SOLUTION: If the diagonal of a swuare is 42.5cm long, find I) the perimeter II) the area Of the square

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Question 1120603: If the diagonal of a swuare is 42.5cm long, find
I) the perimeter
II) the area
Of the square
Found 3 solutions by addingup, ikleyn, MathTherapy:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
A square divided by a diagonal becomes two 45-45-90 triangles. this is a special type of triangle. in this triangle, the two short sides are
= sqrt(hypotenuse)
= sqrt(42.5) = 6.52
I) The perimeter of the square: 6.52 x 4 = 26.08
II) The area: Side^2 = 6.52 x 6.52 = 42.51

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
(I)  If "a" is the side of the square, then, according to Pythagoras,


        a%5E2+%2B+a%5E2 = 42.5%5E2,

        2a%5E2 = 42.5%5E2

        a%5E2 = 42.5%5E2%2F2 = 903.125  =====>  a = sqrt%28903.125%29 = 30.05 cm.

        Then the perimeter of the square is  4a = 4*30.05 cm = 120.2 cm.


(II)  The area of the square is  a%5E2 = 30.05%5E2 = 903.0025 square centimeters.

The solution by the other tutor is absolutely incorrect, and deserves the lowest point.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
If the diagonal of a swuare is 42.5cm long, find
I) the perimeter
II) the area
Of the square
People like adding up don't have a clue about certain math problems. 

In addition, can the areas of 2 triangles that form a square have pretty much the same value as the square's hypotenuse: 42.51 cm2 and 42.5 cm?

How can the perimeter of a square be SHORTER than the length of its hypotenuse? 

The square does form two 45-45-90 special triangles, but the matrix%281%2C3%2C+hypotenuse%2C+%22=%22%2C+S+%2A+sqrt%282%29%29, with one side of the square being S

We can then say that: 

You should now be able to find both the perimeter and area of the square.