Area

Chapter at a Glance

This chapter details the geometric principles of area measurement for 2D figures, starting with basic rectangles and squares, and extending to triangles, parallelograms, rhombuses, and trapeziums. Students learn how area is defined by packing non-overlapping unit squares, and why perimeter is not a reliable indicator of area. The chapter outlines dissection and reflection techniques to derive formulas, split complex polygons into triangles, and optimize perimeters between parallel lines. Finally, it addresses real-world area measurements, conversions between metric and imperial systems, and traditional Indian land measurement units.

Key Definitions & Terminology

• Area: The measure of the size of a 2D region, quantified by the number of unit squares that cover it without gaps or overlaps.
• Perimeter: The total distance around the boundary of a 2D shape.
• Altitude (Height): The perpendicular distance from a vertex of a triangle or parallel side of a quadrilateral to its opposite side (base).
• Median of a Triangle: A line segment joining a vertex of a triangle to the midpoint of the opposite side.
• Dissection: The process of slicing a geometric figure into pieces and rearranging them to form a different shape of equal area.
• Parallelogram: A quadrilateral with two pairs of parallel sides.
• Rhombus: A parallelogram with four sides of equal length, where the diagonals bisect each other at right angles.
• Trapezium: A quadrilateral with at least one pair of parallel sides.
• Representative Fraction (for Area): A scale converting measured map areas to actual ground areas.

Formulas, Rules & Properties

• Area of a Rectangle:
• For length $l$ and width $w$:
  $$\text{Area} = l \cdot w$$
• Area of a Triangle:
• For base $b$ and altitude $h$:
  $$\text{Area} = \frac{1}{2} \cdot b \cdot h$$
• Diagonals and Medians Area Properties:
• A median divides a triangle into two triangles of equal area.
• The diagonals of a rectangle divide it into four triangles of equal area.
• Area of a Parallelogram:
• For base $b$ and height $h$:
  $$\text{Area} = b \cdot h$$
• Area of a Rhombus:
• For diagonals $d_1$ and $d_2$:
  $$\text{Area} = \frac{1}{2} \cdot d_1 \cdot d_2$$
• Area of a Trapezium:
• For parallel sides $a$ and $b$, and height $h$:
  $$\text{Area} = \frac{1}{2} \cdot h \cdot (a + b)$$
• Area of a Quadrilateral:
• For a diagonal of length $d$ and perpendicular heights $h_1$ and $h_2$ from the opposite vertices:
  $$\text{Area} = \frac{1}{2} \cdot d \cdot (h_1 + h_2)$$
• Imperial-Metric Area Conversions:
• $$1\text{ in} = 2.54\text{ cm} \quad \rightarrow \quad 1\text{ in}^2 = 6.4516\text{ cm}^2$$
• $$1\text{ ft} = 12\text{ in} \quad \rightarrow \quad 1\text{ ft}^2 = 144\text{ in}^2 \approx 929.03\text{ cm}^2 \approx 0.0929\text{ m}^2$$
• $$1\text{ acre} = 43,560\text{ ft}^2$$
• $$1\text{ km}^2 = 1,000,000\text{ m}^2$$

Core Concepts & Topics

• Perimeter is Not Area:
• Rectangles can have equal perimeters but different areas. E.g., a $7\text{ cm} \times 4\text{ cm}$ rectangle (perimeter $22\text{ cm}$, area $28\text{ cm}^2$) and an $8\text{ cm} \times 3\text{ cm}$ rectangle (perimeter $22\text{ cm}$, area $24\text{ cm}^2$).
• An object with a larger perimeter can have a smaller area than one with a smaller perimeter. E.g., a $10\text{ cm} \times 1\text{ cm}$ rectangle (perimeter $22\text{ cm}$, area $10\text{ cm}^2$) vs. a $4\text{ cm} \times 4\text{ cm}$ square (perimeter $16\text{ cm}$, area $16\text{ cm}^2$).
• Triangles between Parallel Lines with a Common Base:
• Triangles sharing a base $BC$ and having their opposite vertex on a line parallel to $BC$ have identical areas because their base and height are constant.
• Minimum Perimeter Principle: Using a mirror reflection of the base across the parallel line, the triangle with the minimum perimeter is the isosceles triangle, where the two non-base sides are equal.
• Polygon Dissection into Triangles:
• The area of any complex polygon (quadrilateral, pentagon, hexagon, etc.) can be found by dividing it into non-overlapping triangles and summing their individual areas.
• Local Indian Units of Area:
• Traditional units of land area include bigha, gaj, katha, dhur, cent, and ankanam, which vary by region.

Worked Examples

• Determining Sidelengths from Areas (Page 150 Q1):
• Problem: Find the missing sidelengths in a grid containing four adjacent rectangles:
  • Rectangle 1: $\text{Area} = 14\text{ in}^2$, height $= 7\text{ in}$.
  • Rectangle 2: $\text{Area} = 35\text{ in}^2$, height $= 7\text{ in}$.
  • Rectangle 3: $\text{Area} = 28\text{ in}^2$, width $= 4\text{ in}$.
  • Rectangle 4: $\text{Area} = 21\text{ in}^2$, width $= 3\text{ in}$.
• Solution:
  • Width of Rectangle 1 $= 14 \div 7 = 2\text{ in}$.
  • Width of Rectangle 2 $= 35 \div 7 = 5\text{ in}$.
  • Height of Rectangle 3 $= 28 \div 4 = 7\text{ in}$.
  • Height of Rectangle 4 $= 21 \div 3 = 7\text{ in}$.
• Finding the Altitude of a Triangle (Page 155):
• Problem: In $\triangle ABC$, the sides are $AB = 5$, $BC = 6$, and $AC = 4$. The altitude from $A$ to $BC$ is $AX = 5$. Find the length of the altitude $BY$ from $B$ to $AC$.
• Solution: Calculate the area of the triangle using the two base-altitude pairs:
  $$\text{Area} = \frac{1}{2} \cdot AX \cdot BC = \frac{1}{2} \cdot 5 \cdot 6 = 15\text{ sq. units}$$
  $$\text{Area} = \frac{1}{2} \cdot BY \cdot AC = \frac{1}{2} \cdot BY \cdot 4 = 2BY$$
  $$2BY = 15 \rightarrow BY = 7.5\text{ units}$$
• Area of a Quadrilateral (Page 160 Q1):
• Problem: Find the area of quadrilateral $ABCD$ where diagonal $AC = 22\text{ cm}$, and perpendicular offsets are $BM = 3\text{ cm}$ and $DN = 3\text{ cm}$.
• Solution:
  $$\text{Area} = \frac{1}{2} \cdot AC \cdot (BM + DN) = \frac{1}{2} \cdot 22 \cdot (3 + 3) = 11 \cdot 6 = 66\text{ cm}^2$$
• Area of a Rhombus (Page 169 Q1):
• Problem: Find the area of a rhombus whose diagonals are $20\text{ cm}$ and $15\text{ cm}$.
• Solution:
  $$\text{Area} = \frac{1}{2} \cdot d_1 \cdot d_2 = \frac{1}{2} \cdot 20 \cdot 15 = 150\text{ cm}^2$$
• Area of a Trapezium (Page 169 Q3):
• Problem (ii): Find the area of a trapezium with parallel sides $24\text{ m}$ and $36\text{ m}$, and height $14\text{ m}$.
• Solution:
  $$\text{Area} = \frac{1}{2} \cdot h \cdot (a + b) = \frac{1}{2} \cdot 14 \cdot (24 + 36) = 7 \cdot 60 = 420\text{ m}^2$$

Practical Activities & Experiments

• Visualizing Perimeter-Area Disconnections: Cut a grid sheet into several rectangles, each having a perimeter of $20\text{ cm}$ (e.g., $9 \times 1$, $8 \times 2$, $7 \times 3$, $6 \times 4$, $5 \times 5$). Count the squares within each rectangle to find their areas and graph the results, showing that a square ($5 \times 5$) maximizes the area for a fixed perimeter.
• Cardboard Trapezium Parallelogram Dissection: Cut out two identical trapeziums from cardboard with parallel sides $10\text{ cm}$ and $6\text{ cm}$, and height $5\text{ cm}$. Rotate one by $180^\circ$ and align its slanted side with the other to form a parallelogram. Measure the base ($16\text{ cm}$) and height ($5\text{ cm}$) of the parallelogram to show that the area of one trapezium is exactly half of the parallelogram: $\frac{1}{2} \cdot 5 \cdot 16 = 40\text{ cm}^2$.