must be equal: angle BAC = angle BCA. You are told that angle BAC = 40, so
you can infer that angle BCA = 40. Since the interior angles of a triangle sum
to 180, y + 40 + 40 = 180 → y = 100.- 9√ 3 To get the area of ABC, you will need to determine the length of the base.
Since BD is the altitude of an equilateral triangle, BCD is a 30-60-90 triangle.
The altitude is opposite the 60-degree angle, and DC is opposite the 30-degree
angle. Since BD = 3√ 3 , DC =
3√ 30
√3 = 3. If DC = 3, then AC = 3 (since triangles
ABD and BDC are identical). Thus AC = 6. The area of triangle ABC is:
1
2 × b × h
=^12 × 6 × 3√ 3
=^12 × 18√ 3
= 9√ 3 - A Since triangle ABC has a leg of 6 and a hypotenuse of 10, side BC = 8 (this is
a 6-8-10 triangle). If BC = 8, then BD = 8 – 5 = 3. AB^2 + BD^2 = AD^2. Substitute:
62 + 3^2 = AD^2
36 + 9 = AD^2
45 = AD^2
√ 45 = AD^2
3√ 5 = AD - C Since angle BAD = angle ABD, triangle ABD is an isosceles right triangle.
Let x = the length of each leg of triangle ABD. The area will be (x)( 2 x). Solve
for x. Triangle BDC is a right triangle with a leg of 5 and hypotenuse of 13. It is
thus a 5-12-13 triangle. Side BD = 12. Substitute 12 for x to solve for the area of
triangle ABD: (12)(12) 2 = 72.
60°
33 DBA C30°
33√CHAPTER 13 ■ GEOMETRY 39304-GRE-Test-2018_313-462.indd 393 12/05/17 12:04 pm