Dimensions - rectangle

One side of the rectangle is 12 cm longer than the other. If we reduce each dimension by a third, the area of the rectangle will be reduced by 60 cm². Determine the dimensions of the rectangle.

Final Answer:

a =  18 cm
b =  6 cm

Step-by-step explanation:

$$\begin{gathered} a=12+b \\ S=a\cdot b \\ (a-a/3)\cdot (b-b/3)=S-60 \\ \frac{2}{3}a\cdot \frac{2}{3}b=S-60 \\ \frac{4}{9}\cdot (ab)=S-60 \\ \frac{4}{9}\cdot S=S-60 \\ 4/9\cdot S=S-60 \\ 5S=540 \\ S=\frac{540}{5}=108 \\ S=108 \\ S=(12+b)\cdot b \\ 108=(12+b)\cdot b \\ -b^2-12b+108=0 \\ {b}^2+12b-108=0 \\ p=1;q=12;r=-108 \\ D=q^2-4pr=12^2-4\cdot 1\cdot (-108)=576 \\ D>0 \\ {b}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{-12\pm \sqrt{576}}{2} \\ {b}_{1,2}=\frac{-12\pm 24}{2} \\ {b}_{1,2}=-6\pm 12 \\ {b}_1=6 \\ {b}_2=-18 \\ b=b_1=6 \\ a=12+b=12+6=18 \\ \text{ Verifying Solution: } \\ {S}_2=a\cdot b=18\cdot 6=108{\text{cm}}^2 \\ x=a-a/3=18-18/3=12\text{cm} \\ y=b-b/3=6-6/3=4\text{cm} \\ {S}_3=x\cdot y=12\cdot 4=48{\text{cm}}^2 \\ \Delta S=S_2-S_3=108-48=60{\text{cm}}^2 \end{gathered}$$

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