a The distance between V_2 and O(0,0) is 4. Substitute these values into the Distance Formula and form an equation.
B
b Note that the length of the height is determined by the y-coordinate of V_2.
C
c The area of the octagon is 8 times greater than the area of △ V_1OV_2.
A
a (2.8,2.8)
B
b 5.6 square units
C
c 44.8 square units
Practice makes perfect
a We know that V_2 lies on the line y=x, which means that its x- and y-coordinates are equal, so they can be written as V_2(x,x). It is also given that the radius of the octagon is 4.
Let's use this information to find the coordinates of V_2. We will use the Distance Formula.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
The origin of the octagon is O(0,0). Let's substitute ( x_1, y_1) with ( 0, 0), ( x_2, y_2) with ( x, x), and since the distance between O and V_2 is 4, we can substitute d with 4.
We conclude that the coordinates of V_2 are (2.8,2.8).
b Now, let's use the coordinates of V_2 to find the area of △ V_1OV_2. We will use the following formula.
A=1/2bh
Here, b is the base and h is the height of the triangle. Let's examine the diagram to find the values of b and h for △ V_1OV_2.
As we can see, the distance between O and V_1 is a radius of the octagon, so it is 4 units long. Hence, b=4. What about the height?
The length of the triangle's height is determined by the y-coordinate of V_2. Thus, h=2.8. Finally, we know all the needed information to calculate the area of △ V_1OV_2.
c Analyzing the given diagram, we can see that the octagon consists of 8 similar triangles like △ V_1OV_2. Therefore, the area of the octagon is 8 times greater than the area of the triangle.
\begin{aligned}
A_\text{octagon}=8A_\text{triangle}
\end{aligned}
From Part B, we know that A_\text{triangle}={\color{#0000FF}{5.6}} square units. Let's substitute this value into the above equation and find the area of the octagon.
\begin{aligned}
A_\text{octagon}=8({\color{#0000FF}{5.6}})=44.8\text{ units}^2
\end{aligned}
