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The following solution is based on the work of
Ben and Sarah. Well done to both of you.
.
The boundary of the holly leaf can be divided into sections. There
are two sections made up from arcs of a circle like the first
diagram (above) and there are eight arcs like the second diagram
above. So that boundary can be calculated by finding the sum of the
lengths of these arcs
The length of an arc can be calculated by the following
formula:
$l$ = $rx$ ($x$ measured in radian)
$\angle APB$= 360degrees - $\angle PAC$ - $\angle PBC$ -
$\angle ACB$ = (360 - 90 - 90 - 125) degrees = 55 degrees
P*B*D*Q* is a rectangle, so $\angle Q$*$P$*$B$* = 90
degrees
180 degrees correspond to $\pi $, so 55 degrees can be
converted into 55$\pi$/180 or 11$\pi$/36; and 90 degrees are
equivalent to 90$\pi$/180 or $\pi$/2
So the length of the arc subtending the angle APB is:
11$\pi$r/36, and that subtending the angle B*P*Q* is $\pi$r/2
Hence the length of the boundary of the yellow area is:
4 $\times$ 11$\pi$r/36 + 2 $\times$ $\pi$r/2 =
20$\pi$r/9
The parameter of the holy leaf is
formed by 8 arcs subtended by angle ACB and 2 semi-cirles
.
If $\angle ACB$ = 125 degrees, the parameter of the holy leaf
is:
8 $\times$ (5 $\times$125$\pi$/180) + 2 $\times$ 5 $\times$ $\pi$ =
340$\pi$/9 = 118.68 cm
If$\angle ACB$ = 135 degrees, the parameter of the holy leaf
is:
8 $\times$ (5 $\times$ 135$\pi$/180) + 2 $\times$ 5 $\times$ $\pi$
= 40$\pi$ = 125.66 cm > 118.68cm
The area of the holy leaf is
formed by 8 blue and 2 red areas.
Blue area = 2 $\times$ area of APC - area of the sector
corresponding to the angle ACB
Area of APC = $\frac{1}{2}$ AP.AC
Because $\tan$ ($\angle ACP$) = AP/AC, AP = AC. $tan \angle
ACP$
Also$\angle ACP$ =$\angle ACB$/2
Hence area of APC = $\frac{1}{2}$ (AC. $tan (\angle ACB$/2)).AC
=$\frac{1}{2}$ ($r^2$ $tan (\angle ACB$/2))
Area of the sector subtended by x degrees = $\pi$ $r^2$ $\times$
($x/360$)
So the area of the sector subtended by angle ACB = $\pi$ $r^2$
$\times$ ($\angle ACB$ $/360$)
So the blue area = 2$\times$ $\frac{1}{2}$ ($r^2$ $tan (\angle
ACB$/2)) -$\pi$ $r^2$ $\times$ ($\angle ACB$ $/360$)
Red area = area of rectangle P*B*D*Q* - area of semi-circle =
AP.(2r) -$\pi$ $r^2$/2 =
AC. $tan (\angle ACB$/2)$\times$(2r) - $\pi$ $r^2$/2 = $r^2$ $tan
(\angle ACB$/2) -$\pi$ $r^2$/2
So the area of the holy leaf is:
8 $\times$ ($r^2$ $tan (\angle ACB$/2) - $\pi$ $r^2$ $\times$
($\angle ACB$ $/360$)) + 2 $\times$ ($r^2$ $tan (\angle ACB$/2) -
$\pi$ $r^2$/2)
= 12$r^2$ $tan (\angle ACB$/2) - $\pi$ $r^2$ ($\angle ACB/45$ +
1)
If$\angle ACB$ = 125 degrees, the area of the holy leaf is:
12 $\times$ $5^2$ $tan (125$/2) - $\pi$ $5^2$ ( 125/45 + 1) =279.59
$cm^2$
If$\angle ACB$ = 135 degrees, the area of the holy leaf is:
12 $\times$ $5^2$ $tan (135$/2) - $\pi$ $5^2$ ( 135/45 + 1) =
410.10 $cm^2$ > 279.59 $cm^2$
From our findings, some
conclusions can be drawn as follows:
The boundary of the yellow area is
greater than 2r$\pi$, which means the surface of the holy leaf has
negative curvature.
As the angle ACB increases
($\angle ACB$ < 180 degrees), the curvature of the surface will
increase; both parameters and areas of the holy leaf will also
increase.