Step-by-step explanation:
Let a, b, c be the measures of the interior angles and x, y, z be the measures of the exterior angles of the triangle. Where x and adjacent to a, y is adjacent to b and z is adjacent to c.
By interior angle sum postulate of a triangle:
a + b + c = 180°... (1)
Therefore, by remote interior angle theorem:
x = b + c.... (2)
y = a + c..... (3)
z = a + b.... (4)
Adding equations (2), (3) & (4)
x + y + z = b + c + a + c + a + b
x + y + z = 2a + 2b + 2c
x + y + z = 2(a + b + c)... (5)
From equations (1) & (5)
![x + y + z = 2 \times 180 \degree \\ x + y + z = 360 \degree \\ x + y + z = 4 \times 90\degree \\ x + y + z = 4 \: right \: angles](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20z%20%3D%202%20%5Ctimes%20180%20%5Cdegree%20%5C%5C%20x%20%2B%20y%20%2B%20z%20%3D%20360%20%5Cdegree%20%5C%5C%20%20x%20%2B%20y%20%2B%20z%20%3D%204%20%5Ctimes%20%2090%5Cdegree%20%5C%5C%20x%20%2B%20y%20%2B%20z%20%3D%204%20%5C%3A%20right%20%5C%3A%20angles)
Thus, the sum of exterior angles so formed is equal to four right angles.
Proved.
Answer:
Is there any more to the problem but BC could be the answer it depends on what your supposed to be looking for
Step-by-step explanation:
Answer:
, ![SA=785\ m^{2}](https://tex.z-dn.net/?f=SA%3D785%5C%20m%5E%7B2%7D)
Step-by-step explanation:
Part 1) Find the lateral area of the given prism
we know that
The lateral area of the prism is equal to
![LA=PL](https://tex.z-dn.net/?f=LA%3DPL)
where
P is the perimeter of the triangular base
L is the length of the prism
<em>Find the perimeter P</em>
![P=5+6+11.21=22.21\ m](https://tex.z-dn.net/?f=P%3D5%2B6%2B11.21%3D22.21%5C%20m)
we have
![L=34\ m](https://tex.z-dn.net/?f=L%3D34%5C%20m)
substitute
![LA=(22.21)(34)=755\ m^{2}](https://tex.z-dn.net/?f=LA%3D%2822.21%29%2834%29%3D755%5C%20m%5E%7B2%7D)
Part 2) Find the surface area of the given prism
we know that
The surface area of the prism is equal to
![SA=2B+LA](https://tex.z-dn.net/?f=SA%3D2B%2BLA)
where
B is the area of the triangular base
LA is the lateral area of the prism
<em>Find the area B</em>
![B=(1/2)(5)(6)=15\ m^{2}](https://tex.z-dn.net/?f=B%3D%281%2F2%29%285%29%286%29%3D15%5C%20m%5E%7B2%7D)
we have
![LA=755\ m^{2}](https://tex.z-dn.net/?f=LA%3D755%5C%20m%5E%7B2%7D)
substitute
![SA=(2)(15)+755=785\ m^{2}](https://tex.z-dn.net/?f=SA%3D%282%29%2815%29%2B755%3D785%5C%20m%5E%7B2%7D)