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08 Jan 2020, 01:00
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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:
65%
(hard)
Question Stats:
67% (02:57) correct
33%
(02:30)
wrong
based on 102
sessions
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The following figure shows line\( l: y = 2x\). If the area of \(BOC\) is \(8\), what is the area of triangle \(AOB\)?
1.8ps.png [ 7.31 KiB | Viewed 3109 times ]
A. \(4\)
B. \(6\)
C. \(8\)
D. \(11\)
E. \(13\)
Attachment:
1.8ps.png [ 7.31 KiB | Viewed 3109 times ]
A. \(4\)
B. \(6\)
C. \(8\)
D. \(11\)
E. \(13\)
08 Jan 2020, 14:29
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Attachment:
PS.png [ 5.21 KiB | Viewed 2978 times ]
area of triangle BOC = \(\frac{1}{2}\)*BX*OC
8 = \(\frac{1}{2}\)*BX*4
BX = 4
now Y-coordinate of point B = 4
X-coordinate ---putting the value of y in equation y= 2x
we get X-coordinate = 2---OX = 2
now in triangle ACO and BCX
\(\frac{CO}{AO}\) = \(\frac{CX}{BX}\) (since both triangles are similar)
\(\frac{4}{AO}\) = \(\frac{2}{4}\)
AO = 8
area of triangle ACO = \(\frac{1}{2}\)*4*8 = 16
area of triangle AOB = area of triangle ACO - area of triangle BOC
area of triangle AOB = 16 - 8 = 8
C is the answer
10 Jan 2020, 02:04
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Since the area of the triangle \(OBC\) is \(8\) and the length of the base \(OC\) is \(4\), and the area of a triangle = \(\frac{1}{2}bh\), we have:
\(8 = \frac{1}{2}(4)h\)
\(8 = 2h\)
\(h = 4.\)
The height of the triangle is \(4\), and the y-coordinate of point \(B\) is \(4.\)
Since point \(B\) is on the line \(y = 2x\), we have \(4 = 2x\) or \(x = 2.\)
Then we have point \(B(2, 4).\)
We can then determine the equation of the line passing through \(B (2, 4)\) and \(C (4, 0).\) We first determine the slope of the line using the equation
(y1 - y2) / (x1 - x2)
\(\frac{(4 - 0) }{ (2 - 4)}\)
\(\frac{4 }{ -2}\)
Slope = \(m = -2.\)
Then, using the point \((4, 0)\) and \(m = 4\) we get:
\(y = mx + b\)
\(0 = -2(4) + b\)
\(b = 8\)
The line passing through \(B\) and \(C\) is \(y = -2x + 8.\)
Point \(A\) has an x-coordinate of \(0\), so \(y = -2(0) + 9\), or \(y = 8.\)
Then we have point \(A(0,8).\)
The triangle \(AOB\) has base \(OA = 8\), and the height equals \(2\) since the x-coordinate of \(B\) is \(2.\) Then the area of triangle \(OAB\) is \((\frac{1}{2})*8*2 = 8.\)
Therefore, C is the answer.
Answer: C
Since the area of the triangle \(OBC\) is \(8\) and the length of the base \(OC\) is \(4\), and the area of a triangle = \(\frac{1}{2}bh\), we have:
\(8 = \frac{1}{2}(4)h\)
\(8 = 2h\)
\(h = 4.\)
The height of the triangle is \(4\), and the y-coordinate of point \(B\) is \(4.\)
Since point \(B\) is on the line \(y = 2x\), we have \(4 = 2x\) or \(x = 2.\)
Then we have point \(B(2, 4).\)
We can then determine the equation of the line passing through \(B (2, 4)\) and \(C (4, 0).\) We first determine the slope of the line using the equation
(y1 - y2) / (x1 - x2)
\(\frac{(4 - 0) }{ (2 - 4)}\)
\(\frac{4 }{ -2}\)
Slope = \(m = -2.\)
Then, using the point \((4, 0)\) and \(m = 4\) we get:
\(y = mx + b\)
\(0 = -2(4) + b\)
\(b = 8\)
The line passing through \(B\) and \(C\) is \(y = -2x + 8.\)
Point \(A\) has an x-coordinate of \(0\), so \(y = -2(0) + 9\), or \(y = 8.\)
Then we have point \(A(0,8).\)
The triangle \(AOB\) has base \(OA = 8\), and the height equals \(2\) since the x-coordinate of \(B\) is \(2.\) Then the area of triangle \(OAB\) is \((\frac{1}{2})*8*2 = 8.\)
Therefore, C is the answer.
Answer: C
