Answer:
Option B
Step-by-step explanation:
we know that
It appears as though the overlap is slightly smaller if we place the sprinklers at the two opposite corners of the lawn'
We will compute the total area covered by each placement in order to confirm this.
step 1
Find the area covered if Selina puts the two sprinklers at opposite ends of a 12-meter side of the lawn.
see the attached figure N 1 to better understand the problem
We'll decompose the area covered by the two sprinklers into the isosceles triangle ABX and two circular sectors (one in each circle).
Find the area of triangle ABX
Applying the Pythagorean Theorem find the height of triangle ABX (side XM)
[tex]AX^2=XM^2+AM^2[/tex]
we have
[tex]AX=8.4\ m\\AM=12/2=6\ m[/tex]
[tex]8.4^2=XM^2+6^2[/tex]
[tex]XM^2=34.56\\XM=5.88\ m[/tex]
The area of triangle ABX is about
[tex]A=\frac{1}{2}(12)(5.88)= 35.28\ m^2[/tex]
Find the area of the two sector areas
Find the measure of angle ∠XAM
[tex]m\angle XAM=tan^{-1} (5.88/6)=44.4^o[/tex]
so
The central angle of each sector is equal to
[tex]90^o-44.4^o=45.6^o[/tex]
applying proportion
[tex]\frac{(3.14) (8.4)^2}{360^o}=\frac{x}{45.6^o}\\\\x= 28.06\ m^2[/tex]
so
The area of the two sectors is equal to
[tex]28.06(2)=56.12\ m^2[/tex]
The total area covered is approximately
[tex]35.28+56.12=91.4\ m^2[/tex]
step 2
Find the area covered if Selina puts the two sprinklers at opposite corners of a 12-meter side of the lawn.
see the attached figure N 2 to better understand the problem
we'll decompose the area covered into the rhombus APBQ and four circular sectors.
Find the area of the rhombus APBQ
we have that
Applying Pythagorean theorem
[tex]AB=15\ m[/tex] -----> [tex]AB^2=12^2+9^2[/tex]
so
[tex]AX=7.5\ m[/tex]
[tex]AP=8.4\ m[/tex]
Applying Pythagorean theorem find the length side PX
[tex]PX^2=8.4^2-7.5^2\\PX=3.78\ m[/tex]
The area of the rhombus is equal to
[tex]A=\frac{1}{2}(15)(2*3.78)= 56.7\ m^2[/tex]
Find the area of the four sectors
Find the measure of angle ∠PAX
[tex]m\angle PAX=tan^{-1} (3.78/7.5)=26.75^o[/tex]
The measure of angle ∠PAQ is two times the measure of angle ∠PAX
so
[tex]m\angle PAQ=2(26.75^o)=53.5^o[/tex]
The central angle of the four sectors is equal to
[tex]2(90^o-53.5^o)=73^o[/tex]
applying proportion
[tex]\frac{(3.14) (8.4)^2}{360^o}=\frac{x}{73^o}\\\\x= 44.93\ m^2[/tex]
The total area covered is approximately
[tex]56.7+44.93=101.63\ m^2[/tex]
step 3
Compare the areas
[tex]101.63 > 91.44[/tex]
therefore
The strategy of placing the sprinklers at opposite corners of the lawn is the best strategy
