Answer:

Step-by-step explanation:
Let us consider the equation 
For a quadratic equation in a standard form,
, the axis of symmetry is the vertical line
.
Here in this case we have, 
Putting the values we get,

We can see that the axis of symmetry is x=3 and the graph is giving minimum at x=3.
Therefore, the required equation is
. Refer the image attached.

![\sf \left[\begin{array}{cc}\sf 4&\sf 6\\ \sf 5 &\sf 8 \\ \sf 3 &\sf -2\end{array}\right]-\left[\begin{array}{cc}\sf 2&\sf 3\\ \sf 1 &\sf 4 \\ \sf -5&\sf3\end{array}\right]](https://tex.z-dn.net/?f=%5Csf%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%204%26%5Csf%206%5C%5C%20%5Csf%205%20%26%5Csf%208%20%5C%5C%20%5Csf%203%20%26%5Csf%20-2%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%26%5Csf%203%5C%5C%20%5Csf%201%20%26%5Csf%204%20%5C%5C%20%5Csf%20-5%26%5Csf3%5Cend%7Barray%7D%5Cright%5D)
Just substract corresponding terms
![\\ \sf\longmapsto \left[\begin{array}{cc}\sf 2 &\sf 3\\ \sf 4&\sf4\\ \sf 8&\sf -5\end{array}\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%202%20%26%5Csf%203%5C%5C%20%5Csf%204%26%5Csf4%5C%5C%20%5Csf%208%26%5Csf%20-5%5Cend%7Barray%7D%5Cright%5D)
Option B
Answer:
84,000,108
Step-by-step explanation:
It's obviously none of the above because the true area will have a square root of 3 in it because of the equilateral triangle. They want us to find the approximate area. When I was in school the difference between an approximation and the exact answer was greatly stressed but that seems to be fading of late.
That's a rectangle less an equilateral triangle. The area of an equilateral triangle is
which is easily derived but I won't bother here.

Answer: b