14. What is the length of a diagonal of rectangle EFGH? Show your work

The length of the diagonal of the rectangle is of [tex]\sqrt{58}[/tex] units.

Step-by-step explanation:

Pythagorean theorem:

The square of the hypothenuse h is equal to the sum of the squares of sides a and b, that is:

[tex]h^2 = a^2 + b^2[/tex]

Diagonal of a rectangle:

The diagonal is the hypothenuse, while the sides are a and b.

In this question:

The height has coordinates are y = 9 and y = 12, so the height is [tex]a = 12 - 9 = 3[/tex]

The length has coordinates at [tex]x = -6, x = -13[/tex], so the length is [tex]b = -6 - (-13) = 7[/tex]

Diagonal:

[tex]d^2 = a^2 + b^2[/tex]

[tex]d^2 = 3^2 + 7^2[/tex]

[tex]d^2 = 9 + 49[/tex]

[tex]d^2 = 58[/tex]

[tex]d = \sqrt{58}[/tex]

The length of the diagonal of the rectangle is of [tex]\sqrt{58}[/tex] units.

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

[tex]\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}[/tex]

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

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step-by-step explanation: