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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 m5\text{ m} and a width of 4 m4\text{ m}, as shown.

Figure 0

Rob adds two additional 2 m2\text{ m} by 4 m4\text{ m} plots side by side next to the
initial garden, as shown.

Figure 1

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.

Figure 2

If the width of the path is 1 m1\text{ m}, what is the total combined area of the garden and the
path?
Noah adds nn additional 2 m2\text{ m} by 4 m4\text{ m} plots to the previous version
of the garden (in part (b)), and then continues the 1 m1\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 150m2150 \text{m}^2,
determine the value of nn.

Figure 3

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Solution

Solution 1:

The length of the expanded garden is (5+2×2) m=9 m(5+2\times2)\text{ m}=9\text{ m}, and the
width is 4 m4 \text{ m}.

Thus, the total area of the expanded garden is 9 m×4 m=36 m29\text{ m}\times4\text{ m}=36\text{ m}^2.

Solution 2:

The area of the original 5 m5 \text{ m} by 4 m4 \text{ m} garden
is 5 m×4 m=20 m25\text{ m}\times4\text{ m}=20\text{ m}^2.

Each additional 2 m2 \text{ m} by
4 m4 \text{ m} plot has area 2 m×4 m=8 m22\text{ m}\times4\text{ m}=8\text{ m}^2,
and so the total area of the expanded garden is (20+2×8) m2=36 m2(20+2\times8)\text{ m}^2=36\text{ m}^2.
Solution 1:

The combined garden and path has length 9 m+1 m=10 m9\text{ m}+1\text{ m}=10\text{ m}, and
width (4+2×1) m=6 m(4+2\times1)\text{ m}=6\text{ m}.

Thus, the area of the garden and the path is 10 m×6 m=60 m210\text{ m}\times6\text{ m}=60\text{ m}^2.

Solution 2:

Consider splitting the path into three rectangles, as shown.

Figure 4

Each of the rectangles above and below the garden has dimensions
9 m9\text{ m} by 1 m1\text{ m}, and thus each has area 9 m×1 m=9 m29\text{ m}\times1\text{ m}=9\text{ m}^2.

The remaining section of the path has height(4+2×1) m=6 m(4+2\times1)\text{ m}=6\text{ m} and
width 1 m1\text{ m}, and thus has area
6 m×1 m=6 m26\text{ m}\times1\text{ m}=6\text{ m}^2. The area of the expanded garden is 36 m2^2, and so the total combined area of the
garden and the path is (36+2×9+6) m2=60 m2(36+2\times9+6)\text{ m}^2=60\text{ m}^2.
Solution 1:

Each of the new plots has length 2 m2 \text{ m}, and so nn plots
increase the 9 m9\text{ m} length of
the garden by 2n m2n\text{ m}.

Thus, the combined length of the garden and the path is (9+2n+2×1) m=(2n+11) m(9+2n+2\times1)\text{ m}=(2n+11)\text{ m}. The combined width of the garden and the path is (4+2×1) m=6 m(4+2\times1)\text{ m}=6\text{ m}.

Thus in m2\text{m}^2, the total
combined area of the garden and the path is 6×(2n+11)6\times(2n+11).

Solving 6×(2n+11)=1506\times(2n+11)=150, we get
2n+11=1506=252n+11=\frac{150}{6}=25 or 2n=142n=14, and so n=7n=7.

Solution 2:

Consider splitting the combined area of the garden and path into
three rectangles, as shown.

Figure 5

Each of the rectangles to the left and right of the garden has height
(4+2×1) m=6 m(4+2\times1)\text{ m}=6\text{ m},
width 1 m1\text{ m}, and thus each has
area 6 m×1 m=6 m26\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 6 m6\text{ m}.

Each of the new plots has length 2 m2 \text{ m}, and so nn plots
increase the 9 m9\text{ m} length of
the garden by 2n m2n\text{ m}.

Thus, the length of this remaining rectangle is (2n+9) m(2n+9)\text{ m}.

Measured in m2\text{m}^2, the total
combined area of the garden and the path is 2×6+6×(2n+9)2\times6+6\times(2n+9) or 12+6×(2n+9)12+6\times(2n+9).

Solving 12+6×(2n+9)=15012+6\times(2n+9)=150, we
get 6×(2n+9)=1386\times(2n+9)=138 or 2n+9=1386=232n+9=\frac{138}{6}=23 or 2n=142n=14, and so n=7n=7.

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