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]