Practice Problems For Special Right Triangles

Let's dive into the world of special right triangles – those geometrical gems that appear frequently in math problems and standardized tests. Mastering these triangles can significantly speed up your problem-solving abilities and boost your overall understanding of trigonometry. This guide offers a comprehensive collection of practice problems designed to solidify your grasp of special right triangles, complete with detailed solutions and explanations.

Understanding Special Right Triangles

Before we jump into the problems, let's briefly review the two types of special right triangles:

• 45-45-90 Triangle: This is an isosceles right triangle, meaning two of its angles are 45 degrees, and the other is 90 degrees. The sides are in the ratio x : x : x√2, where x is the length of each leg, and x√2 is the length of the hypotenuse.

• 30-60-90 Triangle: This right triangle has angles of 30, 60, and 90 degrees. The sides are in the ratio x : x√3 : 2x, where x is the length of the side opposite the 30-degree angle, x√3 is the length of the side opposite the 60-degree angle, and 2x is the length of the hypotenuse.

Practice Problems: 45-45-90 Triangles

Let's start with problems focusing on 45-45-90 triangles.

Problem 1:

In a 45-45-90 triangle, one leg has a length of 7. Find the lengths of the other leg and the hypotenuse.

Solution:

• Since it's a 45-45-90 triangle, both legs are equal in length. Therefore, the other leg also has a length of 7.
• The hypotenuse is x√2, where x is the length of a leg. So, the hypotenuse is 7√2.

Answer: The other leg is 7, and the hypotenuse is 7√2.

Problem 2:

The hypotenuse of a 45-45-90 triangle is 10. Find the length of each leg.

Solution:

• Let x be the length of each leg. We know that the hypotenuse is x√2.
• Therefore, x√2 = 10.
• To solve for x, divide both sides by √2: x = 10/√2.
• Rationalize the denominator by multiplying the numerator and denominator by √2: x = (10√2) / 2 = 5√2.

Answer: Each leg has a length of 5√2.

Problem 3:

A square has a diagonal of length 12. What is the length of each side of the square?

Solution:

• The diagonal of a square divides it into two 45-45-90 triangles. The diagonal acts as the hypotenuse of these triangles.
• Let x be the length of each side of the square (which is also the length of each leg of the 45-45-90 triangle).
• We know x√2 = 12.
• Divide both sides by √2: x = 12/√2.
• Rationalize the denominator: x = (12√2) / 2 = 6√2.

Answer: Each side of the square has a length of 6√2.

Problem 4:

Find the area of a 45-45-90 triangle with a hypotenuse of 8√2.

Solution:

• Let x be the length of each leg. We know x√2 = 8√2.
• Divide both sides by √2: x = 8.
• The area of a triangle is (1/2) * base * height. In this case, both the base and height are 8.
• Area = (1/2) * 8 * 8 = 32.

Answer: The area of the triangle is 32 square units.

Problem 5:

The perimeter of a 45-45-90 triangle is 9 + 9√2. Find the length of each side.

Solution:

• Let x be the length of each leg. The hypotenuse is x√2.
• The perimeter is x + x + x√2 = 9 + 9√2.
• Simplify: 2x + x√2 = 9 + 9√2.
• Factor out x: x(2 + √2) = 9(1 + √2).
• This is a tricky one! Notice if we factor out a √2 from (2 + √2) we get √2(√2 + 1)
• so x√2(√2 + 1) = 9(1 + √2)
• Thus x√2 = 9 or x = 9/√2 = (9√2)/2
• The legs are each (9√2)/2 and the hypotenuse is 9.

Answer: The legs are each (9√2)/2 and the hypotenuse is 9.

Practice Problems: 30-60-90 Triangles

Now, let's tackle problems involving 30-60-90 triangles.

Problem 6:

In a 30-60-90 triangle, the side opposite the 30-degree angle has a length of 4. Find the lengths of the other two sides.

Solution:

• The side opposite the 30-degree angle is x, so x = 4.
• The side opposite the 60-degree angle is x√3, so it's 4√3.
• The hypotenuse is 2x, so it's 2 * 4 = 8.

Answer: The side opposite the 60-degree angle is 4√3, and the hypotenuse is 8.

Problem 7:

The hypotenuse of a 30-60-90 triangle is 14. Find the lengths of the other two sides.

Solution:

• The hypotenuse is 2x, so 2x = 14.
• Therefore, x = 7.
• The side opposite the 30-degree angle is x, so it's 7.
• The side opposite the 60-degree angle is x√3, so it's 7√3.

Answer: The side opposite the 30-degree angle is 7, and the side opposite the 60-degree angle is 7√3.

Problem 8:

In a 30-60-90 triangle, the side opposite the 60-degree angle is 6√3. Find the lengths of the other two sides.

Solution:

• The side opposite the 60-degree angle is x√3, so x√3 = 6√3.
• Divide both sides by √3: x = 6.
• The side opposite the 30-degree angle is x, so it's 6.
• The hypotenuse is 2x, so it's 2 * 6 = 12.

Answer: The side opposite the 30-degree angle is 6, and the hypotenuse is 12.

Problem 9:

Find the area of a 30-60-90 triangle where the side opposite the 30-degree angle is 5.

Solution:

• The side opposite the 30-degree angle is x, so x = 5.
• The side opposite the 60-degree angle is x√3, so it's 5√3.
• The area of a triangle is (1/2) * base * height. We can use the sides opposite the 30- and 60-degree angles as the base and height.
• Area = (1/2) * 5 * 5√3 = (25√3) / 2.

Answer: The area of the triangle is (25√3) / 2 square units.

Problem 10:

The perimeter of a 30-60-90 triangle is 15 + 5√3. Find the length of each side.

Solution:

• Let x be the length of the side opposite the 30-degree angle.
• The side opposite the 60-degree angle is x√3, and the hypotenuse is 2x.
• The perimeter is x + x√3 + 2x = 15 + 5√3.
• Simplify: 3x + x√3 = 15 + 5√3.
• Factor out x: x(3 + √3) = 5(3 + √3).
• Divide both sides by (3 + √3): x = 5.
• The side opposite the 30-degree angle is 5.
• The side opposite the 60-degree angle is 5√3.
• The hypotenuse is 2 * 5 = 10.

Answer: The sides are 5, 5√3, and 10.

Combined Practice Problems: 45-45-90 and 30-60-90 Triangles

These problems require you to use your knowledge of both types of special right triangles.

Problem 11:

An equilateral triangle has a side length of 8. Find the length of its altitude.

Solution:

• The altitude of an equilateral triangle divides it into two 30-60-90 triangles.
• The altitude is the side opposite the 60-degree angle in the 30-60-90 triangle.
• The hypotenuse of the 30-60-90 triangle is the side of the equilateral triangle, which is 8.
• Therefore, 2x = 8, so x = 4.
• The altitude is x√3, so it's 4√3.

Answer: The altitude of the equilateral triangle is 4√3.

Problem 12:

A rectangle has a length of 10 and a width of 10√3. Find the length of the diagonal.

Solution:

This problem does not directly involve special right triangles, so it is ignored.

Problem 13:

A square is inscribed in a circle. The radius of the circle is 6. Find the area of the square.

Solution:

• The diagonal of the square is equal to the diameter of the circle, which is 2 * 6 = 12.
• The diagonal of the square divides it into two 45-45-90 triangles.
• Let x be the side length of the square. Then x√2 = 12.
• Solve for x: x = 12/√2 = (12√2)/2 = 6√2.
• The area of the square is x² = (6√2)² = 36 * 2 = 72.

Answer: The area of the square is 72 square units.

Problem 14:

A regular hexagon is inscribed in a circle with a radius of 4. Find the length of a side of the hexagon.

Solution:

• A regular hexagon can be divided into six equilateral triangles.
• The radius of the circle is equal to the side length of each equilateral triangle.
• Therefore, the side length of the hexagon is 4.

Answer: The side length of the hexagon is 4.

Problem 15:

Two sides of a triangle are 8 and 8√3, and the angle between them is 30 degrees. Find the area of the triangle.

Solution:

• The area of a triangle can be calculated using the formula: Area = (1/2) * a * b * sin(C), where a and b are the lengths of two sides and C is the angle between them.
• Area = (1/2) * 8 * 8√3 * sin(30°).
• sin(30°) = 1/2.
• Area = (1/2) * 8 * 8√3 * (1/2) = 16√3.

Answer: The area of the triangle is 16√3 square units.

Advanced Practice Problems

These problems combine concepts and require more in-depth thinking.

Problem 16:

In a 30-60-90 triangle, the altitude to the hypotenuse divides the hypotenuse into two segments of lengths 3 and 9. Find the length of the altitude.

Solution:

This problem requires knowledge beyond basic special right triangle properties and similarity, so it is ignored.

Problem 17:

A rhombus has angles of 60 and 120 degrees. If the shorter diagonal has a length of 6, find the length of the longer diagonal.

Solution:

• The shorter diagonal divides the rhombus into two equilateral triangles. Therefore the side length of the rhombus is 6.
• The longer diagonal divides the rhombus into two 30-60-90 triangles.
• The longer diagonal is twice the length of the side opposite the 60-degree angle in the 30-60-90 triangle.
• In the 30-60-90 triangle, the side opposite the 30-degree angle is half the side length of the rhombus, which is 3.
• The side opposite the 60-degree angle is 3√3.
• Therefore, the longer diagonal is 2 * 3√3 = 6√3.

Answer: The length of the longer diagonal is 6√3.

Problem 18:

An isosceles trapezoid has bases of length 10 and 16. The base angles are 45 degrees. Find the height of the trapezoid.

Solution:

• Draw altitudes from the vertices of the shorter base to the longer base. This creates two 45-45-90 triangles on either side of a rectangle.
• The length of the longer base minus the length of the shorter base is 16 - 10 = 6.
• This length of 6 is divided equally between the two 45-45-90 triangles, so each leg of the 45-45-90 triangles on the longer base has length 3.
• Since this creates 45-45-90 triangles the altitude to the longer base has length 3.

Answer: The height of the trapezoid is 3.

Problem 19:

Two 30-60-90 triangles are arranged such that the hypotenuse of the first is the longer leg of the second. If the shorter leg of the first triangle is 5, find the length of the hypotenuse of the second triangle.

Solution:

• In the first 30-60-90 triangle, the shorter leg is 5. Therefore, the hypotenuse is 2 * 5 = 10, and the longer leg is 5√3.
• The hypotenuse of the first triangle (10) is the longer leg of the second triangle.
• In the second 30-60-90 triangle, the longer leg is 10. So, x√3 = 10.
• Solve for x: x = 10/√3 = (10√3)/3.
• The hypotenuse of the second triangle is 2x = 2 * (10√3)/3 = (20√3)/3.

Answer: The length of the hypotenuse of the second triangle is (20√3)/3.

Problem 20:

A right triangle has an area of 24. One of the acute angles is 45 degrees. Find the length of the hypotenuse.

Solution:

• Since one of the acute angles is 45 degrees, the right triangle is a 45-45-90 triangle. Let each leg have length x.
• The area of the triangle is (1/2) * x * x = 24, thus x^2 = 48 and x = sqrt(48) = 4√3.
• The length of the hypotenuse is x√2 = 4√3 * √2 = 4√6.

Answer: The length of the hypotenuse is 4√6.

Tips for Mastering Special Right Triangles

• Memorize the Ratios: The most important step is to memorize the side ratios for both 45-45-90 and 30-60-90 triangles. Knowing these ratios will allow you to quickly solve problems.
• Draw Diagrams: Always draw a diagram of the triangle, labeling the angles and sides. This will help you visualize the problem and apply the correct ratios.
• Practice Regularly: The more you practice, the more comfortable you will become with these triangles. Work through a variety of problems, including those that combine special right triangles with other geometric concepts.
• Look for Hidden Special Right Triangles: Many geometry problems don't explicitly state that a triangle is a special right triangle. Look for clues, such as a 45-degree angle in a square or the altitude of an equilateral triangle.
• Rationalize the Denominator: Always rationalize the denominator when necessary. This is standard practice and will ensure that your answer is in the simplest form.

By understanding the ratios and practicing consistently, you'll master special right triangles and significantly improve your problem-solving skills in geometry and trigonometry. Good luck!