Six squares are inscribed in an 11 by 13 rectangle. Find the shaded area.
11 * 13 = 143 u2
112 / 9 = 13.44444444 u2
80.667 u2
143 - 80.667 = 62.33333333 u2 area of shaded region
Six squares are inscribed in an 11 by 13 rectangle. Find the shaded area.
Let P2=(x2y2) Let P4=(x4y4)
P1=a(sin(φ)cos(φ))P2=P1+3a(cos(φ)−sin(φ))P2=a(sin(φ)cos(φ))+3a(cos(φ)−sin(φ))P2=a(sin(φ)+3cos(φ)cos(φ)−3sin(φ))x2=a(sin(φ)+3cos(φ))|x2=11a(sin(φ)+3cos(φ))=11a=11(sin(φ)+3cos(φ))(1)
P3=2a(cos(φ)−sin(φ))P4=P3−3a(sin(φ)cos(φ))P4=2a(cos(φ)−sin(φ))−3a(sin(φ)cos(φ))P4=a(2cos(φ)−3sin(φ)−2sin(φ)−3cos(φ))|y4|=a(2sin(φ)+3cos(φ))|y4=13a(2sin(φ)+3cos(φ))=13a=13(2sin(φ)+3cos(φ))(2)
(1):a=11(sin(φ)+3cos(φ))(2):a=13(2sin(φ)+3cos(φ))a=11(sin(φ)+3cos(φ))=13(2sin(φ)+3cos(φ))11(sin(φ)+3cos(φ))=13(2sin(φ)+3cos(φ))11(2sin(φ)+3cos(φ))=13(sin(φ)+3cos(φ))22sin(φ)+33cos(φ)=13sin(φ)+39cos(φ)22sin(φ)−13sin(φ)=39cos(φ)−33cos(φ)9sin(φ)=6cos(φ)tan(φ)=23
a= ?
a=11(sin(φ)+3cos(φ))sin(φ)=tan(φ)√1+tan2(φ)|tan(φ)=23sin(φ)=23√1+49sin(φ)=2√1313cos(φ)=1√1+tan2(φ)|tan(φ)=23cos(φ)=1√1+49cos(φ)=3√1313a=112√1313+3∗3√1313a=112√1313+9√1313a=1111√1313a=13√13a=√13
The shaded area:
The shaded area=11∗13−6a2|a=√13The shaded area=11∗13−6∗13The shaded area=5∗13The shaded area=65