Let Janet's age to be `j` , Kim's age to be `k` , and Sue's age to be `s` .

The square of Janet's age is 400 more than the square of the sum of Kim's and Sue's ages. This information can be represented algebraically as:

`j^{2} = (k+s)^{2} + 400`

Kim's and Sue's ages total 10 less than Janet's age, that is:

`(k+s) + 10 = j`

It then means that `j^{2}` is also `((k+s) + 10)^{2}` from the second equation

We substitute this value of `j^{2}` in the first equation to give

`(k+s)^{2} + 20(k+s) + 100 = (k+s)^{2} + 400`

The `(k+s)^{2}` in both sides of the equation cancel out to yield

`20(k+s) = 300`

So that `(k+s) = 15`

and `j = 15 + 10 = 25`

The square of Janet's age is `(25)^{2} = 625`

However, we need further information in order to calculate Kim and Sue's ages. The square of the sum of Kim and Sue's age is `(k +s)^{2} = 15^{2} = 225`

Let Janet's age be "*x*".

Kim+Sue's ages = *x - 10*

The square of Janet is `x^2.`

The square of the sum of Kim & Sue's ages = `(x-10)^2`

Since Janet's is 400 more,

`x^2 = (x-10)^2 + 400`

`x^2 = x^2-20x+100+400`

`20x = 500`

`x = 25`

Janets is 25 years old, and Kim and Sue's ages together total 15.

**The square of Kim's age is 625. The square of the sum of Kim and Sues ages is 225.**