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!!)

Answer:
A = (L2 n)/[4 tan (180/n)] n =
Step-by-step explanation:
n=number of sides
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 =
[tex] \frac{1}{4} \sqrt{5(5 + 2 \sqrt{5}) } {s}^{2} [/tex]
[tex] \frac{1}{4} \sqrt{5(5 + 2 \sqrt{5}) } {1}^{2} [/tex]
= 43.01 mi²
16) HEXAGON
perimeter = 6s
18 = 6s
s = 3 cm
ar of hexagon =
[tex] \frac{3 \sqrt{3} }{2} \times {s}^{2} [/tex]
[tex] \frac{3 \sqrt{3} }{2} \times {3}^{2} [/tex]
= 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 =
[tex]2( 1 + \sqrt{2} ) {s}^{2} [/tex]
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