Step by step illustration on how to find surface area of triangular prism using Pythagoras theorem to find the hypotenuse involved.

1. Finding: Surface Area of
Triangular Prism 2
- Mr Kim

2. Please view:
Surface Area of Triangular Prism 1
Before you view this lesson

3. 12 m
15 m
9.4 m

4. 12 m
15 m
9.4 m
Find the Area of all 5 Faces
of this Prism

5. 12 m
15 m
9.4 m
Front Area is

6. 12 m
15 m
9.4 m
Front Area is

2
1

7. 12 m
15 m
9.4 m
Front Area is
1215
2
1


8. 12 m
15 m
9.4 m
Since the Back is the
same as the Front…
1215
2
1


9. 12 m
15 m
9.4 m
21215
2
1

Times by 2

10. 12 m
15 m
9.4 m
21215
2
1

Now let’s find the
Left Side

11. 12 m
15 m
9.4 m
 21215
2
1
Add that on

12. 12 m
15 m
9.4 m
 21215
2
1
The Left side is a
Rectangle

13. 12 m
15 m
9.4 m
We have…
 21215
2
1

14. 12 m
15 m
9.4 m
 21215
2
1
We have…

15. 12 m
15 m
9.4 m
 21215
2
1
But not

16. 12 m
15 m
9.4 m
 21215
2
1
We have to find
that length

17. 12 m
15 m
9.4 m
 21215
2
1
We will call that
unknown length x
x

18. 12 m
15 m
9.4 m
x
To find x we need to use
Pythagoras theorem using
the Front Triangle
 21215
2
1

19. 12 m
15 m
9.4 m
x
To find x we need to use
Pythagoras theorem using
the Front Triangle
 21215
2
1

20. 12 m
15 m
9.4 m
x 21215
2
1

21. 12 m
15 m
9.4 m
x
So if we use Pythagoras
theorem we will get…
 21215
2
1

22. 12 m
15 m
9.4 m
x
So if we use Pythagoras
theorem we will get…
x
 21215
2
1

23. 12 m
15 m
9.4 m
x
2
x
squared
 21215
2
1

24. 12 m
15 m
9.4 m
x
2
x
Equals
 21215
2
1

25. 12 m
15 m
9.4 m
x
152
x
Equals
 21215
2
1

26. 12 m
15 m
9.4 m
x
22
15x
squared
 21215
2
1

27. 12 m
15 m
9.4 m
x
Now, see how x is the
longest side of the triangle
so we…
22
15x
 21215
2
1

28. 12 m
15 m
9.4 m
x
 22
15x
Now, see how x is the
longest side of the triangle
so we PLUS
 21215
2
1

29. 15 m
9.4 m
x
121522
x
12 m
 21215
2
1
12

30. 15 m
9.4 m
x
222
1215 x
squared
12 m
 21215
2
1

31. 12 m
15 m
9.4 m
x
222
1215 x
so x equals…
x
 21215
2
1

32. 12 m
15 m
9.4 m
x
222
1215 x
xsquare root…
 21215
2
1

33. 12 m
15 m
9.4 m
x
222
1215 x
xof the whole thing
 21215
2
1

34. 12 m
15 m
9.4 m
x
222
1215 x
22
1215 xof the whole thing
 21215
2
1

35. 12 m
15 m
9.4 m
x
222
1215 x
Because the opposite
of squared…
22
1215 x
 21215
2
1

36. 12 m
15 m
9.4 m
x
222
1215 x
22
1215 xis square root
 21215
2
1

37. 12 m
15 m
9.4 m
x
222
1215 x
Putting this in the
calculator gives…
22
1215 x
 21215
2
1

38. 12 m
15 m
9.4 m
x
222
1215 x
Putting this in the
calculator gives…
22
1215 x
2.19x
 21215
2
1

39. 12 m
15 m
9.4 m
222
1215 x
So the unknown length
is 19.2 m
22
1215 x
2.19x
 21215
2
1 19.2 m

40. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
Now let’s find the
Area of the Left Side
 21215
2
1

41. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
Now let’s find the
Area of the Left Side
 21215
2
1

42. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
2.194.9 
 21215
2
1
Now let’s find the
Area of the Left Side

43. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
2.194.9 
Now find the Area of
the Right Side
 21215
2
1

44. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
 2.194.9
 21215
2
1
Add that on

45. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
So the Area of the
Right Side is
 2.194.9
 21215
2
1

46. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
12
 2.194.9
 21215
2
1
So the Area of the
Right Side is

47. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
So the Area of the
Right Side is
12
 2.194.9
 21215
2
1

48. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
If it’s 9.4 m there
12
 2.194.9
 21215
2
1

49. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
12
9.4 m
 2.194.9
 21215
2
1
It’s 9.4 m there as
well

50. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
4.912
9.4 m
 2.194.9
 21215
2
1
It’s 9.4 m there as
well

51. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
4.912
9.4 m
And finally, find the
Bottom Area
 2.194.9
 21215
2
1

52. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
 4.912
9.4 m
 2.194.9
 21215
2
1
Add that on

53. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
9.4 m
Bottom Area
 4.912
 2.194.9
 21215
2
1

54. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
9.4 m
154.9 
 4.912
 2.194.9
 21215
2
1
Bottom Area

55. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
9.4 m
154.9 
 4.912
 2.194.9
 21215
2
1
Putting this altogether in
the calculator gives…

56. 12 m
15 m
9.4 m
19.2 m
222
1215 x
22
1215 x
2.19x
9.4 m
2
3.614154.9 m
 4.912
 2.194.9
 21215
2
1
Putting this altogether in
the calculator gives…
Our Final Answer!