diketahui suatu persegi panjang abcd dengan ab=(2p-4) dan bc=(m+2). kelilingnya adalah 60 cm. p+m = 20 cm. luas persegi panjang adalah

ab = (2p-4)
bc = (m+2)

p+m = 20 ==> p = 20 - m

K = 2(2p-4) + 2(m+2)
60 = 4p - 8 + 2m + 4
60 = 4(20-m) + 2m - 4
60 = 80 - 4m + 2m - 4
2m = 16
m = 8

p + m = 20
p = 12

ab = 2(12) - 4 = 20
bc = 8 + 2 = 10

maka
L = 20 x 10 = 200 cm²

K = 2p + 2l
60= 2(2p-4) + 2(m+2)
60= (4p-8) + (2m+4)
60= 4p+2m-4
4p + 2m = 60+4
4p+2m = 64

p+m = 20 (×2)
4p+2m = 64 (×1)
__________ -
Menjadi :
2p+2m = 40
4p+2m = 64
___________ -
-2p = -24
p = -24/-2 = 12
berarti, m = p+m = 20
12 + m = 20
m = 20 -12
m = 8
Jadi, panjang = 2p-4 = 2×12-4 = 20
dan lebar = m+2 = 8+2 = 10

Luas = p×l
= 20×10 = 200

Maaf kalau salah.