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 \(\displaystyle \overline{BD}\) is \(\displaystyle \frac{4\sqrt{3}}{3}\)

Note that triangles \(\displaystyle \dpi{100} \small ACD\) and \(\displaystyle \dpi{100} \small BCD\) are both special, 30-60-90 right triangles. Looking specifically at triangle \(\displaystyle \dpi{100} \small ACD\), because we know that segment \(\displaystyle \overline{AD}\) has a length of 4, we can determine that the length of segment \(\displaystyle \overline{CD}\) is 2 using what we know about special right triangles. Then, looking at triangle \(\displaystyle \dpi{100} \small BCD\) now, we can use the same rules to determine that segment \(\displaystyle \overline{BD}\) has a length of \(\displaystyle \frac{4}{\sqrt{3}}\)

which simplifies to \(\displaystyle \frac{4\sqrt{3}}{3}\).

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 \(\displaystyle c\) must always be the hypotenuse:

\(\displaystyle a^2 + b^2 = c^2\) State the theorem.

\(\displaystyle a^2 = 32^2 - 4^2\) Substitute your variables.

\(\displaystyle a^2 = 1008\) Simplify.

\(\displaystyle a = \sqrt{1008} = \sqrt{144\cdot 7} = 12\sqrt7\)

Thus, the ramp covers \(\displaystyle 12\sqrt7\textup{ feet}\) 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.

\(\displaystyle \frac{36}{x}=\frac{9}{6}\)

Now, solve for \(\displaystyle x\) by cross-multiplying.

\(\displaystyle 9x=216\)

\(\displaystyle x=24\)

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 \(\displaystyle 30-60-90\) triangles. Therefore, you can create a ratio to help you find \(\displaystyle h\).

The ratio of the small base to the height is the same as \(\displaystyle 1:\sqrt{3}\).  Therefore, you can write the following equation:

\(\displaystyle \frac{0.5b}{h}=\frac{1}{\sqrt{3}}\)

This means that \(\displaystyle 0.5b\sqrt{3}=h\).

Now, the area of a triangle can be written:

\(\displaystyle A=\frac{1}{2}bh\), and based on our data, we can replace \(\displaystyle h\) with \(\displaystyle \frac{b\sqrt{3}}{2}\).  This gives you:

\(\displaystyle 56.25\sqrt{3}=\frac{1}{2}*b*\frac{b\sqrt{3}}{2}=\frac{b^2\sqrt{3}}{4}\)

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

\(\displaystyle 56.25\sqrt{3}=\frac{b^2\sqrt{3}}{4}\)

Solve for \(\displaystyle b\). Begin by multiplying each side by \(\displaystyle 4\):

\(\displaystyle 225\sqrt{3}=b^2\sqrt{3}\)

Divide each side by \(\displaystyle \sqrt3\):

\(\displaystyle b^2 = 225\)

Finally, take the square root of both sides. This gives you \(\displaystyle b=15\). Therefore, the perimeter is \(\displaystyle 15*3 = 45\).

An equilateral triangle has 3 equal length sides.

Therefore the perimeter equation is as follows,

\(\displaystyle P=3s\).

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

Thus the answer is:

\(\displaystyle 48=3s\)

\(\displaystyle \frac{48}{3}=s\)

\(\displaystyle 16 cm\)

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 \(\displaystyle 5x-1=3x+13\), add \(\displaystyle 1\) to both sides giving you \(\displaystyle 5x=3x+14\). Subtract \(\displaystyle 3x\) from both sides, leaving \(\displaystyle 2x=14\). Finally divide both sides by \(\displaystyle 2\), so you're left with \(\displaystyle x=7\). Plug \(\displaystyle 7\) back in for \(\displaystyle x\)  into either of the equations so that you get a side length of \(\displaystyle 34\). To find the perimeter, multiply the side length \(\displaystyle \left ( 34\right )\), by \(\displaystyle 3\), giving you \(\displaystyle 102\).