Geometry and Trigonometry Flashcards

In the xy-plane, a circle has equation x² + y² − 6x + 4y − 3 = 0. What is the distance from the center of the circle to the x-axis?

Answer: 2

Complete the square for both variables: (x² − 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4, giving (x − 3)² + (y + 2)² = 16. The center is (3, −2). The distance from any point to the x-axis is the absolute value of its y-coordinate, so the distance is |−2| = 2. The radius (4) and the x-coordinate (3) are common distractors.

In triangle PQR, PQ = 5, QR = 7, and the area of the triangle is 15. What is the sum of all possible values of PR²?

Answer: 148

Area = ½ · PQ · QR · sin Q = ½(5)(7)sin Q = 15, so sin Q = 6/7. Then cos Q = ±√(1 − 36/49) = ±√13/7, giving two valid triangles. By the Law of Cosines: PR² = PQ² + QR² − 2(PQ)(QR)cos Q = 25 + 49 − 70(±√13/7) = 74 ∓ 10√13. The two possible values are 74 − 10√13 and 74 + 10√13. Their sum is 148.

A circle is inscribed in a regular hexagon with side length 4 (tangent to all six sides). What is the ratio of the area of the inscribed circle to the area of the hexagon?

Answer: π√3 / 6

The apothem (inradius) of a regular hexagon with side s is r = s√3/2. With s = 4, r = 2√3. The inscribed circle's area is π(2√3)² = 12π. The hexagon's area is (3√3/2)s² = (3√3/2)(16) = 24√3. The ratio is 12π / (24√3) = π/(2√3) = π√3/6 after rationalizing.

Given that tan θ = −4/3 and θ is in the second quadrant, what is the exact value of sin(2θ)?

Answer: −24/25

In the second quadrant, sin θ > 0 and cos θ < 0. From tan θ = −4/3 and a reference triangle with opposite = 4, adjacent = 3, hypotenuse = 5: sin θ = 4/5 and cos θ = −3/5. Applying the double-angle identity: sin(2θ) = 2 sin θ cos θ = 2(4/5)(−3/5) = −24/25. The distractor 24/25 ignores the sign of cos θ; −7/25 is actually cos(2θ) = cos²θ − sin²θ = 9/25 − 16/25.

Two sectors each have an arc length of 20π. The first sector has a central angle of 5π/6 radians and the second has a central angle of 2π/3 radians. What is the positive difference between their areas?

Answer: 60π

From arc length = rθ, solve for each radius: r₁ = 20π ÷ (5π/6) = 24 and r₂ = 20π ÷ (2π/3) = 30. Sector area = ½r²θ, so Area₁ = ½(24²)(5π/6) = ½(576)(5π/6) = 240π and Area₂ = ½(30²)(2π/3) = ½(900)(2π/3) = 300π. The positive difference is 300π − 240π = 60π.

The function y = a · sin(bx + c) has a maximum value of 2, a period of π, and passes through the point (0, √3). If b > 0 and 0 < c < π/2, what is the value of c?

Answer: π/3

The amplitude a equals the maximum value, so a = 2. The period is 2π/b = π, giving b = 2. Substituting the point (0, √3): 2 · sin(2·0 + c) = √3, so sin(c) = √3/2. Since 0 < c < π/2, the solution is c = π/3. The distractor π/6 gives sin(π/6) = 1/2, so y = 1 ≠ √3; π/4 gives y = √2 ≠ √3.