SME
Grade 10 Math · Problem type
Perimeter and area on the grid
Use this for “Find the perimeter and area of the triangle A(0, 0), B(6, 0), C(3, 4).”
Chart
This is an example. Type your own values to start a new problem.
xA
x of A
example
yA
y of A
example
xB
x of B
example
yB
y of B
example
xC
x of C
example
yC
y of C
example
AB
Side AB
worked out (not typed)
BC
Side BC
worked out (not typed)
CA
Side CA
worked out (not typed)
P
Perimeter
worked out (not typed)
K
Area
worked out (not typed)
Formulas
AB = |xB − xA|
6 = |6 − 0|
BC
=
√(
(xC − xB)
2
+
(yC − yB)
2
)
5
=
√(
(3 − 6)
2
+
(4 − 0)
2
)
CA
=
√(
(xA − xC)
2
+
(yA − yC)
2
)
5
=
√(
(0 − 3)
2
+
(0 − 4)
2
)
P
=
AB
+
BC
+
CA
16 = 6 + 5 + 5
K
=
|xB
−
xA|
×
|yC
−
yA|
÷
2
12 = |6 − 0| × |4 − 0| ÷ 2
Assumptions
•
Each side’s length comes from the distance formula.
•
Side AB is level, so it is the base and the height is how far C is above or below it.
•
Area = ½ × base × height.
Step-by-step
WE KNOW
xA = 0, yA = 0, xB = 6, yB = 0, xC = 3, yC = 4
FIND
side AB (AB), side BC (BC), side CA (CA), perimeter (P), area (K)
Step 1 · Find side AB (AB)
Use AB = |xB − xA|
AB is level: its length is the change across.
AB = |6 − 0|
AB = 6
Step 2 · Find side BC (BC)
Use BC = √((xC − xB)² + (yC − yB)²)
The distance formula for BC: the change across and the change up are the legs of a right triangle.
BC
=
√(
(3 − 6)
2
+
(4 − 0)
2
)
BC
=
√(
(−3)
2
+
4
2
)
BC
=
√
9 + 16
BC
=
√
25
BC = 5
Step 3 · Find side CA (CA)
Use CA = √((xA − xC)² + (yA − yC)²)
The distance formula for CA: the change across and the change up are the legs of a right triangle.
CA
=
√(
(0 − 3)
2
+
(0 − 4)
2
)
CA
=
√(
(−3)
2
+
(−4)
2
)
CA
=
√
9 + 16
CA
=
√
25
CA = 5
Step 4 · Find perimeter (P)
Use P = AB + BC + CA
Add the three sides.
P
=
6
+
5
+
5
P
=
16
Step 5 · Find area (K)
Use K = |xB − xA| × |yC − yA| ÷ 2
Half of base times height: the base along the level side, the height straight up to C.
K
=
|6
−
0|
×
|4
−
0|
÷
2
K
=
6
×
4
2
K
=
24
2
K
=
12
Check
Put the numbers back in. Both sides should match.
6 = |6 − 0| ✓
5
=
√(
(3 − 6)
2
+
(4 − 0)
2
)
✓
5
=
√(
(0 − 3)
2
+
(0 − 4)
2
)
✓
16 = 6 + 5 + 5 ✓
12 = |6 − 0| × |4 − 0| ÷ 2 ✓
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