Mathematics · Oct / Nov 2024 · Paper 11 · Question 5

Perimeter =5 + 5 + 6 = 16 (km)

1

1

2

2

Attempts

r

θ

=

×

5

×

1.2

2

2

1

1

2

2

Area AOP =

×

×

5

×

1.2

=

3.75 km

*

4

2

Sets 1

××

5

OP

×

sin1.2

=

3.75

OP

=

...

2

OP =

1.6...

2

2

2

AP =

5

+

"1.6.."

−××

2 5 "1.6.." cos1.2

×

AP =

4.7km

or

4700m

(a)

M1: Attempts

5 1.2

r

θ

= ×

A1: Achieves 16 (km)

(b)

1

1

(or 1

1

2

2

M1: Attempts

r

θ

=

×

5

×

1.2

rl =

××

5 6

) May be implied by Area of sector =

2

2

2

2

15 seen.

1

1

3

1

2

2

M1: Attempts

×

×

5

×

1.2

or

×

×

5

×

1.2

or other full method to find the area for

4

2

4

2

one of the two regions.

A1*: Achieves given answer, including units, showing all steps with no errors and full

accuracy kept throughout. Score A0 if rounded values were used during working.

Approaches by verification are acceptable. The first M1 will apply the same way, a clear

attempt at the sector area formula will be needed to determine the area.

The second M may then be gained for attempting

Then for the A mark the calculations need to both be correct (giving 15), and a

conclusion made: E.g. (so) Area R2 = 3.75 km2.

(c)

M1: Sets 1

××

5

OP

×

sin1.2

=

3.75

OP

=

...

There are some longer methods to find OP, so

2

this may be scored for any complete method. E.g finds area of triangle APB from

1 5

2 sin1

difference of area triangle OAB

=

.2

and 3.75, the uses ratio OP:PB is same

2

Let X be base of perpendicular to OA though P, then

1

"1.5"

5

XP

3.75

XP

1.5

OX

so

××

=

=

=

2

tan1.2

5(c) Alt

"1.5"

AX =

5

−

=

...

tan1.2

AX =

4.416...

2

2

2

AP =

"4.416..."

+

"1.5"

=

...

AP =

4.7km

or

4700m

x =

2

1

1/x type shape

2