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26 Jan 2022, 08:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
63% (02:41) correct
37%
(02:52)
wrong
based on 131
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What is the area of region enclosed by \(|x - 5| + |y - 5| = 5\) ?
A. \(5 \sqrt{2}\)
B. \(20\sqrt{2}\)
C. \(20 \sqrt{5}\)
D. \(25 \sqrt{10}\)
E. \(50\)
Are You Up For the Challenge: 700 Level Questions
A. \(5 \sqrt{2}\)
B. \(20\sqrt{2}\)
C. \(20 \sqrt{5}\)
D. \(25 \sqrt{10}\)
E. \(50\)
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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26 Jan 2022, 08:33
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|x - 5| + |y - 5| = 5
X & Y can not be negative, otherwise the sum of |x - 5| and |y - 5| will be more than 5
Hence X and Y can be only Positive.
|X-5|+|Y-5| = 5
Put X = 0
Y will be 5,
Put Y = 0
X will be 5,
From here we know the the difference between x and y is 5 unit
Also x+y = 15 (|X-5|+|Y-5| = 5)
So if, one of x and y is 10 other will be 5
Hence we have four pairs,
(0,5), (5,0), (10,5), &(5,10) making square.
In which (0,5) and (10,5) lies on same line, and
(5,0) & (5,10) on other
(Intersecting each other at 5,5)
Area of the enclosed area = 4*1/2*5*5 (as 1 right angle triangle in that square consist of 5 height and 5 breadth)
Area = 50
Answer is E
Posted from my mobile device
X & Y can not be negative, otherwise the sum of |x - 5| and |y - 5| will be more than 5
Hence X and Y can be only Positive.
|X-5|+|Y-5| = 5
Put X = 0
Y will be 5,
Put Y = 0
X will be 5,
From here we know the the difference between x and y is 5 unit
Also x+y = 15 (|X-5|+|Y-5| = 5)
So if, one of x and y is 10 other will be 5
Hence we have four pairs,
(0,5), (5,0), (10,5), &(5,10) making square.
In which (0,5) and (10,5) lies on same line, and
(5,0) & (5,10) on other
(Intersecting each other at 5,5)
Area of the enclosed area = 4*1/2*5*5 (as 1 right angle triangle in that square consist of 5 height and 5 breadth)
Area = 50
Answer is E
Posted from my mobile device
26 Jan 2022, 09:44
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Bunuel
The equation \(|x - 5| + |y - 5| = 5\) is a linear pair.
So, we will have 4 lines, and FOUR vertices.
Let us look for the value of when x=5, as |x-5|=|5-5|=0
\( |y - 5| = 5......y-5=5 \ \ y=10........5-y=5 \ \ y=0\).....Thus, (5,0) and (5,10)
Now, let us look for the value of when y=5, as |y-5|=|5-5|=0
\( |x - 5| = 5......x-5=5 \ \ x=10........5-x=5 \ \ x=0\).....Thus, (0,5) and (10,5)
So, we will have a RHOMBUS with diagonals of 10 ( black and blue). Figure attached
Attachment:
new1.png [ 25.96 KiB | Viewed 4621 times ]
