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Iteru [2.4K]
1 year ago
9

the members of the gardening group plan to build a walkway through the garden as formed by the IHOP potenuse of each of the four

triangles in the drawing that way The Gardener's will be able to access all sections of the garden calculate the length of the entire walkway to the nearest hundredth of a yard show your work​

Mathematics
1 answer:
aliya0001 [1]1 year ago
8 0

Answer:

50.24 yards

Step-by-step explanation:

We will use the pythagorean theorem to find hypotenuse of each of the 4 triangles.

Pythagorean Theorem is Hypotenuse^2=OneLeg^2+AnotherLeg^2

For Spinach area, two of the legs are 8 and 6. Solving for hypotenuse:

hyp^2=6^2+8^2\\hyp^2=100\\hyp=10

For watermelon area, two of the legs are 12 and 8. Solving for hypotenuse:

hyp^2=12^2+8^2\\hyp^2=208\\hyp=\sqrt{208}

in red peppers area, the hypotenuse is already given as 15. But we need to find one leg since that will be the leg of the triangle in tomatoes area. Hypotenuse is 15 and one leg is 12, so the other leg is:

15^2=12^2+eg^2\\15^2-12^2=leg^2\\leg=9

So, in tomatoes area, we have one leg 9 and another leg 6. Solving for hypotenuse:

hyp^2=9^2+6^2\\hyp^2=117\\hyp=\sqrt{117}

Adding all the four hypotenuses, we get the length of the walkway:

Length of Walkway = 10+\sqrt{208} +15+\sqrt{117}\\ =50.24 yards

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Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]

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x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]

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x = \frac{1}{14}(2 \times 3 - 1 \times -22)

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Replace the second column with \left[\begin{array}{c}2&-22\end{array}\right] to calculate the value of y

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\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]

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Read more about matrices at:

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