Using Pythagorean Theorem, we can solve for the length of leg x:

x² + 2² = (√8)² = 8

Now we solve for x:

x² + 4 = 8

x² = 8 – 4

x² = 4

x = 2

Use the Pythagorean Theorem. The sum of both legs squared equals the hypotenuse squared. 

The length of segment is

Note that triangles  and  are both special, 30-60-90 right triangles. Looking specifically at triangle , because we know that segment has a length of 4, we can determine that the length of segment is 2 using what we know about special right triangles. Then, looking at triangle  now, we can use the same rules to determine that segment has a length of

which simplifies to .

In this case, we are already given the length of the hypotenuse of the right triangle, but the Pythagorean formula still helps us. Plug and play, remembering that  must always be the hypotenuse:

 State the theorem.

 Substitute your variables.

 Simplify.

Thus, the ramp covers  of horizontal distance.

Similar triangles are proportional.

Base₁ / Height₁ = Base₂ / Height₂

6 / 9 = 20 / Height₂

Cross multiply  and solve for Height₂

6 / 9 = 20 / Height₂

6 * Height₂=  20 * 9

Height₂=  30

The points define a 3-4-5 right triangle.  Its area is A = 1/2bh = ½(3)(4) = 6.  The scale factor (SF) of the new triangle is 3.  The area of the new triangle is given by A_new = (SF)² x (A_old) =

3² x 6 = 9 x 6 = 54 square units (since the units are not given in the original problem).

NOTE:  For a volume problem:  V_new = (SF)³ x (V_old).

Start by drawing out the light poles and their shadows.

In this case, we end up with two similar triangles. We know that these are similar triangles because the question tells us that these poles are on a flat surface, meaning angle B and angle E are both right angles. Then, because the question states that the shadow cast by both poles are at the same time of day, we know that angles C and F are equivalent. As a result, angles A and D must also be equivalent.

Since these are similar triangles, we can set up proportions for the corresponding sides.

Now, solve for  by cross-multiplying.

Recall that from any vertex of an equilateral triangle, you can drop a height that is a bisector of that vertex as well as a bisector of the correlative side. This gives you the following figure:

Notice that the small triangles within the larger triangle are both  triangles. Therefore, you can create a ratio to help you find .

The ratio of the small base to the height is the same as .  Therefore, you can write the following equation:

This means that .

Now, the area of a triangle can be written:

, and based on our data, we can replace  with .  This gives you:

Now, let's write that a bit more simply:

Solve for . Begin by multiplying each side by :

Divide each side by :

Finally, take the square root of both sides. This gives you . Therefore, the perimeter is .

An equilateral triangle has 3 equal length sides.

Therefore the perimeter equation is as follows,

.

So divide the perimeter by 3 to find the length of each side.

Thus the answer is:

Since the triangle is equilateral, the base and the legs are equal, so the first step is to set the two equations equal to each other. Start with , add  to both sides giving you . Subtract  from both sides, leaving . Finally divide both sides by , so you're left with . Plug  back in for   into either of the equations so that you get a side length of . To find the perimeter, multiply the side length , by , giving you .