Circle O is inscribed is square ABCD, and at the same time, is circumscribed about square PQRS. WHich are is the smaller area, the region inside the inscribed circle minus the area of square PQRS or the are inside square ABCD minus the area of the inscribed circle? The length of the side of square ABCD measures two feet.
Draw you diagram so that the vertices of square ABCD and square PQRS are in corresponding order, that is AB is || to PQ
Let M be the midpoint of AB and N be the midpoint of PQ, and let O be the centre of the circle
Joine AC, which will pass through PR
AM =1, and OM = 1
area of ABCD = 4, area of circle = π(1^2) = π
area of region between outer square and circle = 4-π
= appr .8584
PO = 1, let ON = x
x^2 + x^2 = 1^2
2x^2 = 1
x^2 = 1/2
x = 1/√2
so ON = 1/√2 ---> PN = 1/√2 ---> PQ = 2/√2
area of inside square = (2/√2)^2= 2
difference between circle and inside square = π - 2
= appr 1.1416
BTW, this is the approach Archimedes took in finding an approximation to π .
He took successive differences in areas between
incribed and circumsribed squares, octogons, 16-sided polygons, 32-sided polygons, etc.
To determine which area is smaller, let's first find the areas of the regions inside the inscribed circle minus the area of square PQRS and inside square ABCD minus the area of the inscribed circle.
Area of region inside the inscribed circle minus the area of square PQRS:
1. The inscribed circle has a diameter equal to the side length of square ABCD, which is 2 feet. Therefore, its radius is 1 foot.
2. The area of the inscribed circle is given by A1 = πr^2 = π(1^2) = π square feet.
3. The side length of square PQRS is equal to the diameter of the inscribed circle, which is 2 feet. Thus, its area is A2 = (2)^2 = 4 square feet.
4. Subtracting the area of square PQRS from the area of the inscribed circle, we get A1 - A2 = π - 4 square feet.
Area of region inside square ABCD minus the area of the inscribed circle:
1. The side length of square ABCD is given as 2 feet. Therefore, its area is A3 = (2)^2 = 4 square feet.
2. The area of the inscribed circle, as previously calculated, is A1 = π square feet.
3. Subtracting the area of the inscribed circle from the area of square ABCD, we have A3 - A1 = 4 - π square feet.
Comparing the areas:
1. We have A1 - A2 = π - 4 square feet for the region inside the inscribed circle minus the area of square PQRS.
2. We also have A3 - A1 = 4 - π square feet for the area inside square ABCD minus the area of the inscribed circle.
Since π is approximately 3.14, we can see that π - 4 is negative, while 4 - π is positive. Therefore, the area inside square ABCD minus the area of the inscribed circle is smaller.
In conclusion, the area inside square ABCD minus the area of the inscribed circle is the smaller area.