Given that the rope is 5 m long, calculate the percentage of Bill’s field that Gruff is able to graze. Give your answer correct to the nearest integer.

Bill replaces Gruff’s rope with another, this time of length $a,{\text{ }}4 < a < 10$, so that Gruff can now graze exactly one half of Bill’s field.

Show that $a$ satisfies the equation

$${a^2}\arcsin \left( {\frac{4}{a}} \right) + 4\sqrt {{a^2} - 16}  = 40.$$

Find the value of $a$.

EITHER

area of triangle $= \frac{1}{2} \times 3 \times 4\;\;\;( = 6)$     A1

area of sector $= \frac{1}{2}\arcsin \left( {\frac{4}{5}} \right) \times {5^2}\;\;\;( = 11.5911 \ldots )$     A1

OR

$\int_0^4 {\sqrt {25 - {x^2}} {\text{d}}x}$     M1A1

THEN

total area $= 17.5911 \ldots {\text{ }}{{\text{m}}^2}$     (A1)

percentage $= \frac{{17.5911 \ldots }}{{40}} \times 100 = 44\%$     A1

[4 marks]

METHOD 1

area of triangle $= \frac{1}{2} \times 4 \times \sqrt {{a^2} - 16}$     A1

$\theta  = \arcsin \left( {\frac{4}{a}} \right)$     (A1)

area of sector $= \frac{1}{2}{r^2}\theta  = \frac{1}{2}{a^2}\arcsin \left( {\frac{4}{a}} \right)$     A1

therefore total area $= 2\sqrt {{a^2} - 16}  + \frac{1}{2}{a^2}\arcsin \left( {\frac{4}{a}} \right) = 20$     A1

rearrange to give: ${a^2}\arcsin \left( {\frac{4}{a}} \right) + 4\sqrt {{a^2} - 16}  = 40$     AG

METHOD 2

$\int_0^4 {\sqrt {{a^2} - {x^2}} {\text{d}}x = 20}$     M1

use substitution $x = a\sin \theta ,{\text{ }}\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = a\cos \theta$

$\int_0^{\arcsin \left( {\frac{4}{a}} \right)} {{a^2}{{\cos }^2}\theta {\text{d}}\theta  = 20}$

$\frac{{{a^2}}}{2}\int_0^{\arcsin \left( {\frac{4}{a}} \right)} {(\cos 2\theta  + 1){\text{d}}\theta  = 20}$     M1

${a^2}\left[ {\left( {\frac{{\sin 2\theta }}{2} + \theta } \right)} \right]_0^{\arcsin \left( {\frac{4}{a}} \right)} = 40$     A1

${a^2}\left[ {(\sin \theta \cos \theta  + \theta } \right]_0^{\arcsin \left( {\frac{4}{a}} \right)} = 40$

${a^2}\arcsin \left( {\frac{4}{a}} \right) + {a^2}\left( {\frac{4}{a}} \right)\sqrt {\left( {1 - {{\left( {\frac{4}{a}} \right)}^2}} \right)}  = 40$     A1

${a^2}\arcsin \left( {\frac{4}{a}} \right) + 4\sqrt {{a^2} - 16}  = 40$     AG

[4 marks]

solving using ${\text{GDC}} \Rightarrow a = 5.53{\text{ cm}}$     A2

[2 marks]

Total [10 marks]