+1005 We can eliminate choice 50%, since BD splits the square in half, but the shaded doesn't fill it up.
So our answer is less than 50. We remove the right traingle at the bottom which has an area of 4*1/2 = 2. 2/16 = 1/8
That's another 12.5%.
Oops we missed the small one! We over counted that one.
First, 50-12.5 = 37.5, but we need to add that, and the size appears to be 2.5%, since it's a similar triangle of the shaded, but like 16 times smaller
also to clarify, the small triangle is probably 2.5% since it's probably no more than a third of the right triangle there, but if it's even 1 third, it would be 4 percent and STILL around 40.
I'm very sure it's 40 but it's more of a gut feeling than hard proof.
+129899 Correct, Hugo...."gut feelings" are good......except if you're hungry !!!!
+1005 Haha, but mine was more of guesswork and your answer is definitely the best.
+129899 Call the intersection of BD and AQ , M
Then triangle BMA is similar to triangle DMQ
Since the base of BMA is 4 times that of DMQ.....so is its altitude
Therefore
Altitude BMA + Altitude of DMQ = 4
Alltitude of BMA + (1/4)Altitude of BMA = 4
(5/4) Altitude BMA = 4
Altitude of BMA = ( 4 *4/5) = 16/5
Area of BMA = (1/2)(4)(16/5) = 32/5
Ara of the square = 16
(32/5) / 16 =
32 / 80 =
2 / 5 =
40%
+1005 I'm not going to pretend I know what that means but that looks like good proof there!
