Question

find the surface area of this triangular prism. be sure to include the correct unit in your answer.

Explanation:

Step1: Find the area of the triangular bases

The formula for the area of a triangle is \(A=\frac{1}{2}bh\). Here, \(b = 24\) cm and \(h = 10\) cm.
\(A_{triangle}=\frac{1}{2}\times24\times10= 120\) \(cm^{2}\). Since there are 2 triangular bases, \(A_{triangles}=2\times120 = 240\) \(cm^{2}\).

Step2: Find the area of the rectangular faces

There are three rectangular faces.

• For the face with dimensions \(22\) cm and \(26\) cm: \(A_{1}=22\times26 = 572\) \(cm^{2}\)
• For the face with dimensions \(24\) cm and \(26\) cm: \(A_{2}=24\times26=624\) \(cm^{2}\)
• For the face with dimensions \(26\) cm and \(10\) cm (hypotenuse of the base triangle is calculated using Pythagoras \(a=\sqrt{24^{2}+10^{2}}=\sqrt{576 + 100}=\sqrt{676}=26\) cm, which is consistent with the given side). But we can also use the formula for the lateral surface area of a prism \(LSA=(a + b+ c)h\), where \(a = 22\), \(b = 24\), \(c = 26\) (sides of the base triangle) and \(h = 26\) (length of the prism). \(LSA=(22 + 24+26)\times26=(72)\times26 = 1872\) \(cm^{2}\). But another way: sum of the three rectangles \(A_{1}+A_{2}+A_{3}\) (where \(A_{3}\) is the rectangle with sides \(22\) and \(26\) (already calculated \(A_{1} = 572\), \(A_{2}=624\), and the third rectangle with sides \(22\) and \(26\) (wait no, the three rectangles: if the base triangle has sides \(22\), \(24\), \(26\) and height of prism \(26\). The three rectangles: \(22\times26\), \(24\times26\), \(26\times26\) (no, wait the formula for surface area of triangular prism \(SA=2A_{base}+(a + b + c)h\). Here \(a = 22\), \(b = 24\), \(c = 26\) (sides of base triangle), \(h = 26\) (length of prism), \(A_{base}=120\)).

\(SA=2\times120+(22 + 24+26)\times26\)
\(SA = 240+72\times26\)
\(SA=240 + 1872\)

Answer:

\(2112\space cm^{2}\)