Question #606a2

2 Answers
Gió
Jul 25, 2015

I found: #(dA)/(dt)=-27(cm^2)/s#

Explanation:

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When #A=27sqrt(3)cm^2# then using (1) #A=sqrt(3)/4l^2# you get:
#27sqrt(3)=sqrt(3)/4l^2#
#color(red)(l=10.4cm)#
Now you use (2) and get the change in the area as:
#(dA)/(dt)=2sqrt(3)/4xx(10.4)xx(-3)=-27(cm^2)/s#

Tutor B. · Jim H
Jul 25, 2015

#27# cm^2/s

Explanation:

#A = (1/2) a*a*sin60 =( (sqrt3)/4)a^2#

#(dA)/dt = ( (sqrt3)/2)a((da)/dt)#

If #A = 27(sqrt3#) => #a = 6(sqrt3)#

Then, #(dA)/dt =( (sqrt3)/2)6(sqrt3)(3) " cm^2/s" = 27 " cm^2/s"#

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