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03 Feb 2014, 01:13
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C
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Difficulty:
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88% (01:03) correct
12%
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In the figure above, what is the area of triangular region BCD ?
(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)
Attachment:
Untitled.png [ 11.64 KiB | Viewed 36583 times ]
03 Feb 2014, 01:13
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SOLUTION
In the figure above, what is the area of triangular region BCD ?
(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)
Using Pythagorean theorem \(BD=\sqrt{4^2+4^2}=4\sqrt{2}\).
The area of triangle BCD = \(\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}\).
Answer: C.
In the figure above, what is the area of triangular region BCD ?
(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)
Using Pythagorean theorem \(BD=\sqrt{4^2+4^2}=4\sqrt{2}\).
The area of triangle BCD = \(\frac{1}{2}*BD*BC=\frac{1}{2}*4\sqrt{2}*4=8\sqrt{2}\).
Answer: C.
03 Feb 2014, 03:30
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Attachment:
untitled.PNG [ 15.91 KiB | Viewed 32852 times ]
In the figure above, what is the area of triangular region BCD ?
(A) \(4\sqrt{2}\)
(B) 8
(C) \(8\sqrt{2}\)
(D) 16
(E) \(16\sqrt{2}\)
Sol: In the given question Triangle ABD is isosele with sides AB=AD=4 and therefore BD=4\(\sqrt{2}\)
Area of triangle BCD is 1/2*4*4\(\sqrt{2}\)
600 level is okay
Ans is C
