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$ABCD$ is a rectangle with its vertices at $(2, -2), (8, 4), (4, 8)$ and $(- 2, 2)$ taken in order. Length of its diagonal is:
$ABCD$ is a rectangle with its vertices at $(2, -2), (8, 4), (4, 8)$ and $(- 2, 2)$ taken in order. Length of its diagonal is:
a) $4 \sqrt{2}$
b) $6 \sqrt{2}$
a) $4 \sqrt{2}$
b) $6 \sqrt{2}$
c) $4 \sqrt{26}$
d) $2 \sqrt{26}$
This Question has 1 answers.
We are given a rectangle $ABCD$ with its vertices:
$ A(2, -2), B(8,4), C(4,8), D(-2,2) $
To find the length of its diagonal, we use the distance formula:
$ \displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $
Taking diagonal $AC$:
$ \displaystyle AC = \sqrt{(4 - 2)^2 + (8 - (-2))^2} $
$ \displaystyle = \sqrt{(2)^2 + (10)^2} $
$ \displaystyle = \sqrt{4 + 100} $
$ \displaystyle = \sqrt{104} $
$ \displaystyle = 2\sqrt{26} $
Thus, the length of the diagonal is $ 2\sqrt{26} $ units.
$ A(2, -2), B(8,4), C(4,8), D(-2,2) $
To find the length of its diagonal, we use the distance formula:
$ \displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $
Taking diagonal $AC$:
$ \displaystyle AC = \sqrt{(4 - 2)^2 + (8 - (-2))^2} $
$ \displaystyle = \sqrt{(2)^2 + (10)^2} $
$ \displaystyle = \sqrt{4 + 100} $
$ \displaystyle = \sqrt{104} $
$ \displaystyle = 2\sqrt{26} $
Thus, the length of the diagonal is $ 2\sqrt{26} $ units.
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