## terricherri 4 years ago Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 9-inch perimeter A = sq. in Help me my answer needs to be in square root.

1. terricherri

i have 9/3sqrt4 but thats not right

2. id21

P=9in a=9/3=3in High of triangle (all corners are 60 degrees):$h=a \times \sin60^o=a \sqrt3 /2$ $Area=a \times h/2=a^2\sqrt3/2=9\sqrt{3}/2}$

3. id21

$Area=a \times h/2=a^2 \sqrt3/2=9\sqrt{3}/2$

4. terricherri

so its 9sqrt3/2

5. Harkirat

in case of an equilateral triangle, perimeter = 3 * side so side = perimeter/3 so each side = 9 in/3 = 3 in root(3) Area of an equilateral triangle = --------- * s² where s = length of side 4 root(3) 9 root(3) so area = --------- * (3cm)² = ---------- cm² 4 4

6. id21

Oh, yes, I miss one divider 2. Answer is area is $9\sqrt3/4$ in^2

7. id21

$Area=a \times h/2=a^2\sqrt3/4=9\sqrt{3}/4$

8. terricherri

Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 6-inch apothem A = sq. in. for this one i havwe 144sqrt2

9. terricherri

does anyone know that

10. id21

I get $108\sqrt3$

11. id21

must check it again

12. id21

Apothem is 1/3 rd from high, right? Then h=3*6=18 in

13. id21

$Area=a \times h /2=|a=2h/\sqrt3|=2h^2/(\sqrt3 \times 2)=324/\sqrt3=108\sqrt3$