A square is inscribe in a circle with a diameter of 40 cm and the centre C. Four lines are drawn between the midpoints of the adjacent sides of the square to form a smaller square. What is the perimeter of the smaller square to the nearest tenth of a centimetre?   cm

Step 1: Find the side of the larger square

The diameter of the circle is the diagonal of the square. We can use the Pythagorean theorem to find the side of the square. The theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this case, the hypotenuse is the diameter of the circle (40 cm) and the two sides are equal (since it's a square), let's call them 'a'. So, we have:

a^2 + a^2 = 40^2 2a^2 = 1600 a^2 = 800 a = sqrt(800) = 28.3 cm (to the nearest tenth)

Step 2: Find the side of the smaller square

The side of the smaller square is the diagonal of the larger square. So, we can use the Pythagorean theorem again:

a^2 + a^2 = b^2 2(28.3)^2 = b^2 b = sqrt(2*(28.3)^2) = 40 cm

Step 3: Find the perimeter of the smaller square

The perimeter of a square is four times the length of one side. So, the perimeter of the smaller square is:

4 * 40 = 160 cm

So, the perimeter of the smaller square is 160 cm to the nearest tenth of a centimetre.