Geometry — Semester B
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 12A-12EHard

A circle is given by the equation x² + y² − 10x + 6y + 18 = 0. From the external point P(8, 1), a tangent line is drawn to the circle, touching it at point T. What is the length of tangent segment PT?

A1 unit
B5 units
C3 units
D√41 ≈ 6.40 units
Explanation
Complete the square: x² − 10x + y² + 6y = −18 becomes (x − 5)² + (y + 3)² = −18 + 25 + 9 = 16, so the center is O(5, −3) and r = 4. Distance formula: OP = √((8 − 5)² + (1 + 3)²) = √(9 + 16) = 5. Draw radius OT; by the tangent–radius theorem OT ⊥ PT, so △OTP is right with hypotenuse OP. Pythagorean Theorem: PT = √(OP² − r²) = √(25 − 16) = √9 = 3 units. The tempting wrong answer is 5 units: that is OP itself, the distance from the external point to the CENTER, taken as if it were the tangent length — but the tangent reaches only the circle, not the center. Choosing 1 unit comes from subtracting lengths (5 − 4) instead of using the Pythagorean relation on the squares, and √41 comes from adding r² instead of subtracting it.
Question 2 of 10
TEKS 1A-1GMedium Calc Word Diagram
A kite is flying at the end of a 200-foot string. The string makes a 55° angle with the ground. How high above the ground is the kite? Round to the nearest foot. (sin 55° ≈ 0.819) h = ?55°200 ft
A115 feet
B164 feet
C186 feet
D141 feet
Explanation
📌 Step 1: Identify the trig ratio
We know the hypotenuse (200 ft) and want the opposite side (height).
Use sine: sin = opposite / hypotenuse

📌 Step 2: Set up and solve
sin(55°) = h / 200
0.819 = h / 200
h = 200 × 0.819 = 163.8

📌 Answer:164 feet

💡 Tip: Angle of elevation from ground = angle between the string and the horizontal, NOT the vertical.
Question 3 of 10
TEKS 11A-11DHard Calc Word

An open rectangular tank has a base measuring 24 cm by 18 cm and contains water. A solid metal cube is lowered in until it is completely submerged, resting flat on the bottom and touching no wall, and the water level rises exactly 4 cm without overflowing. What is the total surface area of the cube?

A720 cm²
B72 cm²
C864 cm²
D144 cm²
Explanation
By the displacement principle, a fully submerged solid pushes up a layer of water whose volume equals the solid's volume. That layer is a rectangular prism: V = 24 · 18 · 4 = 1728 cm³. Since the solid is a cube, s³ = 1728, so s = ∛1728 = 12 cm. A cube has 6 congruent square faces, so SA = 6s² = 6 · 144 = 864 cm². The most tempting wrong answer is 720 cm², which comes from counting only 5 faces because the cube 'sits on the bottom' — surface area counts every face of the solid regardless of what it rests against. Answering 144 cm² stops at a single face, and 72 cm² comes from using 6s instead of 6s².
Question 4 of 10
TEKS 10A-10BHard Calc

A cube has edge length 12 cm. A plane is passed through the three vertices that are each connected by an edge to one chosen corner of the cube (the corner itself is cut off). What is the exact area of the resulting cross section?

A36√3 cm²
B72 cm²
C72√3 cm²
D144 cm²
Explanation
The three vertices adjacent to one corner are not joined by edges of the cube — each pair lies on a common face and is joined by a face diagonal. By the Pythagorean Theorem the face diagonal is √(12² + 12²) = 12√2 cm, so the cross section is an equilateral triangle with all three sides 12√2. Using A = (√3/4)s²: A = (√3/4)(12√2)² = (√3/4)(288) = 72√3 ≈ 124.7 cm². The tempting wrong value 36√3 comes from putting the edge length 12 into (√3/4)s² — that treats the three cut vertices as if they were one edge apart. The value 72 comes from imagining a right triangle with two legs of 12 and taking ½·12·12, which is the area of half a face, not the slanted cross section. The value 144 is the area of a whole face of the cube.
Question 5 of 10
TEKS 13A-13EEasy Calc Word

In how many ways can 5 students be arranged in a line?

A25
B120
C60
D720
Explanation
📌 Step 1: Understand permutations
Arranging n objects in a line uses n! (n factorial).

📌 Step 2: Calculate 5!
5! = 5 × 4 × 3 × 2 × 1

📌 Step 3: Solve step by step
5 × 4 = 20
20 × 3 = 60
60 × 2 = 120
120 × 1 = 120

💡 Key difference:
Permutation = order matters (arranging in a line)
Combination = order doesn't matter (choosing a team)
• Formula: P(n,r) = n!/(n-r)! and C(n,r) = n!/(r!(n-r)!)
Question 6 of 10
TEKS 12A-12EEasy Calc Word

A central angle of a circle is 90°. What is the measure of the intercepted arc?

A270°
B180°
C45°
D90°
Explanation
📌 Step 1: Recall the central angle-arc relationship
A central angle equals its intercepted arc.

📌 Step 2: Apply
Central angle = 90° → Arc = 90°

💡 Key relationships in circles:
• Central angle = intercepted arc
• Inscribed angle = ½ × intercepted arc
• Two inscribed angles intercepting the same arc are equal
Question 7 of 10
TEKS 11A-11DMedium Calc Word

A regular hexagon has a side length of 6 cm and an apothem of 5.2 cm. What is its area?

A108.0 cm²
B78.0 cm²
C93.6 cm²
D62.4 cm²
Explanation
📌 Step 1: Recall the area formula for a regular polygon
A = ½ × perimeter × apothem

📌 Step 2: Find the perimeter
Perimeter = 6 sides × 6 cm = 36 cm

📌 Step 3: Calculate the area
A = ½ × 36 × 5.2 = 18 × 5.2 = 93.6 cm²

💡 What's an apothem? The distance from the center to the midpoint of a side. It's always perpendicular to the side.
Question 8 of 10
TEKS 3A-3DMedium Calc Word Diagram
Trapezoid WXYZ has vertices W(−4, 2), X(−1, 2), Y(0, −1), and Z(−5, −1). It is translated 6 units right, then reflected over the x-axis. What are the coordinates of W′? xy-4-1252-1WXYZ+6 right, reflect x-axis
AW′(−4, −2)
BW′(2, −2)
CW′(10, −2)
DW′(2, 2)
Explanation
Step 1: Translate 6 right → add 6 to x: W(−4,2) → (−4+6, 2) = (2, 2).
Step 2: Reflect over x-axis → negate y: (2, 2) → (2, −2).
So W′ = (2, −2).
Question 9 of 10
TEKS 13A-13EMedium Calc Word Diagram
A dart is thrown randomly at the square board below. The board has a side length of 20 cm and contains a shaded circle with a radius of 8 cm. What is the probability that the dart lands inside the shaded circle? (Use π ≈ 3.14) 20 cmr = 8
A78.5%
B40.0%
C50.2%
D62.8%
Explanation
Area of square = 20² = 400 cm².
Area of circle = πr² = 3.14 × 8² = 3.14 × 64 = 200.96 cm².
P = circle/square = 200.96/400 ≈ 0.5024 ≈ 50.2%.
Question 10 of 10
TEKS 10A-10BMedium Calc Word

A rectangular prism has dimensions 4 × 6 × 10. If only the height (10) is halved, what happens to the volume?

AIt is reduced by 1/8
BIt is reduced by 1/3
CIt is reduced by 1/4
DIt is halved
Explanation
📌 Step 1: Name the rule
Volume of a rectangular prism is V = length × width × height. Each dimension appears exactly once, so scaling one dimension by a factor k scales the volume by that same factor k.

📌 Step 2: Use this question's numbers
Original: V = 4 × 6 × 10 = 240
Only the height changes, 10 → 5:
New: V = 4 × 6 × 5 = 120

📌 Step 3: Compare
120/240 = 1/2, so the new volume is exactly half the original — 240 cubic units becomes 120 cubic units.

📌 Step 4: Check the other fractions both ways
The remaining options name the fractions 1/3, 1/4, and 1/8. Whether you read those as "the volume shrinks to that fraction" or as "the volume shrinks by that much of itself," none of them produce 120. Reading them as shrinks-to gives 80, 60, and 30. Reading them as shrinks-by gives 160, 180, and 210. The computed volume is 120, which matches only the halving statement.

💡 The tempting error: reaching for 1/8. That factor belongs to a different scenario — halving *all three* dimensions, where 2 × 3 × 5 = 30, and (1/2)(1/2)(1/2) = 1/8 because each halved dimension contributes its own factor. This stem changes one dimension only, so there is exactly one factor of 1/2. Count how many dimensions actually changed before you scale: one changed dimension → factor k, two → k², all three → k³. And when a choice describes a change as a fraction, compute the resulting volume yourself and compare numbers rather than matching the fraction by sight.

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