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by neerajkumar1_1 » Fri Jun 18, 2010 6:29 am
Image


In the figure, what is the area of triangular region BCD ?
A. 4 root2
B. 8
C. 8 root2
D. 16
E. 16 root2


OA:C
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by kvcpk » Fri Jun 18, 2010 6:43 am
neerajkumar1_1 wrote:Image


In the figure, what is the area of triangular region BCD ?
A. 4 root2
B. 8
C. 8 root2
D. 16
E. 16 root2


OA:C
BD = 4 sqrt(2)

Area of BCD = 1/2 * 4 * 4sqrt(2)
= 8 root(2)
C
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by amising6 » Fri Jun 18, 2010 7:53 am
kvcpk wrote:
neerajkumar1_1 wrote:Image


In the figure, what is the area of triangular region BCD ?
A. 4 root2
B. 8
C. 8 root2
D. 16
E. 16 root2


OA:C
in triangle BAD
according to pythogerous theorem
ba^2+ad^2= bd ^2
4^2+4^2=bd^2
root32=bd

now in triangle bcd area=1/2*b*h
=1/2*4*root 32
2 root 32 (root 32 = root (16*2) which is equal to 4root2)
2*4 root2
8 root2 answer



BD = 4 sqrt(2)

Area of BCD = 1/2 * 4 * 4sqrt(2)
= 8 root(2)
C
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by selango » Fri Jun 18, 2010 8:37 am
For 45-45-90 tri sides are in 1:1:sqrt2 ratio.

So BD is 4sqrt2

area of right tri=1/2*(product of sides containing 90)

=1/2*4*4sqrt2=8sqrt2
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