99 of 100: Penultimate Puzzle

3 x 3 = 9

Let's extend the two sides of the big square through A. Then we get a cross with A in the middle that cuts the small square into 4 congruent pieces (rotated by multiples of 90° around A), one of them being the green one. The area is then a quarter of the small square, so 9.

(or)

The location of point B is quite irrelevant. Extending the sides of the blue square shows that the yellow square is divided into four congruent part, each with area $\frac{6^2}4 = \boxed{9}.$

The green area is made up of two right triangles, DBC and ABD.

We know the base and height of DBC, so its area is $\frac{5}{2}$ \ .

The line BD, the shared hyponenteuse, is sqrt(26). As ABD is a right isosceles triangle, we know AB=AD= sqrt (13).

Area of ABD is $\frac{13}{2}$ , So the entire green area is $\frac{5}{2}$ + $\frac{13}{2}$ = $\frac{18}{2}$ = 9.

Simply rotate the blue square so the sides pass through the corners of the yellow square. Then it's clear that the green section is 1/4 of the yellow square = (6*6)/4 = 9

Bisect the yellow square. The divides the quadrilateral into a trapezoid and a right triangle. Since the point is at the center of the square, the legs of the triangle are 3 and 2. Also, the angle on this line is complementary to the other angle in the triangle and the adjacent angle in the trapezoid. This means that the angles are congruent, so the trapezoid is part of a similar triangle, where one of the side lengths is 3. The other leg, by proportions, is 4.5. The side of the trapezoid is 3 units down, which is 1/3 of the way upwards, so its length is 1/3 of 3, which is 1. This means that the bases of the trapezoid are 1 and 3, and the height is 3. The sum of the areas of the triangle and the trapezoid are therefore 2•3/2+3(1+3)/2, or 3+3•4/2 or 3+3•2 or 3+6 or 9.

For any two similar squares attached in the same way, such that the side length of the larger square is greater than AC, and given any angle of rotation, the quadrilateral formed will have an area one quarter the area of the smaller square.

If you imagine rotating the blue triangle from point A, you get a green square with side length 3, and if you solve for that area, you get 9.

Let's label the fourth point of the quadrilateral D. The area of $ABCD = ABC + ACD = \frac{5\cdot 3}{2} + \frac{3\cdot |CD|}{2} = \frac{3(5+|CD|)}{2}$ . And because the answer is an integer, $|CD+5|$ must be divisible by two. This means that $|CD|$ is 1, 3 or 5 because otherwise it would go beyond bounds. A good guess that it's the smallest of the three, so the answer is $\frac{3(5+1)}{2} = 9$

xD

green region has two triangles. the smaller one is equal with the triangle in yellow region under the digonal and above the side of the bigger aquare.

=> similar triangles with two angles and the side in between equal as the half of the diagonal of the small square.

this results in base of 1 with height 3 => 1st triangle inside the green region. the other triangle is a simple triangle with height 3 and base 5.

area: 9

Slope of AB= 2/3, So that of AB= 3/2. Area= upper triangle+ lower rectangle + lower triangle = 1/2 3 2+ 3 1+ 1/2 2*3 = 3+3+3 =9

If you consider the rotation of the blue square about point A, there are two configurations that would obviously make the green are 1/4 of the smaller square- 9, the point when the sides are perpendicular and the point when the the blue squares sides intersect the vertices. Noticing this leads us to consider if the area changes for any rotation, and drawing in a horizontal and vertical line through point A, it is simple to consider that for any rotation the small yellow triangle will be congruent to the small green triangle. Thus any rotation covers 1/4 of the smaller square- 9 (also true for any blue square with sides longer than half the yellow's diagonal).

To throw some coordinate geometry at the problem: if we center the plane at $A=(0,0)$ , such that the smaller square has horizontal and vertical sides, then $B=(3,2)$ and $C=(3,-3)$ . Let $D$ be the fourth corner of the green quadrilateral. Then since $\overline{AD}\perp\overline{AB}$ , the slopes of these two lines are negative reciprocals. Thus $D=(2,-3)$ . Now we have enough information to determine the area of the quadrilateral: $Area(\square ABCD) = Area(\triangle ABC) + Area(\triangle ABD) = 0.5*3*5 +0.5*3*1 = 9$

This generalizes: if $B = (3,x)$ then $D=(x,-3)$ then the area of the quadrilateral is $Area(\square ABCD) = Area(\triangle ABC) + Area(\triangle ABD) = 0.5*3*(3+x) +0.5*3*(3-x) = 9$

Set the coordinate axes such that the coordinates of the square (-3. -3),

C (3, -3), (3, 3), and (-3, 3). Therefore the center of this square lies at (0, 0).

Then A (0, 0) and B (3, 2). The slope of AB is:

m = $\frac{2}{3}$ .

The slope of the side of the largest square is m’ = - $\frac{3}{2}$ and theequation of this side that passes through A is y = - $\frac{3}{2}$ x. So if we

set y = -3, then x = 2. Therefore the the side of the largest square intersectsthe smallest square at D (2, -3). Then, by distance formula:

AB = $\sqrt{13}$ , AD = $\sqrt{13}$ , and CD = 1

Then the area of polygon ABCD = ½[ AB\times AD + BC \times CD] = ½(13 + 5) = 9.