- E First, draw the diagram:
60°
8 ft30°
20GroundBase of RopeThe rope starts off 8 feet above the ground. To determine its total height from
the ground, solve for the side opposite the 60-degree angle and add 8. Since the
triangle in the diagram is a 30-60-90 triangle, the side opposite the 60-degree
angle will be √ 3 × the side opposite the 30-degree angle. The side opposite
the 30-degree angle is half of the hypotenuse:^202 = 10. The side opposite the
60-degree angle is thus 10√ 3. The rope is thus 8 + 10√ 3 feet off the ground.- 18 Since triangles BDC and EDC are right triangles, calculate the area of
triangle BDC and subtract that from the area of triangle EDC. Calculate the
area of triangle BDC: since ABCD is a square, the diagonal is times the base.
Thus DC = 12. If DC = 12, the area of triangle BDC = 12 × 12 ×^12 = 72.
Calculate the area of triangle EDC: Since ABCD is a square, the diagonal
is √ 2 times the base. Thus DC = 12. If DC = 12, the area of triangle
EDC = 12 × 9 ×^12 = 54. The difference of the areas of triangles BDC and
EDC is 72 – 54 = 18. - B Since ∠BDC is the exterior angle of triangle BAD,
∠BDC = ∠BAD + ∠DBA
2 a = a + ∠DBA
∠DBA = a
If ∠DBA = a, then triangle BAD is isosceles, and BD = AD. Since AD = 6,
BD = 6. Since BD = DC, DC = 6. - D To solve for the perimeter, you need to solve for AB and DC. Triangle ABD
is a 9-12-15 triangle (one of the multiples of a 3-4-5 triangle), so AB = 15.
Tr ia ng le BDC is a 12-16-20 triangle (one of the multiples of a 3-4-5 triangle), so
DC = 16. The perimeter is thus 15 + 20 + 9 + 16 = 60. - 163 This is a challenging question! First, identify that AB = 5, since AB is the
hypotenuse of a 3-4-5 triangle. Next, let DC = x. Since triangle BDC is a right
triangle, you can use the Pythagorean theorem to express BC in terms of x:
x^2 + 16 = (BC)^2
√x^2 + 16 = BC
394 PART 4 ■ MATH REVIEW04-GRE-Test-2018_313-462.indd 394 23/01/18 11:11 AM