HSC Worked Solutions
1
7
C
T
06
10
b
projectmaths.com.au
A rectangular piece of paper PQRS has sides
PQ = 12 cm and PS = 13 cm. The point O is
the midpoint of PQ. The points K and M are to
be chosen on OQ and PS respectively, so that
when the paper is folded along KM, the corner
that was at P lands on the edge QR at L. Let
OK = x cm and LM = y cm.
Copy or trace the diagram into your writing
booklet.
(i)
Show that QL2 = 24x.
(ii)
Let N be the point on QR for which MN
is perpendicular to QR. By showing
that ΔQKL ||| ΔNLM, deduce that
6 (6  x)
y=
1
7
C
T
06
(i)
(ii)
10
bx
x
6 (6  x)2
(iii)
Show that the area, A, of ΔKLM is given by A =
(iv)
(v)
Use the fact that 12 ≤ y ≤ 13 to find the possible values of x.
Find the minimum possible area of ΔKLM.
2 x
.
QL2 = (6 + x)2 – (6 – x)2
= 36 + 12x + x2 –[36 – 12x + x2]
= 24x
Let  QLK = 
(straight  )
  MLN = 180 – (90 + )
= 90 - 
  LMN = 180 – (90 + 90 - ) (  sum of Δ)
= 
  MLN =  LMN
and  LQK =  MNL (given)
 ΔQKL ||| ΔNLM (2  s equal)
y
=
6x
y =
=
=
(iii)
1
3
12
(matching sides of sim Δs in propn)
2 4x
1 2(6  x)
2 6x
6(6  x)
6x
6 (6  x)
x
1
 MN  LK
2
1
=
 y  (6 + x)
2
Area =
=
6 (6  x)2
2 x
HSC examination papers © Board of Studies NSW for and on behalf of the Crown in right of State of New South Wales
1
2
3
HSC Worked Solutions
(iv)
projectmaths.com.au
Let y = 12
6 (6  x)
x
Let y = 13
6 (6  x)
= 12
x
6 (6 + x) = 12 x
2
6(6 + x)
36 + 12x + x2
x2 – 12x + 36
(x – 6)2
x
= 13
6 (6 + x) = 13 x
6(6 + x)2 = 169x
6(36 + 12x + x2) = 169x
6x2 + 72x + 216 = 169x
6x2 – 97x + 216 = 0
(3x – 8) (2x – 27) = 0
= 144x
= 24x
=0
=0
=6
x=2
2
1
, 13
3
2
But 0  x  6, therefore, x = 2
As 12 ≤ y ≤ 13, then 2
(v)
A =
2
≤ x ≤ 6.
3
2
.
3
6 (6  x)2
2 x
1
1 
2 x . 6.2(6  x) .1  6 (6  x) .2. x 2
dA
2
=
4x
dx
1
2

1
4 x . 6.(6  x)  6 (6  x)2.x 2
=
4x

=0
1
6 x 2 .(6  x)[4x  (6  x)]
=0
4x
6 (6  x)(3x  6)
4x x
=0
(x + 6)(3x – 6) = 0
x = -6, 2
But as 2
2
≤ x ≤ 6, then x  -6 and x  2
3
The minimum must lie at endpoints: x = 2
Subs x = 2
2
in
3
A =
2
and x = 6.
3
6 (6  x)2
2 x
2
6 (6  2 )2
3
=
2
2 2
3
1
= 56
3
1
2
 The minimum area is 56 units
3
Subs x = 6 in A =
=
6 (6  x)2
2 x
6 (6  6)2
2 6
= 72
* These solutions have been provided by projectmaths and are not supplied or endorsed by the Board of Studies
Board of Studies: Notes from the Marking Centre
HSC examination papers © Board of Studies NSW for and on behalf of the Crown in right of State of New South Wales
HSC Worked Solutions
projectmaths.com.au
Source: http://www.boardofstudies.nsw.edu.au/hsc_exams/
HSC examination papers © Board of Studies NSW for and on behalf of the Crown in right of State of New South Wales