Each side of a square is 20cm. Four points on the sides are joined as shown to form an inner square. What is the minimum area of the inner square?
Can anyone help with this using quadratics? Please!
If the points are located so that each side is divided into sections of length x and (20-x), then the side (s) of the inner square can be calculated
s^2 = x^2 + (20-x)^2
the are of the inner square is thus
a(x) = s^2 = x^2 + (20-x)^2 = 2x^2 - 40x + 400
That is just a parabola with vertex at x = 40/4 = 10
So, the minimum size of the inner square is just 10*10 = 100, or 1/4 the area of the outer square, when its sides are half as long as the outer sides.
Certainly! To find the minimum area of the inner square, we can use the concept of quadratics.
Let's denote the side length of the inner square as 'x'. If we join the midpoints of each side of the outer square, we form a smaller square inside.
The diagonal of the outer square is the hypotenuse of a right-angled isosceles triangle, where the two legs are the side length of the outer square (20 cm). Using the Pythagorean theorem, we can find the length of the diagonal, which is also the diagonal of the inner square.
Using the Pythagorean theorem: (leg)^2 + (leg)^2 = (hypotenuse)^2
(20 cm)^2 + (20 cm)^2 = (diagonal)^2
400 cm^2 + 400 cm^2 = (diagonal)^2
800 cm^2 = (diagonal)^2
Now, the diagonal of the inner square is 'x√2' (since it's a right-angled isosceles triangle). We can equate these two diagonals:
(diagonal of inner square)^2 = (diagonal of outer square)^2
(x√2)^2 = (20 cm)^2
2x^2 = 400 cm^2
x^2 = 200 cm^2
To find the minimum area, we take the square root of both sides:
√(x^2) = √(200 cm^2)
x = √200 cm
Thus, the side length of the inner square is approximately 14.14 cm (rounded to two decimal places).
To find the minimum area of the inner square, we square the side length:
Area of inner square = (side length)^2
Area = (14.14 cm)^2
Area = 200 cm^2
Therefore, the minimum area of the inner square is 200 square centimeters.