Question

Calculate the area a brick region? The length from the base of the rectangle to the top of the arc is called "Sagita". The small side of the bridge into the sagita, `e=21`.

Answer 1

area of brick region `=5323.0 " m"^2`

Explanation:

area of rectangle `ABCD=A_R=153xx53=8109.00 " m"^2`
let `O and r` be the center and the radius of arc `EFG`, respectively.
`r=OH+HF=OH+35`
`DeltaOHE` is a right triangle, by Pythagorean theorem, we get:
`r^2-OH^2=EH^2`
`=> (OH+35)^2-OH^2=55.5^2`
`=> OH^2+70*OH+35^2-OH^2=55^2`
`=> 70*OH+1225=3080.25`
`=> OH=(3080.25-1225)/70~~26.50` m
`=> r=OH+35=26.5+35=61.50` m
let `angleEOH=alpha`
`rsinalpha=55.5, => alpha=sin^-1((55.5)/(61.50))=64.4805^@`
`=> 2alpha=2*64.4805=128.961^@`
area of segment `EFG=A_S=pi*r^2*(2alpha)/(360)-1/2*r^2*sin2alpha`
`=r^2(pi*(2alpha)/360-(sin2alpha)/2)`
`=61.5^2(pi*128.961/360-sin128.961/2)`
`=2786.0 " m"^2`
Hence, area of the brick region `A_B=A_R-A_S`
`=8109.0-2786.0=5323.0 " m"^2`