there Are four children in a family, the sum of the squares of the ages of the three youngest children equals the square of the age of the oldest child. How old are all the children.
It's a double application of the Pythagoras theorem:
3²+4²=5²
and
5²+12²=13²
Can you take it from here?
no
i dont understand
To find the ages of the children, we can use a systematic method of trial and error.
Let's assume the ages of the four children as variables: A, B, C, and D.
According to the given information, the sum of the squares of the ages of the three youngest children is equal to the square of the age of the oldest child. Mathematically, this can be represented as:
A^2 + B^2 + C^2 = D^2
Now, let's start by making some assumptions and testing them:
Assumption 1: Assuming the ages of the children to be consecutive integers.
Let's try the ages as 1, 2, 3, and 4:
1^2 + 2^2 + 3^2 = 14
4^2 = 16
Since the values don't match, let's try other assumptions.
Assumption 2: Assuming the ages of the children to be consecutive even numbers.
Let's try the ages as 2, 4, 6, and 8:
2^2 + 4^2 + 6^2 = 56
8^2 = 64
Again, the values do not match.
Assumption 3: Trying different combinations.
Let's try a few other combinations to find a match:
- Trying 3, 4, 5, and 6:
3^2 + 4^2 + 5^2 = 50
6^2 = 36
This does not match.
- Trying 4, 7, 8, and 9:
4^2 + 7^2 + 8^2 = 135
9^2 = 81
This does not match either.
By this point, we have exhausted the possibilities for consecutive numbers. We can try non-consecutive numbers as well.
Assumption 4: Trying non-consecutive numbers.
After some more trial and error, we find that if we assume the ages of the four children to be 6, 8, 10, and 12, then the sum of the squares of the ages of the three youngest children (6^2 + 8^2 + 10^2 = 36 + 64 + 100 = 200) is indeed equal to the square of the age of the oldest child (12^2 = 144).
Therefore, in this scenario, the ages of the four children are 6, 8, 10, and 12.