Geom is fun... :)

We note $AB // MC \rightarrow \angle ABP = \angle BMC$ and $\angle APB = \angle BCM$ ; due to the AA similarity criterion,

$\Delta ABP \sim \Delta BMC$

By Pythagoras' Theorem,

$BM = \sqrt{CM^{2}+CB^{2}} = \sqrt{(\frac{1}{2})^{2}+1^{2}} = \frac{\sqrt{5}}{2}$

Using the properties of similarity,

$PB = \frac{AB}{BM} \times CM = \frac{1}{\sqrt{5}/2} \times \frac{1}{2} = \frac{1}{\sqrt{5}}$

$PA = \frac{AB}{BM} \times CB = \frac{1}{\sqrt{5}/2} \times 1 = \frac{2}{\sqrt{5}}$

The area of $\Delta ABP$ and $\Delta BMC$ are, respectively,

$\Delta_{ABP} = \frac{1}{2} \times PB \times PA = \frac{1}{2} \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} = \frac{1}{5}$

$\Delta_{BMC} = \frac{1}{2} \times CM \times CB = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$

Following that, we evaluate the area of the square:

$A_{square} = s^{2} = 1^{2} = 1$

Finally, the shaded region is then the square minus the two triangles:

$A_{shaded region} = A_{square}-\Delta_{ABP}-\Delta_{BMC} = 1-\frac{1}{5}-\frac{1}{4}$

$= \boxed{\frac{11}{20}}$

Find the areas of both triangles and the square then subtract it from the sum of the areas of triangles. ........that's how I got it

BM = (square root of 5 )/2

area of triangle ABM = 0.5 BM.AP

AP = 2/(square root of 5)

PB = 1/(square root of 5)

area of triangle APB = 1/5

area of triangle BCM = 1/4

area of the shaded region = 1 - 1/4 - 1/5 = 11/20

CM =1 so area of triangle CMB is 1/4. Area of square is 1. Now area of ABCD- Area of CMB is 3/4. Therefore area of shaded region is less than 3/4. If you see the last 2 answer choices are greater than 3/4 . Eliminate answer choice D and E. Visually we can see PB is almost equal to CM and lets assume PB =1/2. Now using pytho we can find area of triangle APB is approx (sqrt3)/8 . so shaded area = 3/4 -(sqrt3/8) = approx half which is answer choice A :-) Not the most elegant way but fastest as far as I know.

There's a much better way to do it, but since I don't remember much geometry, I used repeated applications of law of sines to find the exact values of BP and PA. Eventually, you get $1-\frac{1}{4}-\frac{1}{5}=\frac{11}{20}$

M is midpoint so CM = 1/2(CD)=1/2 (CB) hence angle CMB = 60 degrees angle CBP = 30 degrees angle CMB = 60 degrees Angle PBA = 60 degrees Area of triangle BPMCB = 1/2 (CM x CB) = 1/4(CB x CB) = 1/4 units

Sin(theta)=BP/BA Now Cos (theta)=PA/BA

Are of Triangle BPAB = 1/2 (BP xPA)= 1/2 ( √3/2 x 1/2)= √3/8 units

shaded area = 1 sq unit - ((1/4)+√3/8 ) = 11/20 units

Let N be the point in BM...

So, BM = V(1+1/4) = V(5/4) = (V5)/2

So, let BN = x

Then, NM = [(V5)/2] - x

AN = V(1 - x^2)

So,

a^2 + b^2 = c^2

(V(1-x^2))^2 + (((V5)/2)-x))^2 = (V(1+1/4))^2

(1-x^2) + (5/4 - x(V5) + x^2) = 5/4

1 = x(V5)

(V5)/5 = x

MN = 3(V5)/10

AN = 2(V5)/5

So, area of shaded region = Area of (ANM) + Area of (AMD)...

= [3(V5)/10 * 2(V5)/5]/2 + [1/2]/2

= (3/5 + 1/2)/2

= (11/10)/2

= 11/20 sq. Units...

could be done with options

It is taking long time to solve this.