Maximize The Area

Let $H=PB\cap QR$ , $PH=x$ . The second Euclid's theorem yields $QH^2=\sqrt{PH\cdot HB}=x(20-x)$ Therefore $[PQR]=PH \cdot QH=x\sqrt{x(20-x)}$ Using AM-GM inequality we get $[PQR]=\sqrt{x^3(20-x)}=3\sqrt{3}\sqrt{\frac{x}{3}\cdot \frac{x}{3}\cdot \frac{x}{3}\cdot(x-20)} \leq 3\sqrt{3}\sqrt{\left(\frac{20}{4}\right)^4}=75\sqrt{3}=129.904$ The maximum is attained when $x=15$ .

Largest area must be an equilateral triangle. So put R=10 in 0.75 x $sqrt{3}$ x $R^{2}$ and get the value, ie 129.903

VERY SIMPLE ... a largest area trianle inside a circle is an equilateral trianle.... so consider PQR AS AN equilateral trianle nd solve ....bingo

Think that, BP and QR intersects at D \ and O is the centre of the circle.\ Add O,R and O,Q\ now DR= $\sqrt { { 10 }^{ 2 }-{ \left( 10-BD \right) }^{ 2 } }$ = $\sqrt { 20.BD-{ BD }^{ 2 } }$ \ $\therefore$ QR=2 $\sqrt { 20.BD-{ BD }^{ 2 } }$ \ and DP=20-BD\ $\therefore$ the area of triangle $\triangle$ PQR= $\frac { 1 }{ 2 } \times 2\sqrt { 20.BD-{ BD }^{ 2 } } \times \left( 20-BD \right)$ \ Now the maximum result is for BD=5, which is "129.9038"

cause QR is parallel to the tangent at P and < angle QOR = x (O is center of circle), so we can get : \n < QOP = < ROP = (360 - x)/2 = 180 - x/2 Area of triangle PQR : = 1/2 QO RO sin(< QOR) + 1/2 QO PO sin(< QOP) + 1/2 PO RO sin(< ROP) = 1/2 10 10 sin(x) + 1/2 10 10 sin(180-x/2) + 1/2 10 10 sin(180-x/2) = 50 sin(x) + 100 sin(x/2)
define the area of triangle PQR in f(x) = 50 sin(x) + 100 sin(x/2)
to get maximum of a function f(x) , f'(x)=0 : f(x) = 50 sin(x) + 100 sin(x/2) f'(x) = 0 50 cos(x) + 50 cos(x/2) = 0 known that cos(x/2) = sqrt((1+cos(x))/2) so cos(x) = -sqrt((1+cos(x))/2) 2 cos^2(x)-cos(x)-1 = 0 (2 cos(x)+1)(cos(x)-1) = 0 x = {0 , 2 pi/3 , 5 pi/3} the reasonable value of x is 2 pi/3, so f(2 pi/3) = 50 sin(2 pi/3) + 100 sin(pi/3) = 75*sqrt(3) = 129.9038

^^V

(1) Take the chord at x from the centre. Hence the expression for the Area is as follows.
(2) The height of the triangle H = 10 + x. The (base/2)^2 = (chord length/2)^2 = 10^2 - x^2..
(3) (Area)^2 = (1/2)^2 (10^2 - x^2.) ( 10 + x )^2.
(4) Differentiating w.r.t. x and setting to zero, we get x =5. {Note that max for area is the same as max for (Area)^2 . I have used this method so the differentiation is easy.}.
(5) H= 10 + 5 = 15;.... base/2 = sqrt( 10^2 - 5^2) = 5(sqrt3)....Area = 129.90.
I give this as another method. Ashu Dablo's is the best method.
I may also mention that given a parameter of a rectangle, square has the max area. Given the total surface area, cube has the max volume. For and open top box, given cardboard area, max. volume is obtain with square base and height = half of the length of the square side. Given surface area of ellipsoid, sphere has max volume. And so on, symmetry gives max. min.