Area and Perimeter - Tenth Grade Geometry
Introduction
Perimeter: The total distance around a closed figure (measured in linear units: cm, m, in, ft)
Area: The amount of space inside a closed figure (measured in square units: cm², m², in², ft²)
Key Difference: Perimeter = 1D measurement (length); Area = 2D measurement (length × width)
Units: Always include correct units in your answer!
Area: The amount of space inside a closed figure (measured in square units: cm², m², in², ft²)
Key Difference: Perimeter = 1D measurement (length); Area = 2D measurement (length × width)
Units: Always include correct units in your answer!
1. Perimeter
Perimeter: The sum of all side lengths of a polygon
General Rule: Add all the sides
Symbol: P or p
Applications: Fencing, framing, borders, boundaries
General Rule: Add all the sides
Symbol: P or p
Applications: Fencing, framing, borders, boundaries
Basic Perimeter Formulas:
Any Polygon:
$$P = \text{sum of all sides}$$
Square:
$$P = 4s$$
where $s$ = side length
Rectangle:
$$P = 2l + 2w = 2(l + w)$$
where $l$ = length, $w$ = width
Triangle:
$$P = a + b + c$$
where $a$, $b$, $c$ are the three sides
Regular Polygon:
$$P = n \times s$$
where $n$ = number of sides, $s$ = side length
Any Polygon:
$$P = \text{sum of all sides}$$
Square:
$$P = 4s$$
where $s$ = side length
Rectangle:
$$P = 2l + 2w = 2(l + w)$$
where $l$ = length, $w$ = width
Triangle:
$$P = a + b + c$$
where $a$, $b$, $c$ are the three sides
Regular Polygon:
$$P = n \times s$$
where $n$ = number of sides, $s$ = side length
2. Area of Rectangles and Squares
Rectangle
Area of Rectangle:
$$A = l \times w$$
or
$$A = \text{length} \times \text{width}$$
Also written as: $A = lw$ or $A = bh$ (base × height)
Perimeter of Rectangle:
$$P = 2l + 2w = 2(l + w)$$
$$A = l \times w$$
or
$$A = \text{length} \times \text{width}$$
Also written as: $A = lw$ or $A = bh$ (base × height)
Perimeter of Rectangle:
$$P = 2l + 2w = 2(l + w)$$
Example: Rectangle area and perimeter
Rectangle has length 12 cm and width 5 cm. Find area and perimeter.
Area:
$$A = l \times w = 12 \times 5 = 60 \text{ cm}^2$$
Perimeter:
$$P = 2(l + w) = 2(12 + 5) = 2(17) = 34 \text{ cm}$$
Rectangle has length 12 cm and width 5 cm. Find area and perimeter.
Area:
$$A = l \times w = 12 \times 5 = 60 \text{ cm}^2$$
Perimeter:
$$P = 2(l + w) = 2(12 + 5) = 2(17) = 34 \text{ cm}$$
Square
Area of Square:
$$A = s^2 = s \times s$$
where $s$ = side length
Also:
$$A = \frac{1}{2}d^2$$
where $d$ = diagonal length
Perimeter of Square:
$$P = 4s$$
$$A = s^2 = s \times s$$
where $s$ = side length
Also:
$$A = \frac{1}{2}d^2$$
where $d$ = diagonal length
Perimeter of Square:
$$P = 4s$$
Example: Square area
Square has side 8 m. Find area and perimeter.
Area:
$$A = s^2 = 8^2 = 64 \text{ m}^2$$
Perimeter:
$$P = 4s = 4(8) = 32 \text{ m}$$
Square has side 8 m. Find area and perimeter.
Area:
$$A = s^2 = 8^2 = 64 \text{ m}^2$$
Perimeter:
$$P = 4s = 4(8) = 32 \text{ m}$$
3. Area of Parallelograms and Triangles
Parallelogram
Base: Any side can be the base
Height: Perpendicular distance from base to opposite side
Key Point: Height must be perpendicular (90°) to the base!
Height: Perpendicular distance from base to opposite side
Key Point: Height must be perpendicular (90°) to the base!
Area of Parallelogram:
$$A = b \times h$$
or
$$A = \text{base} \times \text{height}$$
Perimeter:
$$P = 2(a + b)$$
where $a$ and $b$ are adjacent sides
$$A = b \times h$$
or
$$A = \text{base} \times \text{height}$$
Perimeter:
$$P = 2(a + b)$$
where $a$ and $b$ are adjacent sides
Triangle
Area of Triangle:
$$A = \frac{1}{2}bh = \frac{1}{2} \times \text{base} \times \text{height}$$
Alternative (using sides and angle):
$$A = \frac{1}{2}ab\sin C$$
where $a$, $b$ are two sides and $C$ is included angle
Perimeter:
$$P = a + b + c$$
where $a$, $b$, $c$ are the three sides
$$A = \frac{1}{2}bh = \frac{1}{2} \times \text{base} \times \text{height}$$
Alternative (using sides and angle):
$$A = \frac{1}{2}ab\sin C$$
where $a$, $b$ are two sides and $C$ is included angle
Perimeter:
$$P = a + b + c$$
where $a$, $b$, $c$ are the three sides
Example 1: Parallelogram
Parallelogram has base 10 cm and height 6 cm. Find area.
$$A = bh = 10 \times 6 = 60 \text{ cm}^2$$
Parallelogram has base 10 cm and height 6 cm. Find area.
$$A = bh = 10 \times 6 = 60 \text{ cm}^2$$
Example 2: Triangle
Triangle has base 8 m and height 5 m. Find area.
$$A = \frac{1}{2}bh = \frac{1}{2}(8)(5) = \frac{40}{2} = 20 \text{ m}^2$$
Triangle has base 8 m and height 5 m. Find area.
$$A = \frac{1}{2}bh = \frac{1}{2}(8)(5) = \frac{40}{2} = 20 \text{ m}^2$$
4. Area of Trapezoids
Trapezoid (Trapezium): Quadrilateral with one pair of parallel sides
Bases: The two parallel sides ($b_1$ and $b_2$)
Height: Perpendicular distance between the two bases
Formula Idea: Average of bases × height
Bases: The two parallel sides ($b_1$ and $b_2$)
Height: Perpendicular distance between the two bases
Formula Idea: Average of bases × height
Area of Trapezoid:
$$A = \frac{1}{2}(b_1 + b_2)h$$
or
$$A = \frac{(b_1 + b_2)}{2} \times h$$
In words: Half the sum of the parallel sides times the height
where:
• $b_1$ = length of first base
• $b_2$ = length of second base
• $h$ = height (perpendicular distance between bases)
Perimeter:
$$P = a + b_1 + c + b_2$$
(sum of all four sides)
$$A = \frac{1}{2}(b_1 + b_2)h$$
or
$$A = \frac{(b_1 + b_2)}{2} \times h$$
In words: Half the sum of the parallel sides times the height
where:
• $b_1$ = length of first base
• $b_2$ = length of second base
• $h$ = height (perpendicular distance between bases)
Perimeter:
$$P = a + b_1 + c + b_2$$
(sum of all four sides)
Example: Trapezoid area
Trapezoid has bases 8 cm and 12 cm, height 5 cm. Find area.
$$A = \frac{1}{2}(b_1 + b_2)h$$
$$A = \frac{1}{2}(8 + 12)(5)$$
$$A = \frac{1}{2}(20)(5)$$
$$A = \frac{100}{2} = 50 \text{ cm}^2$$
Trapezoid has bases 8 cm and 12 cm, height 5 cm. Find area.
$$A = \frac{1}{2}(b_1 + b_2)h$$
$$A = \frac{1}{2}(8 + 12)(5)$$
$$A = \frac{1}{2}(20)(5)$$
$$A = \frac{100}{2} = 50 \text{ cm}^2$$
5. Area of Rhombuses
Rhombus: Parallelogram with all four sides equal
Diagonals: Two diagonals that bisect each other at right angles
Two Methods: Using diagonals OR using base and height
Diagonals: Two diagonals that bisect each other at right angles
Two Methods: Using diagonals OR using base and height
Area of Rhombus:
Method 1 (Using Diagonals):
$$A = \frac{1}{2}d_1 \times d_2$$
or
$$A = \frac{d_1 \times d_2}{2}$$
where $d_1$ and $d_2$ are the lengths of the two diagonals
Method 2 (Using Base and Height):
$$A = b \times h$$
where $b$ = base (side) and $h$ = perpendicular height
Perimeter:
$$P = 4s$$
where $s$ = side length
Method 1 (Using Diagonals):
$$A = \frac{1}{2}d_1 \times d_2$$
or
$$A = \frac{d_1 \times d_2}{2}$$
where $d_1$ and $d_2$ are the lengths of the two diagonals
Method 2 (Using Base and Height):
$$A = b \times h$$
where $b$ = base (side) and $h$ = perpendicular height
Perimeter:
$$P = 4s$$
where $s$ = side length
Example: Rhombus area
Rhombus has diagonals 10 cm and 14 cm. Find area.
$$A = \frac{1}{2}d_1 \times d_2$$
$$A = \frac{1}{2}(10)(14)$$
$$A = \frac{140}{2} = 70 \text{ cm}^2$$
Rhombus has diagonals 10 cm and 14 cm. Find area.
$$A = \frac{1}{2}d_1 \times d_2$$
$$A = \frac{1}{2}(10)(14)$$
$$A = \frac{140}{2} = 70 \text{ cm}^2$$
6. Area of Regular Polygons
Regular Polygon: Polygon with all sides equal and all angles equal
Apothem: Perpendicular distance from center to midpoint of any side
Symbol: $a$ (apothem)
Examples: Regular hexagon, regular octagon, regular pentagon
Apothem: Perpendicular distance from center to midpoint of any side
Symbol: $a$ (apothem)
Examples: Regular hexagon, regular octagon, regular pentagon
Area of Regular Polygon:
$$A = \frac{1}{2} \times \text{apothem} \times \text{perimeter}$$
or
$$A = \frac{1}{2}ap$$
or
$$A = \frac{1}{2}a(ns)$$
where:
• $a$ = apothem
• $p$ = perimeter
• $n$ = number of sides
• $s$ = side length
Alternative Form:
$$A = \frac{ns}{2} \times a$$
$$A = \frac{1}{2} \times \text{apothem} \times \text{perimeter}$$
or
$$A = \frac{1}{2}ap$$
or
$$A = \frac{1}{2}a(ns)$$
where:
• $a$ = apothem
• $p$ = perimeter
• $n$ = number of sides
• $s$ = side length
Alternative Form:
$$A = \frac{ns}{2} \times a$$
Example: Regular hexagon
Regular hexagon has side 6 cm and apothem 5.2 cm. Find area.
Step 1: Find perimeter
$$P = ns = 6(6) = 36 \text{ cm}$$
Step 2: Find area
$$A = \frac{1}{2}ap = \frac{1}{2}(5.2)(36)$$
$$A = \frac{1}{2}(187.2) = 93.6 \text{ cm}^2$$
Regular hexagon has side 6 cm and apothem 5.2 cm. Find area.
Step 1: Find perimeter
$$P = ns = 6(6) = 36 \text{ cm}$$
Step 2: Find area
$$A = \frac{1}{2}ap = \frac{1}{2}(5.2)(36)$$
$$A = \frac{1}{2}(187.2) = 93.6 \text{ cm}^2$$
7. Area and Perimeter in the Coordinate Plane
Strategy: Use distance formula and coordinate geometry
Distance Formula: Find lengths of sides
Area Methods: Shoelace formula, divide into simpler shapes, or use standard formulas
Distance Formula: Find lengths of sides
Area Methods: Shoelace formula, divide into simpler shapes, or use standard formulas
Key Formulas for Coordinate Plane:
Distance Formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Midpoint Formula:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Shoelace Formula (for polygon area):
For vertices $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$:
$$A = \frac{1}{2}|x_1y_2 + x_2y_3 + ... + x_ny_1 - (y_1x_2 + y_2x_3 + ... + y_nx_1)|$$
Distance Formula:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Midpoint Formula:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Shoelace Formula (for polygon area):
For vertices $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$:
$$A = \frac{1}{2}|x_1y_2 + x_2y_3 + ... + x_ny_1 - (y_1x_2 + y_2x_3 + ... + y_nx_1)|$$
Example: Rectangle in coordinate plane
Rectangle has vertices A(1, 2), B(5, 2), C(5, 6), D(1, 6). Find area and perimeter.
Find side lengths:
AB = $\sqrt{(5-1)^2 + (2-2)^2} = \sqrt{16} = 4$
BC = $\sqrt{(5-5)^2 + (6-2)^2} = \sqrt{16} = 4$
Area:
$$A = l \times w = 4 \times 4 = 16 \text{ square units}$$
Perimeter:
$$P = 2(4 + 4) = 16 \text{ units}$$
Rectangle has vertices A(1, 2), B(5, 2), C(5, 6), D(1, 6). Find area and perimeter.
Find side lengths:
AB = $\sqrt{(5-1)^2 + (2-2)^2} = \sqrt{16} = 4$
BC = $\sqrt{(5-5)^2 + (6-2)^2} = \sqrt{16} = 4$
Area:
$$A = l \times w = 4 \times 4 = 16 \text{ square units}$$
Perimeter:
$$P = 2(4 + 4) = 16 \text{ units}$$
8. Area and Circumference of Circles
Circle: Set of all points equidistant from a center point
Radius (r): Distance from center to edge
Diameter (d): Distance across circle through center ($d = 2r$)
Circumference: Perimeter of a circle
π (Pi): Approximately 3.14 or $\frac{22}{7}$
Radius (r): Distance from center to edge
Diameter (d): Distance across circle through center ($d = 2r$)
Circumference: Perimeter of a circle
π (Pi): Approximately 3.14 or $\frac{22}{7}$
Circle Formulas:
Circumference (Perimeter):
$$C = 2\pi r$$
or
$$C = \pi d$$
where $r$ = radius, $d$ = diameter
Area:
$$A = \pi r^2$$
Relationship:
$$d = 2r \quad \text{or} \quad r = \frac{d}{2}$$
Circumference (Perimeter):
$$C = 2\pi r$$
or
$$C = \pi d$$
where $r$ = radius, $d$ = diameter
Area:
$$A = \pi r^2$$
Relationship:
$$d = 2r \quad \text{or} \quad r = \frac{d}{2}$$
Example 1: Given radius
Circle has radius 7 cm. Find circumference and area.
Circumference:
$$C = 2\pi r = 2\pi(7) = 14\pi \approx 43.98 \text{ cm}$$
Area:
$$A = \pi r^2 = \pi(7)^2 = 49\pi \approx 153.94 \text{ cm}^2$$
Circle has radius 7 cm. Find circumference and area.
Circumference:
$$C = 2\pi r = 2\pi(7) = 14\pi \approx 43.98 \text{ cm}$$
Area:
$$A = \pi r^2 = \pi(7)^2 = 49\pi \approx 153.94 \text{ cm}^2$$
Example 2: Given diameter
Circle has diameter 10 m. Find area.
Find radius: $r = \frac{d}{2} = \frac{10}{2} = 5$ m
Area:
$$A = \pi r^2 = \pi(5)^2 = 25\pi \approx 78.54 \text{ m}^2$$
Circle has diameter 10 m. Find area.
Find radius: $r = \frac{d}{2} = \frac{10}{2} = 5$ m
Area:
$$A = \pi r^2 = \pi(5)^2 = 25\pi \approx 78.54 \text{ m}^2$$
9. Area of Compound Figures
Compound Figure: Figure made up of two or more simple shapes
Strategy: Break into simpler shapes, find each area, then add or subtract
Common Shapes: Rectangles, triangles, circles, semicircles
Strategy: Break into simpler shapes, find each area, then add or subtract
Common Shapes: Rectangles, triangles, circles, semicircles
Steps to Find Area of Compound Figures:
Step 1: Divide the figure into simpler shapes (rectangles, triangles, circles, etc.)
Step 2: Find the area of each simple shape
Step 3: Add all areas together
Formula:
$$A_{\text{total}} = A_1 + A_2 + A_3 + ...$$
Step 1: Divide the figure into simpler shapes (rectangles, triangles, circles, etc.)
Step 2: Find the area of each simple shape
Step 3: Add all areas together
Formula:
$$A_{\text{total}} = A_1 + A_2 + A_3 + ...$$
Example: L-shaped figure
Figure consists of two rectangles:
• Rectangle 1: 8 cm × 3 cm
• Rectangle 2: 5 cm × 4 cm
Area of Rectangle 1:
$$A_1 = 8 \times 3 = 24 \text{ cm}^2$$
Area of Rectangle 2:
$$A_2 = 5 \times 4 = 20 \text{ cm}^2$$
Total Area:
$$A = 24 + 20 = 44 \text{ cm}^2$$
Figure consists of two rectangles:
• Rectangle 1: 8 cm × 3 cm
• Rectangle 2: 5 cm × 4 cm
Area of Rectangle 1:
$$A_1 = 8 \times 3 = 24 \text{ cm}^2$$
Area of Rectangle 2:
$$A_2 = 5 \times 4 = 20 \text{ cm}^2$$
Total Area:
$$A = 24 + 20 = 44 \text{ cm}^2$$
10. Area Between Two Shapes
Shaded Region: Area between two figures (outer minus inner)
Common Examples: Ring (circle inside circle), frame, border
Method: Subtract smaller area from larger area
Common Examples: Ring (circle inside circle), frame, border
Method: Subtract smaller area from larger area
Area Between Two Shapes:
$$A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}}$$
Common Case - Ring (Annulus):
$$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$$
where $R$ = outer radius, $r$ = inner radius
$$A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}}$$
Common Case - Ring (Annulus):
$$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$$
where $R$ = outer radius, $r$ = inner radius
Example: Ring area
Outer circle has radius 10 cm, inner circle has radius 6 cm. Find shaded area.
Outer area:
$$A_{\text{outer}} = \pi(10)^2 = 100\pi \text{ cm}^2$$
Inner area:
$$A_{\text{inner}} = \pi(6)^2 = 36\pi \text{ cm}^2$$
Shaded area:
$$A = 100\pi - 36\pi = 64\pi \approx 201.06 \text{ cm}^2$$
Outer circle has radius 10 cm, inner circle has radius 6 cm. Find shaded area.
Outer area:
$$A_{\text{outer}} = \pi(10)^2 = 100\pi \text{ cm}^2$$
Inner area:
$$A_{\text{inner}} = \pi(6)^2 = 36\pi \text{ cm}^2$$
Shaded area:
$$A = 100\pi - 36\pi = 64\pi \approx 201.06 \text{ cm}^2$$
11. Area and Perimeter of Similar Figures
Similar Figures: Same shape, different size
Scale Factor (k): Ratio of corresponding sides
Key Relationships: Perimeter ratio = k; Area ratio = k²
Scale Factor (k): Ratio of corresponding sides
Key Relationships: Perimeter ratio = k; Area ratio = k²
Scaling Rules for Similar Figures:
If scale factor = $k$, then:
Perimeter:
$$\frac{P_{\text{new}}}{P_{\text{original}}} = k$$
$$P_{\text{new}} = k \times P_{\text{original}}$$
Area:
$$\frac{A_{\text{new}}}{A_{\text{original}}} = k^2$$
$$A_{\text{new}} = k^2 \times A_{\text{original}}$$
If scale factor = $k$, then:
Perimeter:
$$\frac{P_{\text{new}}}{P_{\text{original}}} = k$$
$$P_{\text{new}} = k \times P_{\text{original}}$$
Area:
$$\frac{A_{\text{new}}}{A_{\text{original}}} = k^2$$
$$A_{\text{new}} = k^2 \times A_{\text{original}}$$
Example: Scaling effects
Original triangle has perimeter 12 cm and area 20 cm². If enlarged by scale factor 3, find new perimeter and area.
New perimeter:
$$P_{\text{new}} = k \times P = 3 \times 12 = 36 \text{ cm}$$
New area:
$$A_{\text{new}} = k^2 \times A = 3^2 \times 20 = 9 \times 20 = 180 \text{ cm}^2$$
Original triangle has perimeter 12 cm and area 20 cm². If enlarged by scale factor 3, find new perimeter and area.
New perimeter:
$$P_{\text{new}} = k \times P = 3 \times 12 = 36 \text{ cm}$$
New area:
$$A_{\text{new}} = k^2 \times A = 3^2 \times 20 = 9 \times 20 = 180 \text{ cm}^2$$
12. Perimeter and Area: Changes in Scale
Effects of Scaling:
If each dimension is multiplied by k:
• Perimeter: Multiplied by $k$
• Area: Multiplied by $k^2$
• Volume (3D): Multiplied by $k^3$
Examples:
• Double dimensions ($k = 2$): Perimeter × 2, Area × 4
• Triple dimensions ($k = 3$): Perimeter × 3, Area × 9
• Half dimensions ($k = \frac{1}{2}$): Perimeter × $\frac{1}{2}$, Area × $\frac{1}{4}$
If each dimension is multiplied by k:
• Perimeter: Multiplied by $k$
• Area: Multiplied by $k^2$
• Volume (3D): Multiplied by $k^3$
Examples:
• Double dimensions ($k = 2$): Perimeter × 2, Area × 4
• Triple dimensions ($k = 3$): Perimeter × 3, Area × 9
• Half dimensions ($k = \frac{1}{2}$): Perimeter × $\frac{1}{2}$, Area × $\frac{1}{4}$
Example: Doubling dimensions
Square has side 5 cm. If side is doubled, how do perimeter and area change?
Original:
Perimeter = $4(5) = 20$ cm
Area = $5^2 = 25$ cm²
New side = 10 cm (doubled):
Perimeter = $4(10) = 40$ cm (× 2)
Area = $10^2 = 100$ cm² (× 4)
Square has side 5 cm. If side is doubled, how do perimeter and area change?
Original:
Perimeter = $4(5) = 20$ cm
Area = $5^2 = 25$ cm²
New side = 10 cm (doubled):
Perimeter = $4(10) = 40$ cm (× 2)
Area = $10^2 = 100$ cm² (× 4)
13. Heron's Formula
Heron's Formula: Find area of triangle when you know all three side lengths
Named After: Hero of Alexandria
Use When: You have SSS (three sides) but no height
Semi-perimeter (s): Half the perimeter
Named After: Hero of Alexandria
Use When: You have SSS (three sides) but no height
Semi-perimeter (s): Half the perimeter
Heron's Formula:
Step 1: Find semi-perimeter
$$s = \frac{a + b + c}{2}$$
Step 2: Calculate area
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
where $a$, $b$, $c$ are the three sides and $s$ is the semi-perimeter
Step 1: Find semi-perimeter
$$s = \frac{a + b + c}{2}$$
Step 2: Calculate area
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
where $a$, $b$, $c$ are the three sides and $s$ is the semi-perimeter
Example: Using Heron's formula
Triangle has sides 5, 6, and 7. Find the area.
Step 1: Find semi-perimeter
$$s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$$
Step 2: Apply Heron's formula
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
$$A = \sqrt{9(9-5)(9-6)(9-7)}$$
$$A = \sqrt{9(4)(3)(2)}$$
$$A = \sqrt{216}$$
$$A = 6\sqrt{6} \approx 14.70 \text{ square units}$$
Triangle has sides 5, 6, and 7. Find the area.
Step 1: Find semi-perimeter
$$s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$$
Step 2: Apply Heron's formula
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
$$A = \sqrt{9(9-5)(9-6)(9-7)}$$
$$A = \sqrt{9(4)(3)(2)}$$
$$A = \sqrt{216}$$
$$A = 6\sqrt{6} \approx 14.70 \text{ square units}$$
Complete Area and Perimeter Formulas
Shape | Area Formula | Perimeter Formula |
---|---|---|
Square | $A = s^2$ | $P = 4s$ |
Rectangle | $A = l \times w$ | $P = 2(l + w)$ |
Triangle | $A = \frac{1}{2}bh$ | $P = a + b + c$ |
Parallelogram | $A = bh$ | $P = 2(a + b)$ |
Trapezoid | $A = \frac{1}{2}(b_1 + b_2)h$ | $P = a + b_1 + c + b_2$ |
Rhombus | $A = \frac{1}{2}d_1d_2$ or $A = bh$ | $P = 4s$ |
Circle | $A = \pi r^2$ | $C = 2\pi r$ or $C = \pi d$ |
Regular Polygon | $A = \frac{1}{2}ap$ (a=apothem, p=perimeter) | $P = ns$ (n=sides, s=side length) |
Special Triangle Area Formulas
Method | Formula | When to Use |
---|---|---|
Base × Height | $A = \frac{1}{2}bh$ | When base and height are known |
Two Sides + Angle | $A = \frac{1}{2}ab\sin C$ | When two sides and included angle known |
Heron's Formula | $A = \sqrt{s(s-a)(s-b)(s-c)}$ | When all three sides known (SSS) |
Coordinates | Shoelace formula | When vertices given as coordinates |
Coordinate Plane Tools
Tool | Formula | Use |
---|---|---|
Distance Formula | $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ | Find side lengths |
Midpoint Formula | $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ | Find center or divide segments |
Slope Formula | $m = \frac{y_2-y_1}{x_2-x_1}$ | Check perpendicular/parallel |
Scaling Effects Summary
Scale Factor (k) | Effect on Side Lengths | Effect on Perimeter | Effect on Area |
---|---|---|---|
2 (double) | × 2 | × 2 | × 4 (2²) |
3 (triple) | × 3 | × 3 | × 9 (3²) |
1/2 (halve) | × 1/2 | × 1/2 | × 1/4 |
k (general) | × k | × k | × k² |
Problem-Solving Strategy
Step | Action | Example |
---|---|---|
1. Identify Shape | What shape(s) are you working with? | Rectangle, triangle, circle, compound |
2. Choose Formula | Select appropriate area/perimeter formula | Rectangle: A = l × w |
3. Identify Values | What measurements are given? | Length = 8, width = 5 |
4. Calculate | Substitute and solve | A = 8 × 5 = 40 |
5. Units | Include correct units | 40 cm² (area) or 26 cm (perimeter) |
Success Tips for Area and Perimeter:
✓ Perimeter = sum of sides (linear units); Area = space inside (square units)
✓ Rectangle: A = l×w, P = 2(l+w); Square: A = s², P = 4s
✓ Triangle: A = ½bh; Parallelogram: A = bh
✓ Trapezoid: A = ½(b₁+b₂)h; Rhombus: A = ½d₁d₂
✓ Circle: A = πr², C = 2πr or πd
✓ Regular polygon: A = ½ap (apothem × perimeter)
✓ Heron's formula: A = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
✓ Compound figures: Break into simple shapes, add areas
✓ Shaded regions: Outer area - inner area
✓ Scale factor k: Perimeter × k, Area × k²
✓ ALWAYS include units: cm, m, cm², m², etc.!
✓ Perimeter = sum of sides (linear units); Area = space inside (square units)
✓ Rectangle: A = l×w, P = 2(l+w); Square: A = s², P = 4s
✓ Triangle: A = ½bh; Parallelogram: A = bh
✓ Trapezoid: A = ½(b₁+b₂)h; Rhombus: A = ½d₁d₂
✓ Circle: A = πr², C = 2πr or πd
✓ Regular polygon: A = ½ap (apothem × perimeter)
✓ Heron's formula: A = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
✓ Compound figures: Break into simple shapes, add areas
✓ Shaded regions: Outer area - inner area
✓ Scale factor k: Perimeter × k, Area × k²
✓ ALWAYS include units: cm, m, cm², m², etc.!