MEASUREMENT OF A CIRCLE.
Proposition 1.
*The area of any circle is equal to a right-angled triangle
in which one of the sides about the right angle is equal to the radius,
and the other to the circumference, of the circle.*

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Let *ABCD *be the given circle, K the triangle described

Then, if the circle is not equal to *K, it *must
be either greater or less.
I. If possible, let the circle be greater than
*K.*

Inscribe a square *ABCD, *bisect the arcs *AB, BC,
CD, DA, *then bisect (if necessary) the halves, and so on, until the
sides of the inscribed polygon whose angular points are the points of division
subtend segments whose sum is less than the excess of the area of the circle
over
Thus the area of the polygon is greater than K.

Let *AE* be any side of it, and *ON *the perpendicular
on *AE *from the centre *0.*

Then* ON* is less than the radius of the circle
and therefore less than one of the sides about the right angle in K. Also
the perimeter of the polygon is less than the circumference of the circle,
i.e. less than the other side about the right angle in *K.*

Therefore the area of the polygon is less than K; which
is inconsistent with the hypothesis

Thus the area of the circle is not *greater than.*

II. If possible,_{ }let the circle be less than
*K*

Circumscribe a square,_{ }and let two adjacent
sides, touching the circle in* E*, *H, *meet in *T*.
Bisect the arcs between adjacent points of contact and draw the tangents
at the points of bisection. Let *A* be the middle point of the arc
*EH*, and *FAG *the tangent at *A.*

Then the angle *TAG *is a right angle.

Therefore
*TG* > *GA*

> *GH*.
It follows that the triangle *FTG* is greater than
half the area *TEAH.*

Similarly, if the arc *AH *be bisected and the tangent
at the point of bisection be drawn, it will cut off from the area *GAH*
more than one-half.

Thus, by continuing the process, we shall ultimately arrive
at a circumscribed polygon such that the spaces intercepted between it
and the circle are together less than the excess of K over the area of
the circle.

Thus the area of the polygon will be less than *K*

Now, since the perpendicular from *0* on any side
of the polygon is equal to the radius of the circle, while the perimeter
of the polygon is greater than the circumference of the circle, it follows
that the area of the polygon is greater than the triangle *K; *which
is impossible.

Therefore the area of the circle is not less than K.

Since then the area of the circle is neither greater nor
less than K, it is equal to it.

**Proposition 2.**
*The area of a circle is to the square on its diameter
as 11 to 14*

[The text of this proposition is not satisfactory, and
Archimedes cannot have placed it before Proposition 3, as the approximation
depends upon the result of that proposition.]

**Proposition 3.**
*The ratio of the circumference of any circle to its diameter
is less than 3 1/7
but greater than 3 10/71.*