Two squares have side lengths that are consecutive odd integers. The total area of the squares is [tex]650 \, \text{cm}^2[/tex]. What is their total perimeter?
Answer (2)
t h e co n sec u t i v e o dd in t e g ers i s : 2 n + 1 an d 2 n + 3 ( n ∈ N ) A re a 1 = ( 2 n + 1 ) 2 = 4 n 2 + 4 n + 1 A re a 2 = ( 2 n + 3 ) 2 = 4 n 2 + 12 n + 9 A re a 1 + A re a 2 = 4 n 2 + 4 n + 1 + 4 n 2 + 12 n + 9 = 8 n 2 + 16 n + 10 8 n 2 + 16 n + 10 = 650 8 n 2 + 16 n + 10 − 650 = 0 8 n 2 + 16 n − 640 = 0 / : 8
n 2 + 2 n − 80 = 0 Δ = 2 2 − 4 ⋅ 1 ⋅ ( − 80 ) = 4 + 320 = 324 ; Δ = 324 = 18 1 = 2 ⋅ 1 − 2 − 18 < 0 ; n 2 = 2 ⋅ 1 − 2 + 18 = 2 16 = 8 2 n + 1 = 2 ⋅ 8 + 1 = 16 + 1 = 17 ( m ) 2 n + 3 = 2 ⋅ 8 + 3 = 16 + 3 = 19 ( m ) t h e t o t a l p er im e t er = 4 ⋅ 17 + 4 ⋅ 19 = 68 + 76 = 144 ( m )
The two squares have side lengths of 17 cm and 19 cm. Their total perimeter is 144 cm. This was found by solving the equation for the areas of the squares based on their side lengths being consecutive odd integers.
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