An equilateral triangle has an area of $64\sqrt{3}$ $\text{cm}^2$. if each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?

Let

ABC> an equilateral triangle

x> length side of the triangle

we know that

if ABC is an equilateral triangle

then

AB=BC=AC> x

and

angle A=angle B=angle C=60 degrees

Applying the law of sines

Area of triangle=(1/2)*a*b*sin C

in this problem

a=b=x cm

C=60 degrees

Area=64/(√3) cm²> 36.95 cm²

A=(1/2)*a*b*sin C> (1/2)*x*x*sin 60°> (1/2)*x²*(√3)/2

64/(√3)=(1/2)*x²*(√3)/2> 64*4=√3*√3*x²> 256=3*x²

x²=256/3> x=16/√3 cm> x=(16√3)/3 cm

if each side of the triangle is decreased by 4 cm

the new side of triangle (x1) is equal to

x1=[(16√3)/3]-4> x1=[16√3-12]/3 cm

the new area (A1) decreased is equal to

A1=(1/2)*x1²*(√3)/2> A1=(√3/4)*[(16√3-12)/3]²> A1=(√3/36)*[(16√3-12)]²

A1=(√3/36)*[(16√3)²-2*(16√3)*12+12²)]> A1=(√3/36)*[(768)-(384√3)+144)]

A1=(√3/36)*[(912)-(384√3)]> A1=11.879 cm²

therefore

the answer is

the area decreased is 11.879 cm²

4

step-by-step explanation:

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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