Math Problem Statement

What is the surface area of a triangular prism given its net dimensions, including two triangular faces with a base of 10 meters, a height of 12 meters, and three rectangular faces with lengths of 20 meters, 21 meters, and 13 meters?

Solution

To solve the surface area of the triangular prism shown in the image, we need to find the area of all the faces in the net and then sum them up.

The net consists of:

  1. Two triangular faces
  2. Three rectangular faces

Step 1: Area of the triangular faces

The triangles are right-angled, and their dimensions are given as:

  • Base = 10 meters
  • Height = 12 meters

The area of one triangle is: Area of triangle=12×base×height=12×10×12=60m2\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 12 = 60 \, \text{m}^2 Since there are two triangular faces: Total area of triangles=60×2=120m2\text{Total area of triangles} = 60 \times 2 = 120 \, \text{m}^2

Step 2: Area of the rectangular faces

There are three rectangles with the following dimensions:

  • One rectangle with length 20 meters and width (equal to the height of the triangle) 12 meters: Area of first rectangle=20×12=240m2\text{Area of first rectangle} = 20 \times 12 = 240 \, \text{m}^2
  • One rectangle with length 21 meters and width (equal to the hypotenuse of the triangle, calculated using Pythagoras theorem): Hypotenuse=102+122=100+144=244=15.62m\text{Hypotenuse} = \sqrt{10^2 + 12^2} = \sqrt{100 + 144} = \sqrt{244} = 15.62 \, \text{m} Area of second rectangle=21×15.62=328.02m2\text{Area of second rectangle} = 21 \times 15.62 = 328.02 \, \text{m}^2
  • One rectangle with length 13 meters and width 20 meters: Area of third rectangle=13×20=260m2\text{Area of third rectangle} = 13 \times 20 = 260 \, \text{m}^2

Step 3: Total surface area

Now, summing up all the areas: Total surface area=120+240+328.02+260=948.02m2\text{Total surface area} = 120 + 240 + 328.02 + 260 = 948.02 \, \text{m}^2

Thus, the surface area of the prism is 948.02 square meters.

Would you like further details or clarifications?

Here are 5 related questions you might find useful:

  1. How is the surface area of other types of prisms calculated?
  2. How does Pythagoras' theorem help in finding the hypotenuse?
  3. What is the difference between volume and surface area in 3D shapes?
  4. How do the dimensions of the net relate to the 3D shape itself?
  5. Can you explain how to find the area of complex polygons?

Tip: Always ensure the units used in your calculations are consistent, especially when dealing with different types of geometric shapes.

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Math Problem Analysis

Mathematical Concepts

Geometry
Surface Area
Nets of 3D Shapes
Right-Angled Triangles

Formulas

Area of a triangle = 1/2 × base × height
Surface Area of a prism = Sum of the areas of all faces
Hypotenuse = √(base^2 + height^2)

Theorems

Pythagoras' Theorem

Suitable Grade Level

Grades 8-10