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Snowcat [4.5K]
3 years ago
5

7. The length of a rectangle is two more than its width. The perimeter of the rectangle is 36 feet. What are the dimensions (len

gth and width) of the rectangle?
*Show work*​
Mathematics
1 answer:
Reika [66]3 years ago
5 0

Step-by-step explanation:

length = x + 2

width = x

x + x + (x+2) + (x+2) = 36 ft

4x + 4 = 36 ft

4x = 32 ft

x = 32 ÷ 4

x = 8 ft

dimensions = length x width

= (x+2) × x

= (8+2) × 8

= 10 ft × 8 ft

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Percentage increase.

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Determine whether parallelogram JKLM with vertices J(-7, -2), K(0, 4), L(9, 2) and M(2, -4) is a rhombus, square, rectangle or a
White raven [17]
Check the picture below

now, we know is a parallelogram, so the diagonals will bisect

looking at the picture, the interior angles are not right-angles, and thus is not a rectangle, and is not a square either, due to the same reason

is it a rhombus?  well, a rhombus, is a parallelogram, that's "slanted" per se, but regardless of how slanted it may be, the sides are all equal

now, we don't need to check the length of both pairs, just one of the segments of each pair, let's check only then JK and KL, the ones in red in the picture, since each pair of segments are twin, if JK = KL, then that means all sides are equal

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
J&({{-7 }}\quad ,&{{ -2}})\quad 
%  (c,d)
K&({{ 0}}\quad ,&{{ 4}})\\\\
K&({{ 0}}\quad ,&{{ 4}})\quad 
%  (c,d)
L&({{ 9}}\quad ,&{{ 2}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
JK=\sqrt{[0-(-7)]^2+[4-(-2)]^2}\implies JK=\sqrt{(0+7)^2+(4+2)^2}
\\\\\\
JK=\sqrt{7^2+6^2}
\\\\\\
KL=\sqrt{(9-0)^2+(2-4)^2}\implies KL=\sqrt{9^2+(-2)^2}

now... another characteristic of a rhombus is, the diagonals, meet at right-angle, that means KM ⟂ JL

and that simply means, if you get the slope of KM and the slope of JL, the product of their slope is -1

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
K&({{ 0}}\quad ,&{{ 4}})\quad 
%   (c,d)
M&({{ 2}}\quad ,&{{ -4}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
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\bf -------------------------------\\\\
\begin{array}{lllll}
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%   (a,b)
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%   (c,d)
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\\\\\\
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slope = {{ m}}= \cfrac{rise}{run} \implies 
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then multiply both slopes, if their product is -1, that means, they're perpendicular to each other, and it IS a rhombus only, then.

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