangle ADC,[tex]|CD|^2+12^2=15^2[/tex][tex]|CD|^2+144=225[/tex][tex]|CD|^2=225-144[/tex][tex]|CD|^2=81[/tex][tex]|CD|=9[/tex]The perimeter is the distance around the figure.The perimeter[tex]=15+20+16+9[/tex][tex]=60[/tex]