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05 Nov 2019, 01:58
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Difficulty:
65%
(hard)
Question Stats:
67% (03:10) correct
33%
(03:20)
wrong
based on 217
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If side BE has length 10 and side AC has length 8, what is the area of the triangle BOC ?
A. \(2\sqrt{3}\)
B. \(4\sqrt{3}\)
C. \(6\sqrt{3}\)
D. \(8\sqrt{3}\)
E. \(12\sqrt{3}\)
Are You Up For the Challenge: 700 Level Questions
Attachment:
triangle BOC.JPG [ 8.97 KiB | Viewed 6167 times ]
firas92
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05 Nov 2019, 05:19
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\(ABC\) and \(BDE\) are both \(30-60-90\) right triangles and so their sides should be in the ratio \(1:\sqrt{3}:2\)
Therefore we have \(AC=8, BC=4, AB=4\sqrt{3}\) and \(BE=10, DE=5, BD=5\sqrt{3}\)
Draw \(OF\) perpendicular to \(BD\)
We now have 2 pairs of similar triangles
1. \(ABC\) & \(OFC\)
2. \(BDE\) & \(BOF\)
From \(ABC\) & \(OFC\)
\(\frac{OF}{FC}=\frac{4\sqrt{3}}{4}\) or \(OF=FC\sqrt{3}\)
From \(BDE\) & \(BOF\)
\(\frac{OF}{BF}=\frac{5}{5\sqrt{3}}\) or \(OF=\frac{BF}{\sqrt{3}}\)
So, \(FC\sqrt{3}=\frac{BF}{\sqrt{3}}\) or \(BF=3FC\)
We also know that \(BF+FC=BC=4\)
So, \(3FC+FC=4\) and \(FC=1\)
Therefore \(OF=\sqrt{3}\)
Area of \(BOC\)\(=\frac{1}{2}*BC*OF=\frac{1}{2}*4*\sqrt{3} = 2\sqrt{3}\)
Answer is (A)
Therefore we have \(AC=8, BC=4, AB=4\sqrt{3}\) and \(BE=10, DE=5, BD=5\sqrt{3}\)
Draw \(OF\) perpendicular to \(BD\)
We now have 2 pairs of similar triangles
1. \(ABC\) & \(OFC\)
2. \(BDE\) & \(BOF\)
From \(ABC\) & \(OFC\)
\(\frac{OF}{FC}=\frac{4\sqrt{3}}{4}\) or \(OF=FC\sqrt{3}\)
From \(BDE\) & \(BOF\)
\(\frac{OF}{BF}=\frac{5}{5\sqrt{3}}\) or \(OF=\frac{BF}{\sqrt{3}}\)
So, \(FC\sqrt{3}=\frac{BF}{\sqrt{3}}\) or \(BF=3FC\)
We also know that \(BF+FC=BC=4\)
So, \(3FC+FC=4\) and \(FC=1\)
Therefore \(OF=\sqrt{3}\)
Area of \(BOC\)\(=\frac{1}{2}*BC*OF=\frac{1}{2}*4*\sqrt{3} = 2\sqrt{3}\)
Answer is (A)
05 Nov 2019, 05:39
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Triangle ACB is 30-60-90 triangle
AC=8
hence, BC=8/2=4
Triangle BOC is 30-60-90
Hence area of BOC= \(\frac{1}{2}*2*2\sqrt{3}\)= \(2\sqrt{3}\)
AC=8
hence, BC=8/2=4
Triangle BOC is 30-60-90
Hence area of BOC= \(\frac{1}{2}*2*2\sqrt{3}\)= \(2\sqrt{3}\)
Bunuel
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