One important concept to recognize in this problem is that the triangles ABD and ECD are similar. Each has a right angle and they share the same angle at point D, meaning that their third angles (BAD and CED, the angles at the upper left of each triangle) must also have the same measure.

With that knowledge, you can use the given side lengths to establish a ratio between the side lengths of the triangles. If BC is 2 and CD is 8, that means that the bottom side of the triangles are 10 for the large triangle and 8 for the smaller one, or a 5:4 ratio.

You’re then told the area of the larger triangle. Knowing that the area is 25 and that area =  Base x Height, you can plug in 10 as the base and determine that the height, side AB, must be 5.

Since you know that the smaller triangle’s height will be  the length of 5, you can then conclude that side EC measures 4, and that is your right answer.

As you unpack the given information, a few things should stand out:

1. Triangles ABC and ADE are similar. They each have a right angle and they each share the angle at point A, meaning that their lower-left-hand angles (at points B and D) will be the same also since all angles in a triangle must sum to 180.  

2. Side BC has to measure 6, as you’re given one side (AC = 8) and the hypotenuse (AB = 10) of a right triangle. You can use Pythagorean Theorem to solve, or you can recognize the 3-4-5 side ratio (which here amounts to a 6-8-10 triangle). 

3. This then allows you to use triangle similarity to determine the side lengths of the large triangle. Since the hypotenuse is 20 (segments AB and BD, each 10, combine to form a side of 20) and you know it’s a 3-4-5 just like the smaller triangle, you can fill in side DE as 12 (twice the length of BC) and segment CE as 8.

So you now know the dimensions of the parallelogram: BD is 10, BC is 6, CE is 8, and DE is 12. The sum of those four sides is 36.

If the perimeter of triangle ABC is twice as long as the perimeter of triangle DEF, and you know that the triangles are similar, that then means that each side length of ABC is twice as long as its corresponding side in triangle DEF. That also means that the heights have the same 2:1 ratio: the height of ABC is twice the length of the height of DEF.

Since the formula for area of a triangle is  Base x Height, you can express the area of triangle DEF as bh and the area of ABC as .  You’ll then see that the areas of ABC to DEF are  and bh, for a ratio of 4:1.

NOTE: It can seem surprising that the ratio isn’t 2:1 if each length of one triangle is twice its corresponding length in the other. But keep in mind that for an area you multiply two lengths together, and go from a unit like “inches” to a unit like “square inches.”  Because each length is multiplied by 2, the effect is exacerbated.

In beginning this problem, it is important to note that the two triangles pictured, ABC and CED, are similar. They each have a right angle and they share the vertical angle at point C, meaning that the angles at A and D must also be congruent and therefore the triangles are similar.

This means that their side lengths will be proportional, allowing you to answer this question. You’re asked to match the ratio of AB to AC, which are the side across from angle C and the hypotenuse, respectively.  In triangle CED, those map to side ED and side CD, so the ratio you want is ED:CD.

The formula for the equation of a circle is:

Where (h,k) is the center of the circle.

 and 

and diameter = 6 therefore radius = 3

The general equation of a circle is , where (h, k) represents the location of the circle's center, and r represents the length of its radius. 

Circle A first has the equation of . This means that its center must be located at (4, –3), and its radius is .

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is .

Now, that we have circle A's new center and radius, we can write its general equation using .

.

.

The answer is .

We need to expand this equation to  and then complete the square.

This brings us to .

We simplify this to .

Thus the radius is 7.

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

This radical cannot be reduced further.

Recall that the general form of the equation of a circle centered at the origin is:

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

Now, the question asks for the positive y-coordinate when .  To solve this, simply plugin for :

Since our answer will be positive, it must be .

Remember that the general equation for a circle with center  and radius  is . 

With that in mind, our original center is at  .

If we move the center  units to the left, that means that we are subtracting  from our given coordinates. 

Therefore, our new center is .