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the school uses about yards of fencing.

Question

the school uses about yards of fencing.

Explanation:

To determine the amount of fencing (perimeter) of the polygon, we first identify the coordinates of the vertices. Let's assume the vertices (from the grid) are:

  • \( A(-1, 6) \)
  • \( B(4, 5) \)
  • \( C(6, 0) \)
  • \( D(4, -5) \)
  • \( E(-1, -6) \)
  • \( F(-4, 0) \)
Step 1: Calculate distance between each pair of consecutive vertices using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Distance \( AB \):

\( A(-1, 6) \), \( B(4, 5) \)
\( d_{AB} = \sqrt{(4 - (-1))^2 + (5 - 6)^2} = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.1 \)

Distance \( BC \):

\( B(4, 5) \), \( C(6, 0) \)
\( d_{BC} = \sqrt{(6 - 4)^2 + (0 - 5)^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.4 \)

Distance \( CD \):

\( C(6, 0) \), \( D(4, -5) \)
\( d_{CD} = \sqrt{(4 - 6)^2 + (-5 - 0)^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.4 \)

Distance \( DE \):

\( D(4, -5) \), \( E(-1, -6) \)
\( d_{DE} = \sqrt{(-1 - 4)^2 + (-6 - (-5))^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.1 \)

Distance \( EF \):

\( E(-1, -6) \), \( F(-4, 0) \)
\( d_{EF} = \sqrt{(-4 - (-1))^2 + (0 - (-6))^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.7 \)

Distance \( FA \):

\( F(-4, 0) \), \( A(-1, 6) \)
\( d_{FA} = \sqrt{(-1 - (-4))^2 + (6 - 0)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.7 \)

Step 2: Sum all distances to find the perimeter.

Perimeter \( = d_{AB} + d_{BC} + d_{CD} + d_{DE} + d_{EF} + d_{FA} \)
\( \approx 5.1 + 5.4 + 5.4 + 5.1 + 6.7 + 6.7 \)
\( = (5.1 + 5.1) + (5.4 + 5.4) + (6.7 + 6.7) \)
\( = 10.2 + 10.8 + 13.4 \)
\( = 34.4 \)

(Note: If the grid has 1 unit = 1 yard, the perimeter is approximately 34 yards. Adjustments may be needed based on exact coordinates, but this is a reasonable estimate.)

Answer:

\(\boldsymbol{34}\) (or a close approximation based on precise vertex coordinates)