15m x2m ;
The\ rectangle:a\ \times\ b\\\\ \left\{\begin{array}{ccc}ab=30\\2a+2b=34&/:2\end{array}\right\\\\\left\{\begin{array}{ccc}ab=30\\a+b=17\end{array}\right\\\\\left\{\begin{array}{ccc}ab=30\\a=17-b\end{array}\right\\\\substitute:\\\\(17-b)b=30
− b 2 + 17 b − 30 = 0 a = − 1 ; b = 17 ; c = − 30 Δ = b 2 − 4 a c Δ = 1 7 2 − 4 ⋅ ( − 1 ) ⋅ ( − 30 ) = 289 − 120 = 169 ; Δ = 169 = 13 b 1 = 2 ⋅ ( − 1 ) − 17 − 13 = − 2 − 30 = 15 ; b 2 = 2 ⋅ ( − 1 ) − 17 + 13 = − 2 − 4 = 2 a 1 = 17 − 15 = 2 ; a 2 = 17 − 2 = 15 A n s w er : T h e rec t an g l e : 15 m × 2 m .
The dimensions of the rectangle are 15 meters and 2 meters. This is derived from solving the area and perimeter equations simultaneously. Both combinations of length and width satisfy the conditions given in the problem.
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