Jacob's Responses in WyzAnt Answers

So I think your problem is that you are confusing how to compute the mean and median which I will cover first. After that I'll explain how to find the other values just in case those are a problem for you as well.

First, we should be clear about what the mean and median are.

The mean is what we typically think of as the average of the numbers. So in this case we would add all the numbers as you have, and get 260 as you have. The next step is to divide by the number of items in the list which in this case is 10. This gives us a mean of 26.

The median is technically the number that splits the data set into its higher and lower halves. When a data set has an odd number of entries it is simply the middle number in an ordered list of the entries.

e.g. the median of the list of numbers {1,2,3,4,5} is 3 because there are 2 entries above and below 3 in the ordered list.

However, if there are an even number of entries it is a little bit harder. We take the ordered list and find the two central values and then take the mean of those values.

e.g. the median of the list of numbers {1,2,3,4,5,6} is found by averaging the two middle terms, 3 and 4, which results in the answer of 3.5. (I got 3.5 by computing 3+4/2)

The extremes are simply the highest and lowest value in the set 

e.g. the extremes of the set {1,2,3,4,5,6,} are 1 and 6

The range is the difference in extremes

e.g. the ranger of the set {1,2,3,4,5,6,} is 6-1=5

The quartiles are the medians of the upper and lower halves of the data. To figure this out we first split the data into its upper and lower halves based on the median.

e.g. the set {1,2,3,4,5,6,7} would be divided into the sets {1,2,3} and {5,6,7}

       the set {1,2,3,4,5,6} would be divided into the sets {1,2,3} and {4,5,6}

       note: the sets must be equal in size so if there are an odd number of entries in the original set we exclude the median.

Next we find the median of each of these sets and those are the quartiles

e.g. {1,2,3,4,5,6,7} -> {1,2,3} and {5,6,7} which gives us the first quartile value of 2 and the third quartile value of 6.

       {1,2,3,4,5,6} -> {1,2,3} and {4,5,6} which gives us the first quartile value of 2 and the third quartile value of 5.

Note: It may seem like the upper number should be the second quartile, but the median technically counts as the second quartile number so the higher bound is the 3rd quartile.

To find the interquartile range we subtract the 1st quartile value from the 3rd quartile value.

e.g. {1,2,3,4,5,6,7} has an interquartile range of 4.

       {1,2,3,4,5,6} has an interquartile range of 3.

I hope this helps you to be able to find all of the statistics your teacher wanted you to find. 

I'm not sure what the actual question is, but if your goal is to understand why this is true, then I think I can help.

First note that in all right triangles the following are true:

a² + b² = c²

Area = (1/2)(a)(b)

Next observe:

a=b if and only if t is isosceles.

This allows the substitution of a for b in the original equations.

a² + a² = c²

Area = (1/2) a²

Solving the top equation for a² yields

a² = (1/2) c²

which when substituted into the area formula gives

Area = (1/2)(1/2) c² = (1/4)c²

I hope that was helpful!