Question
Find the TOTAL surface area of this triangular prism in cm^2.
h^2 = 12cm, L^2 = 5cm, b = 6cm, h^2 = 4cm.
h^2 = 12cm, L^2 = 5cm, b = 6cm, h^2 = 4cm.
Answers
GPT-4o mini
To find the total surface area of a triangular prism, we need to consider the area of the two triangular bases and the area of the three rectangular lateral faces.
Given the dimensions:
- Height (\( h \)) of the triangle = 2 cm (from the given \( h^2 = 4 \, \text{cm} \)), thus \( h = \sqrt{4} = 2 \, \text{cm} \)
- Base (\( b \)) of the triangle = 6 cm
- Length (\( L \)) of the prism = 5 cm (from \( L^2 = 25 \, \text{cm} \)), thus \( L = \sqrt{25} = 5 \, \text{cm} \)
First, we need to find the area of the triangular base. The area \( A \) of a triangle is given by the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h
\]
Substituting in the values:
\[
A = \frac{1}{2} \times 6 \times 2 = \frac{12}{2} = 6 \, \text{cm}^2
\]
Since there are two triangular bases, the total area of the triangular bases is:
\[
\text{Total area of bases} = 2 \times A = 2 \times 6 = 12 \, \text{cm}^2
\]
Next, we calculate the area of the three rectangular lateral faces. The areas of these rectangles are as follows:
1. Rectangle 1 (base \( b \) and height \( L \)):
\[
\text{Area}_1 = b \times L = 6 \times 5 = 30 \, \text{cm}^2
\]
2. Rectangle 2 (side \( a \) and height \( L \)): First, we need to find side \( a \). For an isosceles triangle with base \( b = 6 \, \text{cm} \) and height \( h = 2 \, \text{cm} \), we can calculate \( a \) using the Pythagorean theorem:
The half of the base (from the midpoint to the base edge) is \( \frac{b}{2} = 3 \, \text{cm} \):
\[
a = \sqrt{h^2 + \left(\frac{b}{2}\right)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \, \text{cm}
\]
Now, we can calculate the area of this rectangle:
\[
\text{Area}_2 = a \times L = \sqrt{13} \times 5 = 5\sqrt{13} \, \text{cm}^2
\]
3. Rectangle 3 (side \( a \) and height \( L \)), which is identical to Rectangle 2:
\[
\text{Area}_3 = \text{Area}_2 = 5\sqrt{13} \, \text{cm}^2
\]
Now we can add these areas together to find the total lateral area:
\[
\text{Total lateral area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 30 + 5\sqrt{13} + 5\sqrt{13} = 30 + 10\sqrt{13} \, \text{cm}^2
\]
Finally, the total surface area \( A_{\text{total}} \) of the prism is:
\[
A_{\text{total}} = \text{Total area of bases} + \text{Total lateral area} = 12 + (30 + 10\sqrt{13}) = 42 + 10\sqrt{13} \, \text{cm}^2
\]
Thus, the total surface area of the triangular prism is:
\[
\boxed{42 + 10\sqrt{13} \, \text{cm}^2}
\]
Given the dimensions:
- Height (\( h \)) of the triangle = 2 cm (from the given \( h^2 = 4 \, \text{cm} \)), thus \( h = \sqrt{4} = 2 \, \text{cm} \)
- Base (\( b \)) of the triangle = 6 cm
- Length (\( L \)) of the prism = 5 cm (from \( L^2 = 25 \, \text{cm} \)), thus \( L = \sqrt{25} = 5 \, \text{cm} \)
First, we need to find the area of the triangular base. The area \( A \) of a triangle is given by the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h
\]
Substituting in the values:
\[
A = \frac{1}{2} \times 6 \times 2 = \frac{12}{2} = 6 \, \text{cm}^2
\]
Since there are two triangular bases, the total area of the triangular bases is:
\[
\text{Total area of bases} = 2 \times A = 2 \times 6 = 12 \, \text{cm}^2
\]
Next, we calculate the area of the three rectangular lateral faces. The areas of these rectangles are as follows:
1. Rectangle 1 (base \( b \) and height \( L \)):
\[
\text{Area}_1 = b \times L = 6 \times 5 = 30 \, \text{cm}^2
\]
2. Rectangle 2 (side \( a \) and height \( L \)): First, we need to find side \( a \). For an isosceles triangle with base \( b = 6 \, \text{cm} \) and height \( h = 2 \, \text{cm} \), we can calculate \( a \) using the Pythagorean theorem:
The half of the base (from the midpoint to the base edge) is \( \frac{b}{2} = 3 \, \text{cm} \):
\[
a = \sqrt{h^2 + \left(\frac{b}{2}\right)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \, \text{cm}
\]
Now, we can calculate the area of this rectangle:
\[
\text{Area}_2 = a \times L = \sqrt{13} \times 5 = 5\sqrt{13} \, \text{cm}^2
\]
3. Rectangle 3 (side \( a \) and height \( L \)), which is identical to Rectangle 2:
\[
\text{Area}_3 = \text{Area}_2 = 5\sqrt{13} \, \text{cm}^2
\]
Now we can add these areas together to find the total lateral area:
\[
\text{Total lateral area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 30 + 5\sqrt{13} + 5\sqrt{13} = 30 + 10\sqrt{13} \, \text{cm}^2
\]
Finally, the total surface area \( A_{\text{total}} \) of the prism is:
\[
A_{\text{total}} = \text{Total area of bases} + \text{Total lateral area} = 12 + (30 + 10\sqrt{13}) = 42 + 10\sqrt{13} \, \text{cm}^2
\]
Thus, the total surface area of the triangular prism is:
\[
\boxed{42 + 10\sqrt{13} \, \text{cm}^2}
\]