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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:
75%
(hard)
Question Stats:
50% (02:32) correct
50%
(02:47)
wrong
based on 52
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In the figure below, what is the area of the region shown?
(A) \(8 \sqrt{3}\)
(B) 16
(C) \(12 + 4 \sqrt{3}\)
(D) \(8 + 8 \sqrt{3}\)
(E) \(16 + 8 \sqrt{3}\)
2020-12-22_15-58-07.png [ 18.4 KiB | Viewed 2261 times ]
(A) \(8 \sqrt{3}\)
(B) 16
(C) \(12 + 4 \sqrt{3}\)
(D) \(8 + 8 \sqrt{3}\)
(E) \(16 + 8 \sqrt{3}\)
Attachment:
2020-12-22_15-58-07.png [ 18.4 KiB | Viewed 2261 times ]
29 Dec 2020, 07:39
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Consider \(\triangle ABC\), \(AB^2=AC^2-BC^2=8^2-4^2=48\)
\(AB=4\sqrt{3}\)
Area of \(\triangle ABC =\frac{1}{2}*BC*AB=\frac{1}{2}*4\sqrt{3}*4=8\sqrt{3}\)
We can find the \(\angle BCA = 30^{\circ}\) and \(\angle ACB=60^{\circ}\) using \(\frac{PPB}{HHB}\). Therefore, \(\angle CAD=60^{\circ}\) and \(\angle DCA=30^{\circ}\) which makes both \(\triangle ABC \approx \triangle ADC\) (similar).
Area of \(\triangle ADC =8\sqrt{3}\)
But we don't have any such option \(16\sqrt{3}\).
So, the \(\triangle ADC\) is an isoceles \(\triangle\), therefore, \(AD=DC\)
\(AC^2=AD^2+DC^2=2AD^2=>AD=4\sqrt{2}\)
Area of \(\triangle ADC = \frac{1}{2}*4\sqrt{2}*4\sqrt{2}=16\)
\(Total-Area = 16+8\sqrt{3}\)
IMO Ans E
\(AB=4\sqrt{3}\)
Area of \(\triangle ABC =\frac{1}{2}*BC*AB=\frac{1}{2}*4\sqrt{3}*4=8\sqrt{3}\)
We can find the \(\angle BCA = 30^{\circ}\) and \(\angle ACB=60^{\circ}\) using \(\frac{PPB}{HHB}\). Therefore, \(\angle CAD=60^{\circ}\) and \(\angle DCA=30^{\circ}\) which makes both \(\triangle ABC \approx \triangle ADC\) (similar).
Area of \(\triangle ADC =8\sqrt{3}\)
But we don't have any such option \(16\sqrt{3}\).
So, the \(\triangle ADC\) is an isoceles \(\triangle\), therefore, \(AD=DC\)
\(AC^2=AD^2+DC^2=2AD^2=>AD=4\sqrt{2}\)
Area of \(\triangle ADC = \frac{1}{2}*4\sqrt{2}*4\sqrt{2}=16\)
\(Total-Area = 16+8\sqrt{3}\)
IMO Ans E
29 Dec 2020, 13:13
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Area of Triangle ABC = 1/2 *BC*AB = 1/2*4*4Sqt3= 8*Sqrt(3)
Now, we have Option D and E to pickup.
AC=8 so, AD+DC >AC=8, So, both sides AD and DC will be greater than 4.
Hence, Total Area = 8*Sqrt(3) + 1/2 * (>4)*(>4)
so, Option E.
Now, we have Option D and E to pickup.
AC=8 so, AD+DC >AC=8, So, both sides AD and DC will be greater than 4.
Hence, Total Area = 8*Sqrt(3) + 1/2 * (>4)*(>4)
so, Option E.
