LUANDATO wrote:
Given BD perpendicular to AC, what is the area of triangle BCD?
1) Angle BAD=69°
2) BD=CD
The OA is
B.
I'm really confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
The area of rightangled triangle BCD = 1/2*BD*CD
1) Angle BAD = 69°
No information about BD and CD. Insufficient.
2) BD = CD
Since the traingle ABD is a rightangled traingle, we have BD^2 = AB^2 - AD^2 => BD^2 = 11^2 - 4^2 = 121 - 16 = 105.
We know that the area of rightangled triangle BCD = 1/2*BD*CD = 1/2*BD^2; since BD = CD
The area of the rightangled traingle BCD = 1/2*BD^2 = 1/2*105 = 105/2. Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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