Perimeter and area of basic figures

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MODULE: Planimetry (2D Geometry)

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βœ–οΈ 1. Area formulas for rectangles, parallelograms, and triangles

πŸ“ Area Formulas for Rectangles, Parallelograms, and Triangles

  • Rectangle area = base Γ— height (or length Γ— width).
  • Parallelogram area = base Γ— height (height is perpendicular to base).
  • Triangle area = 12Γ—baseΓ—height\frac{1}{2} \times \text{base} \times \text{height}.
  • The height must always be perpendicular to the base you choose.
  • For triangles, any side can be the base if you use its corresponding height.

Example: A rectangle 8 cm by 5 cm has area 8Γ—5=408 \times 5 = 40 square cm. A triangle with base 6 m and height 4 m has area 12Γ—6Γ—4=12\frac{1}{2} \times 6 \times 4 = 12 square m.

πŸ’‘ Rectangle = full box, Triangle = half a rectangle, Parallelogram = slanted rectangle.

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1. Area formulas for rectangles, parallelograms, and triangles

Area formulas for rectangles, parallelograms, and triangles

Area measures the two-dimensional space enclosed within a boundary, expressed in square units. For basic polygons, area formulas relate linear dimensions to this enclosed region.

Intuition: Area quantifies "how much surface" a shape covers, analogous to counting unit squares that tile the interior.

Core formulas:

  • Rectangle: A=lΓ—wA = l \times w where ll is length and ww is width
  • Parallelogram: A=bΓ—hA = b \times h where bb is base length and hh is perpendicular height (not slant side)
  • Triangle: A=12bhA = \frac{1}{2}bh where bb is base and hh is height perpendicular to that base

Key distinction: For parallelograms and triangles, height must be measured perpendicular to the chosen base, not along a slanted edge.

Example: A triangle with base 8 cm and perpendicular height 5 cm has area A=12(8)(5)=20A = \frac{1}{2}(8)(5) = 20 square cm.

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A rectangle has a length of 1212 units and a width of 55 units. Calculate its area.

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βœ–οΈ 2. Area formula for a trapezoid

πŸͺœ Trapezoid Area Formula

  • Trapezoid area = 12(b1+b2)h\frac{1}{2}(b_1 + b_2)h where b1b_1 and b2b_2 are the two parallel bases.
  • The height hh is the perpendicular distance between the two parallel sides.
  • Add the two bases first, then multiply by height, then divide by 2.
  • Order of b1b_1 and b2b_2 does not matter (addition is commutative).

Example: A trapezoid with bases 7 cm and 5 cm and height 4 cm has area 12(7+5)Γ—4=12Γ—12Γ—4=24\frac{1}{2}(7 + 5) \times 4 = \frac{1}{2} \times 12 \times 4 = 24 square cm.

πŸ’‘ Average the two bases, then multiply by heightβ€”like stacking two triangles.

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2. Area formula for a trapezoid

Area formula for a trapezoid: 12(b1+b2)h\frac{1}{2}(b_1 + b_2)h

A trapezoid is a quadrilateral with exactly one pair of parallel sides called bases. Its area formula averages the two base lengths and multiplies by the perpendicular height.

Intuition: The formula computes the area of an "average rectangle" whose width is the mean of the two bases.

Core rules:

  • b1b_1 and b2b_2 denote the lengths of the two parallel bases
  • hh is the perpendicular distance between the parallel bases
  • Order of bases does not matter since addition is commutative
  • Height must be measured perpendicular to both bases, not along a leg

Derivation insight: A trapezoid can be decomposed into two triangles sharing the same height, yielding 12b1h+12b2h=12(b1+b2)h\frac{1}{2}b_1 h + \frac{1}{2}b_2 h = \frac{1}{2}(b_1 + b_2)h.

Example: A trapezoid with bases 6 m and 10 m and height 4 m has area A=12(6+10)(4)=32A = \frac{1}{2}(6 + 10)(4) = 32 square m.

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A trapezoid has parallel bases of length 88 cm and 1212 cm. The perpendicular distance between these bases is 55 cm. Calculate the area of the trapezoid in square cm.

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βœ–οΈ 3. Perimeter calculations for polygons

πŸ“ Perimeter of Polygons

  • Perimeter = sum of all side lengths of the polygon.
  • For a rectangle, perimeter = 2Γ—(length+width)2 \times (\text{length} + \text{width}).
  • For a square with side ss, perimeter = 4s4s.
  • For any polygon, just add every side length together.
  • Always use the same units for all sides before adding.

Example: A triangle with sides 3 m, 4 m, and 5 m has perimeter 3+4+5=123 + 4 + 5 = 12 m. A rectangle 6 cm by 2 cm has perimeter 2(6+2)=162(6 + 2) = 16 cm.

πŸ’‘ Walk around the shape and count every edgeβ€”that's the perimeter.

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3. Perimeter calculations for polygons

Perimeter calculations for polygons

Perimeter is the total length of all boundary edges of a polygon, measured in linear units. It represents the distance around the figure's exterior.

Intuition: Perimeter answers "how much fencing is needed to enclose this shape" by summing all side lengths.

Core rules:

  • General polygon: P=s1+s2+β‹―+snP = s_1 + s_2 + \cdots + s_n where sis_i are individual side lengths
  • Regular polygon: P=nΓ—sP = n \times s where nn is the number of sides and ss is the common side length
  • Rectangle: P=2l+2wP = 2l + 2w (sum of two lengths and two widths)
  • Perimeter is always expressed in the same linear unit as the side measurements

Critical distinction: Perimeter measures one-dimensional boundary length, while area measures two-dimensional interior space.

Example: A pentagon with sides 3 cm, 5 cm, 4 cm, 6 cm, and 7 cm has perimeter P=3+5+4+6+7=25P = 3 + 5 + 4 + 6 + 7 = 25 cm.

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A polygon has three sides with lengths of 12 meters, 15 meters, and 18 meters. Calculate its perimeter.

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βœ–οΈ 4. Decomposing complex composite figures into basic shapes for area calculation

🧩 Decomposing Composite Figures

  • Break complex shapes into rectangles, triangles, trapezoids, or other basic shapes.
  • Calculate the area of each piece separately using known formulas.
  • Add all the areas together to get the total area.
  • Sometimes you subtract areas (like cutting out a hole from a rectangle).
  • Draw dividing lines clearly to see the basic shapes inside.

Example: An L-shape can be split into two rectangles: one 5 m by 3 m (area 15 square m) and one 4 m by 2 m (area 8 square m), total area = 15+8=2315 + 8 = 23 square m.

πŸ’‘ Slice the puzzle into pieces you already know how to solve.

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4. Decomposing complex composite figures into basic shapes for area calculation

Decomposing complex composite figures into basic shapes for area calculation

Composite figures are shapes formed by combining or subtracting basic geometric figures. Area is computed by partitioning the composite into recognizable components.

Intuition: Break irregular shapes into rectangles, triangles, and other simple forms whose areas can be calculated individually, then sum or subtract as needed.

Core strategy:

  • Identify constituent shapes (rectangles, triangles, trapezoids, circles)
  • Measure necessary dimensions for each component
  • Calculate individual areas using standard formulas
  • Combine by addition for adjoined regions or subtraction for removed sections

Key principle: Multiple valid decompositions exist; choose the partition that minimizes calculation complexity.

Example: An L-shaped figure can be split into two rectangles: one 5 m by 3 m (area 15 square m) and one 4 m by 2 m (area 8 square m), giving total area 15+8=2315 + 8 = 23 square m.

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An L-shaped room is divided into two rectangular sections to calculate its area. The first section measures 55 meters by 44 meters. The second section measures 33 meters by 22 meters. What is the total area of the room in square meters?

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βœ–οΈ 5. Applications: Estimating material requirements for construction and land area in agriculture

πŸ—οΈ Real-World Applications

  • Construction: Calculate floor area to estimate tiles, carpet, or paint needed.
  • Agriculture: Measure land area in acres or hectares for planting or irrigation.
  • Fencing: Use perimeter to determine how much fencing material is required.
  • Material costs: Multiply area or perimeter by cost per unit (e.g., dollars per square meter).
  • Always convert units to match the problem (e.g., feet to yards, meters to kilometers).

Example: A rectangular garden 10 m by 6 m needs fencing around it. Perimeter = 2(10+6)=322(10 + 6) = 32 m. If fencing costs 5 dollars per meter, total cost = 32Γ—5=16032 \times 5 = 160 dollars.

πŸ’‘ Area = how much stuff fills it; Perimeter = how much stuff wraps around it.

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5. Applications: Estimating material requirements for construction and land area in agriculture

Applications: Estimating material requirements for construction and land area in agriculture

Perimeter and area calculations enable practical resource estimation in construction and agriculture. These applications translate geometric measurements into material quantities and land productivity assessments.

Intuition: Area determines coverage needs (paint, flooring, seed), while perimeter determines boundary materials (fencing, trim).

Construction applications:

  • Flooring/roofing: Area calculation determines square footage for material purchase
  • Fencing/molding: Perimeter determines linear footage needed
  • Cost estimation: Multiply area or perimeter by unit price

Agricultural applications:

  • Field area: Determines planting capacity and yield estimates
  • Irrigation planning: Area informs water volume requirements
  • Fencing: Perimeter calculation for livestock enclosures

Example: A rectangular field 80 m by 50 m has area 4000 square m. At 5 grams of seed per square m, total seed requirement is 4000Γ—5=200004000 \times 5 = 20000 grams or 20 kg.

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A rectangular garden is 30 m by 20 m. How many meters of fencing are needed to enclose it completely?

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