Basic Math

Area and perimeter | Tenth Grade

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!

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
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

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:
$$P = 2l + 2w = 2(l + w)$$
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:
$$P = 2(l + w) = 2(12 + 5) = 2(17) = 34 \text{ cm}$$

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:
$$P = 4s$$
Example: Square area

Square has side 8 m. Find area and perimeter.

Area:
$$A = s^2 = 8^2 = 64 \text{ m}^2$$

Perimeter:
$$P = 4s = 4(8) = 32 \text{ m}$$

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!
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

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
Example 1: Parallelogram

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.

$$A = \frac{1}{2}bh = \frac{1}{2}(8)(5) = \frac{40}{2} = 20 \text{ m}^2$$

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
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)
Example: Trapezoid area

Trapezoid has bases 8 cm and 12 cm, height 5 cm. Find area.

$$A = \frac{1}{2}(b_1 + b_2)h$$
$$A = \frac{1}{2}(8 + 12)(5)$$
$$A = \frac{1}{2}(20)(5)$$
$$A = \frac{100}{2} = 50 \text{ cm}^2$$

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
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
Example: Rhombus area

Rhombus has diagonals 10 cm and 14 cm. Find area.

$$A = \frac{1}{2}d_1 \times d_2$$
$$A = \frac{1}{2}(10)(14)$$
$$A = \frac{140}{2} = 70 \text{ cm}^2$$

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
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:
$$A = \frac{ns}{2} \times a$$
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
$$A = \frac{1}{2}ap = \frac{1}{2}(5.2)(36)$$
$$A = \frac{1}{2}(187.2) = 93.6 \text{ cm}^2$$

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
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 $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$:
$$A = \frac{1}{2}|x_1y_2 + x_2y_3 + ... + x_ny_1 - (y_1x_2 + y_2x_3 + ... + y_nx_1)|$$
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:
$$P = 2(4 + 4) = 16 \text{ units}$$

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 $\frac{22}{7}$
Circle Formulas:

Circumference (Perimeter):
$$C = 2\pi r$$

or

$$C = \pi d$$

where $r$ = radius, $d$ = diameter

Area:
$$A = \pi r^2$$

Relationship:
$$d = 2r \quad \text{or} \quad r = \frac{d}{2}$$
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:
$$A = \pi r^2 = \pi(7)^2 = 49\pi \approx 153.94 \text{ cm}^2$$
Example 2: Given diameter

Circle has diameter 10 m. Find area.

Find radius: $r = \frac{d}{2} = \frac{10}{2} = 5$ m

Area:
$$A = \pi r^2 = \pi(5)^2 = 25\pi \approx 78.54 \text{ m}^2$$

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
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:
$$A_{\text{total}} = A_1 + A_2 + A_3 + ...$$
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:
$$A = 24 + 20 = 44 \text{ cm}^2$$

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
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
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:
$$A = 100\pi - 36\pi = 64\pi \approx 201.06 \text{ cm}^2$$

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
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:
$$\frac{A_{\text{new}}}{A_{\text{original}}} = k^2$$
$$A_{\text{new}} = k^2 \times A_{\text{original}}$$
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:
$$A_{\text{new}} = k^2 \times A = 3^2 \times 20 = 9 \times 20 = 180 \text{ cm}^2$$

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 × $\frac{1}{4}$
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)

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
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
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
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
$$A = \sqrt{9(9-5)(9-6)(9-7)}$$
$$A = \sqrt{9(4)(3)(2)}$$
$$A = \sqrt{216}$$
$$A = 6\sqrt{6} \approx 14.70 \text{ square units}$$

Complete Area and Perimeter Formulas

ShapeArea FormulaPerimeter 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

MethodFormulaWhen 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)
CoordinatesShoelace formulaWhen vertices given as coordinates

Coordinate Plane Tools

ToolFormulaUse
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 LengthsEffect on PerimeterEffect 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

StepActionExample
1. Identify ShapeWhat shape(s) are you working with?Rectangle, triangle, circle, compound
2. Choose FormulaSelect appropriate area/perimeter formulaRectangle: A = l × w
3. Identify ValuesWhat measurements are given?Length = 8, width = 5
4. CalculateSubstitute and solveA = 8 × 5 = 40
5. UnitsInclude correct units40 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.!
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