Answer:
The area of the small square is 1 cm^2
Step-by-step explanation:
The large square consist in four identical rectangles and one small square.
Then the area of the small square will be equal to the difference between the area of the large square and the areas of the rectangles.
Because we have 4 equal rectangles, if R is the area of one rectangle, and S is the area of the large square, the area of the small square will be:
area = S - 4*R
We know that the area of the large square is 49 cm^2
Then:
S = 49cm^2
Remember that the area of a square of side length K is:
A = K^2
Then the side length of the large square is:
K^2 = 49 cm^2
K = √(49 cm^2) = 7cm
And we know that the diagonal of one rectangle is 5cm.
Remember that for a rectangle of length L and width W, the diagonal is:
D = √(L^2 + W^2)
Then:
D = √(L^2 + W^2) = 5cm
And for how we construct this figure, we must have that the length of the rectangle plus the width of the rectangle is equal to the side length of the large square, then:
L + W = 7cm
L = (7cm - W)
Replacing this in the diagonal equation, we get:
√((7cm - W)^2 + W^2) = 5cm
(7cm - W)^2 + W^2 = (5cm)^2 = 25cm^2
49cm^2 - 14cm*W + W^2 + W^2 = 25cm^2
2*W^2 - 14cm*W + 49cm^2 = 25cm^2
2*W^2 - 14cm*W + 49cm^2 - 25cm^2 = 0
2*W^2 - 14cm*W + 24cm^2 = 0
We can solve this for W using the Bhaskara's formula, the solutions are:
[tex]W = \frac{-(-14cm) \pm \sqrt{(-14cm)^2 - 4*2*(24cm^2)} }{2*2} = \frac{14cm \pm 2cm}{4}[/tex]
Then we have two solutions, and we only need one (because the length will have the other value)
We can take:
W = (14 cm + 2cm)/4 = 4cm
Then using the equation:
L + W = 7cm
L + 4cm = 7cm
L = 7cm - 4cm = 3cm
L = 3cm
Now remember that the area of one rectangle of length L and width W is:
R = L*W
Then the area of one of these rectangles is:
R = 4cm*3cm = 12cm^2
Now we can compute the area of the small square:
area = S - 4*R = 49cm^2 - 4*12cm^2 = 1cm^2
The area of the small square is 1 cm^2
