Solutions

1. 256 inches²

   The diagonal of a square is ; therefore, the side length of square ABCD is 16. The area of the square is s², or 16², which is 256.

2. 13 miles

   If you draw a rough sketch of the path Biker Bob takes, as shown to the right, you can see that the direct distance from A to B forms the hypotenuse of a right triangle. The short leg (horizontal) is 10 − 5 = 5 miles, and the long leg (vertical) is 9 + 3 = 12 miles. Therefore, you can use the Pythagorean theorem to find the direct distance from A to B:

   

   You might recognize the common right triangle: 5–12–13.

3. 24 miles

   If you draw a rough sketch of the path Biker Bob takes, as shown, you can see that the direct distance from B to C forms the hypotenuse of a right triangle:

   

   You might also recognize this as a multiple of the common 5–12–13 triangle.

4. 200

   An isosceles right triangle is a 45–45–90 triangle, with sides in the ratio of . If the longest side, the hypotenuse, measures , the two other sides each measure 20. Therefore, the area of the triangle is:

   
5. 

   The longest distance between any two posts is the diagonal of the field. If the area of the square field is 400 square meters, then each side must measure 20 meters. Diagonal is , so d is .

6. 10

   If AD = BD = DC, then the three triangular regions in this figure are all isosceles triangles. Therefore, you can fill in some of the missing angle measurements as shown here. Because you know that there are 180° in the large triangle ABC, you can write the following equation:

   

   

7. 7

   If two sides of a triangle are 4 and 10, the third side must be greater than 10 − 4 and smaller than 10 + 4. Therefore, the possible values for x are {7, 8, 9, 10, 11, 12, and 13}. You can draw a sketch to convince yourself of this result:

   

8. inches

   The diagonal of a cube with side s is . Therefore, the longest object Jack could fit inside the box would be inches long.

9. 

   Draw in the height of the triangle (see figure). If triangle ABC is an equilateral triangle, and ABD is a right triangle, then ABD is a 30–60–90 triangle. Therefore, its sides are in the ratio of . If the hypotenuse is 8, then the short leg is 4, and the long leg is . This is the height of equilateral triangle ABC. Find the area of triangle ABC with the formula for area of a triangle:

   

   
10. 

    Think of this star as a large equilateral triangle with sides 12 centimeters long, and three additional smaller equilateral triangles (shaded in the figure below) with sides 4 inches long. Using the same 30–60–90 logic you applied in problem 9, you can see that the height of the larger equilateral triangle is , and the height of the smaller equilateral triangle is .

    Therefore, the areas of the triangles are as follows:

    • Large triangle:

      Small triangles: 

    The total area of three smaller triangles and one large triangle is:

    

    

11. 

    You can calculate the area of the triangle using the side of length 12 as the base:

    

    Next, use the side of length 7 as the base (remember, any side can function as the base, provided that you can find the corresponding height) and write the equation for the area:

    

    Now solve for x, the unknown height:

    

    You could also solve this problem using the Pythagorean theorem, but the process is much harder.

12. (D)

    Although this appears to be a 5 : 12 : 13 triangle, you do not know that it is a right triangle. There is no right angle symbol in the diagram. Remember, don’t trust the picture! Here are a couple of possible triangles:

    

    Therefore, the relationship cannot be determined from the information given.

13. (A)

    Although there seems to be very little information here, the two small triangles that comprise triangle ABC may seem familiar. First, fill in the additional angles in the diagram:

    

    With the additional angles filled in, it is clear that the two smaller triangles are special right triangles: a 45–45–90 triangle and a 30–60–90 triangle. You know the ratios of the side lengths for each of these triangles. For a 45–45–90 triangle, the ratio is . In this diagram, the value of x is 1 (side BD), so AD is 1 and AB is :

    

    For a 30–60–90 triangle, the ratio is . In this diagram, x is 1 (side BD), so DC is and BC is 2:

    

    Now calculate the perimeter of triangle ABC:

    Quantity A	Quantity B
    The perimeter of triangle ABC	5

    Now you need to compare this sum to 5. A good approximation of is 1.4 and a good approximation of is 1.7:

    Quantity A	Quantity B
    	5

    In fact, simply knowing that each square root is greater than 1 would let you conclude that, Quantity A is greater.

    Alternatively, you could use the calculator to compute Quantity A.

14. (C)

    The diagonal of the rectangle is the hypotenuse of a right triangle whose legs are the length and width of the rectangle. In this case, you are given the width and the diagonal. Plug these into the Pythagorean theorem to determine the length:

    

    Label this value on the diagram:

    

    The key to this question is recognizing that each of the triangles is a 30–60–90 triangle. Any time you see a right triangle and one of the sides has a length of or a multiple of , you should check to see whether it is a 30–60–90 triangle. Another clue is a right triangle in which the hypotenuse is twice the length of one of the other sides.

    Now, in addition to the side lengths, you can fill in the values of the angles in this diagram. Angle x is opposite the short leg, which means it has a degree measure of 30. Similarly, 2y is opposite the long leg, which means it has a degree measure of 60:

    • 2y = 60
         y = 30

    Quantity A	Quantity B
    x = 30	y = 30

    Therefore, the two quantities are equal.