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Geometry Difficulty 2.0 Junior Prove it Canada

Three students are helping to expand their school’s garden.
Initially, the garden has a length of $5\text{ m}$ and a width of $4\text{ m}$ , as shown.IMG1 Rob adds two additional $2\text{ m}$ by $4\text{ m}$ plots side by side next to the initial garden, as shown.What is the total area of the expanded garden after Rob adds these two plots? Kirima adds a path around three sides of the previous garden, as shown.If the width of the path is $\$1\text{}$
m}, what is the total combined area of the garden and the path? Noah adds

Figure for this problem

Figure for this problem n $additional$ $2\text{}$ m} $by$ $4\text{}$ m} plots to the previous version of the garden (in part (b)), and then continues the

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Figure for this problem $1\text{}$ m} wide path so that it surrounds the entire garden, as shown. If the total combined area of the garden and the path is

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Figure for this problem 150 $\text{m}^2$ $, determine the value of$ n$.

Solution

Solution 1:

The length of the expanded garden is $(5+2\times2)\text{ m}=9\text{ m}$ , and the width is $4 \text{ m}$ . Thus, the total area of the expanded garden is $\$9\text{ m}\times4\text{ m}=36\text{}$
m}^2. Solution 2: The area of the original

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Figure for this problem 5 $\text{}$
m} $by$ 4 $\text{}$ m} $garden is$ $5\text{ m}\times4\text{ m}=20\text{}$
m}^2 $. Each additional$ 2 $\text{}$ m} $by$ 4 $\text{}$ m} $plot has area$ $2\text{ m}\times4\text{ m}=8\text{}$ m}^2, and so the total area of the expanded garden is

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Figure for this problem $(20+2\times8)\text{ m}^2=36\text{}$
m}^2. Solution 1: The combined garden and path has length

Figure for this problem

Figure for this problem $9\text{ m}+1\text{ m}=10\text{}$ m} $, and width$ $(4+2\times1)\text{ m}=6\text{}$
m}. Thus, the area of the garden and the path is

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Figure for this problem $10\text{ m}\times6\text{ m}=60\text{}$
m}^2. Solution 2: Consider splitting the path into three rectangles, as shown. [[IMAGE0]] Each of the rectangles above and below the garden has dimensions

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Figure for this problem $9\text{}$ m} $by$ $1\text{}$ m}, and thus each has area

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Figure for this problem $9\text{ m}\times1\text{ m}=9\text{}$ m}^2$.

The remaining section of the path has height $(4+2\times1)\text{ m}=6\text{ m}$ and width $1\text{ m}$ , and thus has area $\$6\text{ m}\times1\text{ m}=6\text{}$
m}^2. The area of the expanded garden is 36 m

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Figure for this problem ^2, and so the total combined area of the garden and the path is

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Figure for this problem $(36+2\times9+6)\text{ m}^2=60\text{}$ m}^2. Solution 1: Each of the new plots has length

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Figure for this problem 2 $\text{}$
m} $, and so$ n $plots increase the$ $9\text{}$ m} $length of the garden by$ $2n\text{}$ m}. Thus, the combined length of the garden and the path is

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Figure for this problem $(9+2n+2\times1)\text{ m}=(2n+11)\text{}$
m}. The combined width of the garden and the path is

Figure for this problem

Figure for this problem $(4+2\times1)\text{ m}=6\text{}$ m} $. Thus in$ $\text{m}^2$ , the total combined area of the garden and the path is

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Figure for this problem $6\times(2n+11)$ $. Solving$ $6\times(2n+11)=150$ $, we get$ $2n+11=\frac{150}{6}=25$ $or$ 2n=14 $, and so$ n=7. Solution 2: Consider splitting the combined area of the garden and path into three rectangles, as shown. [[IMAGE1]] Each of the rectangles to the left and right of the garden has height

Figure for this problem

Figure for this problem $(4+2\times1)\text{ m}=6\text{}$ m} $, width$ $1\text{}$ m}, and thus each has area

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Figure for this problem $6\text{ m}\times1\text{ m}=6\text{}$
m}^2. The remaining rectangle, which combines the garden and the remaining sections of the path, also has height

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Figure for this problem $6\text{}$ m}. Each of the new plots has length

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Figure for this problem 2 $\text{}$
m} $, and so$ n $plots increase the$ $9\text{}$ m} $length of the garden by$ $2n\text{}$ m}. Thus, the length of this remaining rectangle is

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Figure for this problem $(2n+9)\text{}$ m} $. Measured in$ $\text{m}^2$ , the total combined area of the garden and the path is

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Figure for this problem $2\times6+6\times(2n+9)$ $or$ $12+6\times(2n+9)$ $. Solving$ $12+6\times(2n+9)=150$ $, we get$ $6\times(2n+9)=138$ $or$ $2n+9=\frac{138}{6}=23$ $or$ 2n=14 $, and so$ n=7$.

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