cannot be the side lengths of a right triangle.does not equal. Also, special right triangleandrules can eliminate all the other choices.

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

x²+ 6²= 10²

Now we solve forx:

x²+ 36 = 100

x²= 100 – 36

x²= 64

x= 8

Also note that this is proportionally a 3/4/5 right triangle, which is very common. Always look out for a side-to-hypoteneuse ratio of 3/5 or 4/5, or a side-to-side ratio of 3/4, in any right triangle, so that you may solve such triangles rapidly.

Using the pythagorean theorem, 8²=7²+x². Solving for x yields the square root of 15, which is 3.9

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

x²+ 2²= (√8)²= 8

Now we solve forx:

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 segmentis

Note that trianglesandare both special, 30-60-90 right triangles. Looking specifically at triangle, because we know that segmenthas a length of 4, we can determine that the length of segmentis 2 using what we know about special right triangles. Then, looking at trianglenow, we can use the same rules to determine that segmenthas 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 thatmust always be the hypotenuse:

State the theorem.

Substitute your variables.

Simplify.

Thus, the ramp coversof 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)²x (A_old) =

3²x 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)³x (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 forby cross-multiplying.