Atoy pyramid has the dimensions shown below. the base of the pyramid i is an equilateral triangle. what is the ared
the base of this pyramid?
9 cm
21 cm

That is funny.  The base isn't even an equilateral triangle, if it was the height that is marked 7 should actually be 3√3≈ 5.2 cm.

Anyway, I'll play along...the area of any triangle is bh/2, using their fictional and incorrect values you have:

A=6*7/2=21cm^2

Now if it REALLY was an equilateral triangle the area would simply be:

A=√(3s^4/16), s is just a side of which are all equal in an equilateral triangle

In this case:

A=√(3*6^4/16) cm^2

A=√243 cm^2

A≈15.59 cm^2

Or we could say because it is an equilateral triangle that h=6√3/2=3√3

And its base is 6 so

A=(6*3√3)/2

A=9√3 cm^2

A≈15.59 cm^2

Yes of course i will indeed you with your problem.