Geometry — Semester B Practice

Question 1 of 10

TEKS 10A-10B Medium Calc Word Diagram

A cylinder has radius 4 and height 10. If the radius is tripled but the height stays the same, by what factor does the volume increase?

A 9

B 3

C 6

D 27

Explanation

📌 Step 1: Recall how dimensional changes affect volume
V = πr²h
When r is multiplied by k (and h stays the same):
New V = π(kr)²h = k²πr²h = k² × original V

📌 Step 2: Apply k = 3
Factor = 3² = 9

📌 Verification:
Original: π(4²)(10) = 160π
New: π(12²)(10) = 1440π
1440π / 160π = 9 ✓

💡 Key rule: Changing ONE linear dimension by factor k changes:
• Length → k
• Area → k²
• Volume → k² (if only radius) or k³ (if all dimensions)

Question 2 of 10

TEKS 3A-3D Easy Calc Word Diagram

Which of the following figures has BOTH reflectional and rotational symmetry?

A B (Regular hexagon)

B A (Scalene triangle)

C D (Arrow)

D C (Parallelogram)

Explanation

📌 Step 1: Check each figure

A (Scalene triangle): No lines of symmetry, no rotational symmetry ✗
B (Regular hexagon): 6 lines of symmetry + rotational symmetry at 60° ✓
C (Parallelogram): No lines of symmetry, rotational symmetry at 180° only → partial ✗
D (Arrow): 1 line of symmetry (vertical) but no rotational symmetry ✗

📌 Answer: B (Regular hexagon)

💡 Tip: All regular polygons have BOTH reflectional AND rotational symmetry. The number of symmetry lines = number of sides.

Question 3 of 10

TEKS 11A-11D Medium Calc Word Diagram

Find the volume of the cone shown below. Round to the nearest tenth. (Use π ≈ 3.14)

A 565.2 cm³

B 452.2 cm³

C 1695.6 cm³

D 339.1 cm³

Explanation

📌 Step 1: Recall the cone volume formula
V = (1/3)πr²h

📌 Step 2: Plug in values
r = 6 cm, h = 15 cm
V = (1/3)(3.14)(36)(15)

📌 Step 3: Calculate step by step
3.14 × 36 = 113.04
113.04 × 15 = 1695.6
1695.6 / 3 = 565.2 cm³

💡 Common mistake: Don't forget to divide by 3! A cone is 1/3 the volume of a cylinder with the same base and height.

Question 4 of 10

TEKS 10A-10B Medium Calc Word Diagram

A plane cuts through a right square pyramid parallel to the base, as shown. What is the shape of the cross section?

A Circle

B Triangle

C Square

D Pentagon

Explanation

When a plane cuts through a pyramid parallel to its base, the cross section is always the same shape as the base.
Since the base is a square, the cross section is also a square (a smaller, similar square).

Question 5 of 10

TEKS 1A-1G Medium Calc Word Diagram

Quadrilateral ABCD has the properties shown below. Which type of quadrilateral is ABCD?

A Trapezoid

B Rhombus

C Parallelogram

D Rectangle

Explanation

A quadrilateral with exactly one pair of parallel sides is a trapezoid.
AB ∥ DC but AB ≠ DC (16 ≠ 22), confirming it is a trapezoid, not a parallelogram.

Question 6 of 10

TEKS 13A-13E Medium 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)

A 78.5%

B 62.8%

C 50.2%

D 40.0%

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 7 of 10

TEKS 13A-13E Medium Calc Word Diagram

The two-way table shows student preferences. If a student is randomly selected from those who prefer Math, what is the probability they are in 10th grade?

A 40%

B 30%

C 20%

D 60%

Explanation

📌 Step 1: Identify the condition
"Given that a student prefers Math" means we only look at the Math column.

📌 Step 2: Find the relevant numbers
Total who prefer Math = 30
10th graders who prefer Math = 12

📌 Step 3: Calculate conditional probability
P(10th | Math) = 12/30 = 2/5 = 0.40 = 40%

💡 Key concept: Conditional probability restricts the sample space. We're not looking at all 60 students — only the 30 who prefer Math.

Question 8 of 10

TEKS 11A-11D Medium Calc Word Diagram

A swimming pool has the shape shown below — a rectangle with a semicircle on each end. Find the total area of the pool. (Use π ≈ 3.14)

A 257.0 m²

B 356.0 m²

C 278.5 m²

D 200.0 m²

Explanation

Rectangle area = 20 × 10 = 200 m².
Two semicircles = one full circle with r = 5: π × 5² = 3.14 × 25 = 78.5 m².
Total = 200 + 78.5 = 278.5 m².

Question 9 of 10

TEKS 12A-12E Medium Calc Word Diagram

A tangent line touches circle O at point T. OT = 5 and the external point P is 13 units from the center O. What is the length of tangent segment PT?

A 12

B 10

C 14

D 8

Explanation

The tangent is perpendicular to the radius at the point of tangency. Using the Pythagorean theorem: PT = √(OP² − OT²) = √(13² − 5²) = √(169 − 25) = √144 = 12.

Question 10 of 10

TEKS 1A-1G Medium 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)

A 164 feet

B 115 feet

C 141 feet

D 186 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.

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