Question 753612
The area of a square is 72 cm. The area of the circle drawn on its diagonal is?
<pre>
I can't tell what is meant by "the circle drawn on its diagonal".


THIS:{{{drawing(200,200,-12,12,-12,12,
triangle(-8,-8,-8,8,-8,0),triangle(-8,-8,8,-8,8,-8),
triangle(8,-8,8,8,8,0),circle(0,0,8),triangle(-8,-8,8,8,8,8),
triangle(8,-8,8,8,8,0),triangle(-8,8,8,8,8,8) )}}}OR THIS:{{{drawing(200,200,-12,12,-12,12,
triangle(-8,-8,-8,8,-8,0),triangle(-8,-8,8,-8,8,-8),
triangle(8,-8,8,8,8,0),circle(0,0,8sqrt(2)),triangle(-8,-8,8,8,8,8),
triangle(8,-8,8,8,8,0),triangle(-8,8,8,8,8,8) )}}}

In either case we calculate s the length of a side of the square, 
using:

    A = s²
   72 = s²
  &#8730;<span style="text-decoration: overline">72</span> = &#8730;<span style="text-decoration: overline">s²</span>
  &#8730;<span style="text-decoration: overline">72</span> = s
&#8730;<span style="text-decoration: overline">36·2</span> = s
  6&#8730;<span style="text-decoration: overline">2</span> = s

If it's the first case we draw a horizontal diameter (in green) which is equal
in length to the side of the square, 6&#8730;<span style="text-decoration: overline">2</span>. and therefore each radius is half of 
that or 3&#8730;<span style="text-decoration: overline">2</span>

{{{drawing(200,200,-12,12,-12,12, green(line(-8,0,8,0)),
triangle(-8,-8,-8,8,-8,0),triangle(-8,-8,8,-8,8,-8),locate(2,0,3sqrt(2)),
locate(-6,0,3sqrt(2)),
triangle(8,-8,8,8,8,0),circle(0,0,8),triangle(-8,-8,8,8,8,8),
triangle(8,-8,8,8,8,0),triangle(-8,8,8,8,8,8) )}}}

So to find the area of the circle we use:

A = <font face="symbol">p</font>r<sup>2</sup>
A = <font face="symbol">p</font>(3&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup>
A = <font face="symbol">p</font>·3<sup>2</sup>·(&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup>
A = <font face="symbol">p</font>·9·2
A = 18<font face="symbol">p</font> cm²
A &#8776; 56.55 cm² 

But if it's the second way, then the diagonal is the diameter
of the circle, d:

{{{drawing(200,200,-12,12,-12,12,
locate(0,-8,6sqrt(2)),locate(8.3,0,6sqrt(2)),locate(.1,.3,d),
triangle(-8,-8,-8,8,-8,0),triangle(-8,-8,8,-8,8,-8),
triangle(8,-8,8,8,8,0),circle(0,0,8sqrt(2)),triangle(-8,-8,8,8,8,8),
triangle(8,-8,8,8,8,0),triangle(-8,8,8,8,8,8) )}}}

We calculate d with the Pythagorean theorem:

d<sup>2</sup> = (6&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup> + (6&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup>
d<sup>2</sup> = 2·(6&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup> 
d<sup>2</sup> = 2·6<sup>2</sup>(&#8730;<span style="text-decoration: overline">2</span>)<sup>2</sup>
d<sup>2</sup> = 2·36·2
d<sup>2</sup> = 144
 d = 12

And the radius is half the diameter, so the radius of the circle
is 6 cm.

To find the area of the circle we use:

A = <font face="symbol">p</font>r<sup>2</sup>
A = <font face="symbol">p</font>6<sup>2</sup>
A = <font face="symbol">p</font>·36
A = 36<font face="symbol">p</font> cm²
A &#8776; 113.1 cm²

The area of the circle in the second case is exactly twice
the area in the first case.  That is to say, the circle
circumscribed about a square has twice the area of the circle
inscribed in the square. 

Edwin</pre>