Computer Science Geometry Word Problems

Geometry Word Problems

Grade 11 Computer Science Geometry Word Problems Advanced 5 Questions 27 Views

Classroom Worksheet

Computer Science Geometry Word Problems

Geometry Word Problems

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Grade: Grade 11

Subject: Computer Science

Topic: Geometry

Time: 20 Minutes

Difficulty: Advanced

Questions: 5

Instructions:
Solve each word problem carefully. Show your reasoning, define any formulas you use, and give your final answer with correct units. Questions increase in difficulty.

Skills Practiced:

Geometry
Computer Science
Critical Thinking
Problem Solving

1Question

A game designer is creating a square tile for a 2D map. Each side of the tile is 18 cm. What is the area of the tile, and how many such tiles are needed to cover a rectangular floor that is 9 m long and 6 m wide?

2Question

A computer screen has a rectangular display with width 48 cm and height 27 cm. A developer wants to place a diagonal line from one corner to the opposite corner for a layout guide. What is the length of the diagonal?

3Question

In a 3D modeling project, a rectangular prism has dimensions 12 cm by 8 cm by 5 cm. How many cubic centimeters of space does it occupy, and what is the total surface area of the prism?

4Question

A circular Wi-Fi sensor is placed in the center of a park. The sensor can detect devices within a radius of 35 meters. If a straight walking path crosses the center of the circle, what is the length of the path inside the detection area, and what is the area covered by the sensor? Use π ≈ 3.14.

5Question

A drone maps a triangular field for a farming app. The field has side lengths of 50 m, 62 m, and 71 m. The app also needs the height from the 71 m side to calculate crop density zones. What is the area of the field, and what is the height to the 71 m side? Round your answers to the nearest tenth.

Answer Key

#	Answer
1	Area of one tile = 18 × 18 = 324 cm². Floor area = 9 m × 6 m = 54 m² = 540,000 cm². Number of tiles = 540,000 ÷ 324 ≈ 1666.67, so 1,667 tiles are needed.
2	Use the Pythagorean theorem: d = √(48² + 27²) = √(2304 + 729) = √3033 ≈ 55.07 cm. The diagonal is about 55.1 cm.
3	Volume = 12 × 8 × 5 = 480 cm³. Surface area = 2(lw + lh + wh) = 2(12×8 + 12×5 + 8×5) = 2(96 + 60 + 40) = 392 cm².
4	The path through the center is the diameter: 2 × 35 = 70 m. Area of detection = πr² = 3.14 × 35² = 3.14 × 1225 = 3846.5 m².
5	Use Heron's formula. Semiperimeter s = (50 + 62 + 71)/2 = 91.5. Area = √[s(s−50)(s−62)(s−71)] = √[91.5 × 41.5 × 29.5 × 20.5] ≈ √2299964.3 ≈ 1516.6 m². Height to the 71 m side = (2 × area) ÷ 71 = 3033.2 ÷ 71 ≈ 42.7 m.