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03 Jun 2018, 12:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
86% (02:21) correct
14%
(02:18)
wrong
based on 123
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In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, y), (4, y), and (4, –1). If the area of WXYZ is 18, what is the length of its diagonal?
A) \(3 \sqrt{2}\)
B) \(3 \sqrt{3}\)
C) \(3 \sqrt{5}\)
D) \(3 \sqrt{6}\)
E) \(3 \sqrt{7}\)
A) \(3 \sqrt{2}\)
B) \(3 \sqrt{3}\)
C) \(3 \sqrt{5}\)
D) \(3 \sqrt{6}\)
E) \(3 \sqrt{7}\)
03 Jun 2018, 22:54
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Bunuel
Attachment:
2018-06-03coordrectangle.png [ 7.06 KiB | Viewed 7135 times ]
Find length of the rectangle.
Two vertices, Y and Z, lie at (–2, –1) and (4, –1)
Subtract the NONequal x-values to find length:
4 - (-2) = 6
L = 6
Find width of the rectangle from area
Area = 18 = L * W
6 * W = 18
W = 3
Length = one leg of a right triangle
Width = other leg of a right triangle
Pythagorean theorem applies regardless of
the other two vertices. (No need to graph W and X.)
Leg1\(^2\) + Leg2\(^2\) = Diagonal, \(d^2\)
\(6^2 + 3^2 = d^2\)
\(36 + 9 = d^2\)
\(d = \sqrt{45}\)
\(d = \sqrt{9 * 5}\)
\(d = 3\sqrt{5}\)
Answer
04 Jun 2018, 00:17
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Solution
Given:
- • In the coordinate plane, rectangle WXYZ has vertices at W (-2, -1), X (-2, y), Y (4, y), and Z (4, -1)
• Area of WXYZ is equal to 18
To find:
- • Length of the diagonal of WXYZ (i.e. either WY or XZ)
Approach and Working:
To find the length of WY or XZ, we need to find the value of y
- • One side of the rectangle = difference of x coordinates = 4 – (-2) = 6 units
• Other side of the rectangle = difference of y coordinates = y – (-1) = y + 1 units
As the area is 18, we can say
- • 6 (y + 1) = 18
Or, y = 2
All the coordinates: W (-2, -1), X (-2, 2), Y (4, 2), and Z (4, -1)
- • Length of WY or XZ = \(\sqrt{[4 – (-2)]^2 + [(-1) – 2]^2}\) = \(\sqrt{36 + 9}\) = \(\sqrt{45}\) = \(3\sqrt{5}\)
Hence, the correct answer is option C.
Answer: C
