how to find the area of all these polygons just given their perimeter, the last one btw says 4 yd, (I will give brainliest answer!!)

11) SQUARE

ar (sq) = s² (s is the side)

and perimeter is 4s

now , perimeter = 4s

28 = 4s

s = 7

now , ar of sq = 49 sq inch

12) dodecagon

perimeter = 12s

48 = 12s ft

s = 4 ft

ar = 3×s²×(2+√3)

ar = 3 × 4² × (2+1.73)

ar = 3×16×3.73

ar = 179.04 sq ft

13) NoNAGON

perimeter = 9s

36 = 9s

s = 4yd

area = 9/4 s²cot(180°/9)

= 98.91 sq yd

14) similar to 11

15)PENTAGON

s = 5 mi

ar =

= 43.01 mi²

16) HEXAGON

perimeter = 6s

18 = 6s

s = 3 cm

ar of hexagon =

= 23.28 cm²

17) Similar to 12)

s = 5mi

ar = 3× 25 × 3.73

ar = 279.75 sq mi

18) OCTAGON

peri meter = 8s

64 = 8s

s = 8 m

ar =

ar = 2(1+1.41)×8²

ar = 2(2.41)×64

= 4.82 × 64

= 308.48 m²

19) SIMILAR TO 15)

s = 4

ar = 27.53

THAT TOOK ME A LOT OF EFFORTS

srry ask agian

step-by-step explanation:

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x