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

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5

send it over

step-by-step explanation: