Question 121582
THE AREA OF THE SEMICIRCLE IS:
PIR^2/2=3.14*1.5^2/2=3.14*2.25/2=7.065/2=3.53 FT^2
23-3.53=19.47 FT^2
SEEING AS THE WIDTH OF THE WINDOW IS 3 THEN:
19.47/3=6.49 FEET IS THE HEIGHT OF THE RECTANGULAR WINDOW.