A rectangle whose perimeter is 80 m has an area of [tex]384 \, \text{m}^2[/tex]. Find the dimensions of the rectangle.

2 a + 2 b = 80 / : 2 ⇒ a + b = 40 ⇒ a = 40 − b a ⋅ b = 384 ⇒ ( 40 − b ) ⋅ b = 384 ⇒ − b 2 + 40 b − 384 = 0 Δ = 4 0 2 − 4 ⋅ ( − 1 ) ⋅ ( − 384 ) = 1600 − 1536 = 64 ⇒ Δ ​ = 8 b 1 ​ = 2 ⋅ ( − 1 ) − 40 − 8 ​ = − 2 − 48 ​ = 24 ⇒ a 1 ​ = 40 − b 1 ​ = 40 − 24 = 16 b 2 ​ = 2 ⋅ ( − 1 ) − 40 + 8 ​ = − 2 − 32 ​ = 16 ⇒ a 2 ​ = 40 − b 2 ​ = 40 − 16 = 24
A n s . T h e d im e n s i o n s o f t h e rec t an g l e : 16 m an d 24 m

Answered by kate200468 | 2024-06-10

Since area of a rectangle = length X width I came up with an answer of 16 X 24 =384 Also, 16+16+24+24 = 80

Answered by missyslone | 2024-06-10

The dimensions of the rectangle are 16 m and 24 m. This can be verified through the perimeter and area formulas used during the problem-solving process. The methods used include solving a quadratic equation derived from the perimeter and area conditions.
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Answered by kate200468 | 2024-12-23