SME
Grade 9 Math · Problem type
Precision: least and greatest possible area
Use this for “A rectangle measures 8 cm by 5 cm to the nearest centimeter. What are the least and greatest possible areas and perimeters?”
Diagram
Measured 8 by 5 cm, each off by up to e = 0.5 cm
Least: (7.5 × 4.5) = 33.75 cm²
Greatest: (8.5 × 5.5) = 46.75 cm²
The true edges lie in the shaded band
This is an example. Type your own values to start a new problem.
(
l
−
e
)
×
(
w
−
e
)
=
Aₘᵢₙ
(
l
+
e
)
×
(
w
+
e
)
=
Aₘₐₓ
u
Measured to the nearest
example · cm
A
Area as measured
calculated · cm²
Pₘᵢₙ
Least possible perimeter
worked out (not typed) · cm
Pₘₐₓ
Greatest possible perimeter
worked out (not typed) · cm
Formulas
e
=
u
2
0.5
=
1
2
A
=
l
×
w
40 = 8 × 5
Aₘᵢₙ
=
(
l
−
e
)
×
(
w
−
e
)
33.75 = (8 − 0.5) × (5 − 0.5)
Aₘₐₓ
=
(
l
+
e
)
×
(
w
+
e
)
46.75 = (8 + 0.5) × (5 + 0.5)
Pₘᵢₙ
=
2
×
(
l
−
e
)
+
2
×
(
w
−
e
)
24 = 2 × (8 − 0.5) + 2 × (5 − 0.5)
Pₘₐₓ
=
2
×
(
l
+
e
)
+
2
×
(
w
+
e
)
28 = 2 × (8 + 0.5) + 2 × (5 + 0.5)
Assumptions
•
A length measured to the nearest u is off by at most half of u, the greatest possible error.
•
The true sides lie between l − e and l + e, so the true area and perimeter lie between their least and greatest values.
•
A finer tool (a smaller u) narrows the range of possible areas.
Step-by-step
WE KNOW
u = 1 cm, l = 8 cm, w = 5 cm
FIND
greatest possible error (e), area as measured (A), least possible area (Aₘᵢₙ), greatest possible area (Aₘₐₓ), least possible perimeter (Pₘᵢₙ), greatest possible perimeter (Pₘₐₓ)
Step 1 · Find greatest possible error (e)
Use e = u ÷ 2
A
reading
to
the
nearest
u
can
be
off
by
up
to
half
of
u
either
way.
e
=
1
2
e
=
0.5
cm
Step 2 · Find area as measured (A)
Use A = l × w
The area from the measured sides.
A
=
8
×
5
A
=
40
cm²
Step 3 · Find least possible area (Aₘᵢₙ)
Use Aₘᵢₙ = (l − e) × (w − e)
The shortest sides the readings allow give the least area.
Aₘᵢₙ = (8 − 0.5) × (5 − 0.5)
Aₘᵢₙ = 7.5 × 4.5
Aₘᵢₙ = 33.75 cm²
Step 4 · Find greatest possible area (Aₘₐₓ)
Use Aₘₐₓ = (l + e) × (w + e)
The longest sides the readings allow give the greatest area.
Aₘₐₓ = (8 + 0.5) × (5 + 0.5)
Aₘₐₓ = 8.5 × 5.5
Aₘₐₓ = 46.75 cm²
Step 5 · Find least possible perimeter (Pₘᵢₙ)
Use Pₘᵢₙ = 2 × (l − e) + 2 × (w − e)
The shortest sides the readings allow give the least perimeter too.
Pₘᵢₙ = 2 × (8 − 0.5) + 2 × (5 − 0.5)
Pₘᵢₙ = 2 × 7.5 + 2 × 4.5
Pₘᵢₙ = 15 + 9
Pₘᵢₙ = 24 cm
Step 6 · Find greatest possible perimeter (Pₘₐₓ)
Use Pₘₐₓ = 2 × (l + e) + 2 × (w + e)
The longest sides the readings allow give the greatest perimeter.
Pₘₐₓ = 2 × (8 + 0.5) + 2 × (5 + 0.5)
Pₘₐₓ = 2 × 8.5 + 2 × 5.5
Pₘₐₓ = 17 + 11
Pₘₐₓ = 28 cm
Check
Put the numbers back in. Both sides should match.
0.5
=
1
2
✓
40 = 8 × 5 ✓
33.75 = (8 − 0.5) × (5 − 0.5) ✓
46.75 = (8 + 0.5) × (5 + 0.5) ✓
24 = 2 × (8 − 0.5) + 2 × (5 − 0.5) ✓
28 = 2 × (8 + 0.5) + 2 × (5 + 0.5) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Units, accuracy and precision in measurement and modeling
Main lesson
Square units and a cost
Use this for “Carpet costs $30 a square yard. What does it cost for a 12 ft by 15 ft room?”
Units in a formula: d = rt
Use this for “A cyclist rides at 18 km per hour for 40 minutes. How far does she go?”
A product to the right significant figures
Use this for “A table is 4.25 m by 3.1 m. Give its area to the right number of significant figures.”
Choosing a level of accuracy
Use this for “Which unit and how much accuracy fit this measurement?”