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08 Jun 2020, 11:47
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C
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In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x=0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?
A. 0
B. 25
C. 50
D. 75
E. 100
Are You Up For the Challenge: 700 Level Questions
A. 0
B. 25
C. 50
D. 75
E. 100
Are You Up For the Challenge: 700 Level Questions
08 Jun 2020, 13:41
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2x + |4y| =20
So , 2x + 4y =20 (y>0)---equation 1
2x-4y=20 (y<0)---equation 2
We will have a close bounded area by linear equations x=0 (y-axis) and above 2 equations
equation 1 will intersect y-axis at (0,5) and x-axis at (10,0)
equation 2 will intersect y-axis at (0,-5) and x-axis at (10,0)
So we have isosceles triangle :- Area will be 1/2* base * Height
A(R)= 1/2 * 10 (Base of triangle or distance between coordinate (0,-5) and (0,5) * 10 (Height (distance between (0,0) and (10,0))
A(R)=50 sq unit
Similarly A(P) will be a iscosceles triangle with coordinate (-5,0) ; (5,0) and (0,10)
A(P) will also be =1/2 * 10*10
A(P)= 50 Sq unit
So |A(P)- 2A (R) | = 50
So , 2x + 4y =20 (y>0)---equation 1
2x-4y=20 (y<0)---equation 2
We will have a close bounded area by linear equations x=0 (y-axis) and above 2 equations
equation 1 will intersect y-axis at (0,5) and x-axis at (10,0)
equation 2 will intersect y-axis at (0,-5) and x-axis at (10,0)
So we have isosceles triangle :- Area will be 1/2* base * Height
A(R)= 1/2 * 10 (Base of triangle or distance between coordinate (0,-5) and (0,5) * 10 (Height (distance between (0,0) and (10,0))
A(R)=50 sq unit
Similarly A(P) will be a iscosceles triangle with coordinate (-5,0) ; (5,0) and (0,10)
A(P) will also be =1/2 * 10*10
A(P)= 50 Sq unit
So |A(P)- 2A (R) | = 50
yashikaaggarwal
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08 Jun 2020, 14:21
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Area of Region R = Area of ∆ABC = 1/2*base*height
Base = distance between coordinate Y axis to Y' axis.
Height = distance between origin and X coordinate on x axis
Putting values of x and y zero in equation 2x + |4y| = 20
One by one.
X = 0
|4y| = 20
+4y = 20 => y = 5
-4y = 20 => y = -5
Distance between 5&-5 = 5+5 = 10 = Base.
Y = 0
2x + |4y| = 20
2x = 20
X = 10 = height
Area of region R = 1/2*10*10 = 50 -------(1)
Area of Region P = Area of ∆PQR = 1/2*base*height
Base = distance between coordinate X axis to X' axis.
Height = distance between origin and Y coordinate on Y axis
Putting values of x and y zero in equation |4x| + 2y = 20
One by one.
Y = 0
|4x| = 20
+4x = 20 => x = 5
-4x = 20 => x = -5
Distance between 5&-5 = 5+5 = 10 = Base.
X = 0
2y + |4x| = 20
2y = 20
Y = 10 = height
Area of region P = 1/2*10*10 = 50 ----------(2)
Putting values from equation 1&2
The value of |A(P) – 2A(R)|
=>|50-2(50)| => |50-100|
=> 50
Answer is C
Posted from my mobile device
Base = distance between coordinate Y axis to Y' axis.
Height = distance between origin and X coordinate on x axis
Putting values of x and y zero in equation 2x + |4y| = 20
One by one.
X = 0
|4y| = 20
+4y = 20 => y = 5
-4y = 20 => y = -5
Distance between 5&-5 = 5+5 = 10 = Base.
Y = 0
2x + |4y| = 20
2x = 20
X = 10 = height
Area of region R = 1/2*10*10 = 50 -------(1)
Area of Region P = Area of ∆PQR = 1/2*base*height
Base = distance between coordinate X axis to X' axis.
Height = distance between origin and Y coordinate on Y axis
Putting values of x and y zero in equation |4x| + 2y = 20
One by one.
Y = 0
|4x| = 20
+4x = 20 => x = 5
-4x = 20 => x = -5
Distance between 5&-5 = 5+5 = 10 = Base.
X = 0
2y + |4x| = 20
2y = 20
Y = 10 = height
Area of region P = 1/2*10*10 = 50 ----------(2)
Putting values from equation 1&2
The value of |A(P) – 2A(R)|
=>|50-2(50)| => |50-100|
=> 50
Answer is C
Posted from my mobile device
