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2 years ago

8

A 45-45-90 triangle has a hypotenuse of length 7. What is the length of one of its legs? If necessary, round your answer to two

decimal places.

1 answer:

harkovskaia [24]2 years ago

4 0

If the legs are legnth x, then the hyptonuse is x√2

x√2=7
divide both sides by √2
x=7/√2
times (√2)/(√2)
(7√2)/2
3.5√2
aprox 4.94974
round
4.95

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Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

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$AB = \sqrt{64 + 16} = \sqrt{80}$

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Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

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$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

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$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

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Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

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Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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