A symmetrical cross with equal arms has an area of 2016 cm^2 and all sides of integer length in centimetres. What is the smallest petimeter the cross can have?
draw such a cross. It consists of 5 squares of side x.
As you recall, the area of each such square is x^2
Now look at the perimeter. It has length 15x.
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To find the smallest possible perimeter of the symmetrical cross, we first need to find the length of the arms and then calculate the perimeter.
Let's assume the length of the arms is 'x' cm. Since the cross is symmetrical, we know that the area of the cross is the sum of the areas of four congruent right-angled triangles formed by the arms.
The area of each right-angled triangle can be calculated using the formula:
Area = (1/2) * base * height
In this case, the base and height of each triangle are equal to 'x' cm, so the formula becomes:
Area = (1/2) * x * x
Now we equate the given area to 2016 cm^2:
2016 = (1/2) * x * x
We can simplify this equation to:
4032 = x^2
To find an integer value for 'x', we need to find the factors of 4032.
The prime factorization of 4032 is: 2 * 2 * 2 * 2 * 2 * 3 * 3 * 7 * 2.
To find the factors, we can combine the prime factors in different ways.
For example:
2 * 2 * 2 * 2 * 2 * 3 * 3 * 7 = 2016
This indicates that 'x' could be equal to 2 * 2 * 2 * 2 * 2 * 3 * 3 * 7, which simplifies to 2^5 * 3^2 * 7.
Now we have the length of one arm, and to find the perimeter, we multiply it by 8 since there are 4 arms in the cross:
Perimeter = 8 * (2^5 * 3^2 * 7)
Simplifying this expression:
Perimeter = 8 * 32 * 9 * 7
Perimeter = 16128 cm
Therefore, the smallest possible perimeter of the symmetrical cross is 16128 cm.