8.2 Perimeter and Area of Quadrilaterals, Polygons, Circles & Shaded Region Problems

Triangles: Area and Dimension Formulations

+-----------------------------------------------------------------------------+
|                         TRIANGLE AREA FORMULATION MATRIX                    |
+-----------------------------------------------------------------------------+
| Triangle Type      | Area Formula                   | Key Geometric Radii   |
+--------------------+--------------------------------+-----------------------+
| General Scalene    | Heron's: √[s(s-a)(s-b)(s-c)]   | r = Δ/s, R = abc/(4Δ) |
| Right-Angled       | (1/2) * base * height          | r = (a+b-c)/2, R = c/2|
| Equilateral        | (√3 / 4) * a²                  | r = a/(2√3), R = a/√3 |
| Isosceles          | (b / 4) * √(4a² - b²)          | h = (1/2) * √(4a² - b²)|
+-----------------------------------------------------------------------------+

1. General Scalene Triangle

• Base & Height: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} ab \sin C$.
• Heron's Formula: When all three sides $a, b, c$ are known:

$$s = \frac{a+b+c}{2} \quad (\text{semi-perimeter})$$ $$\text{Area} (\Delta) = \sqrt{s(s-a)(s-b)(s-c)}$$

2. Equilateral Triangle (Side $a$)

• Height (Altitude): $h = \frac{\sqrt{3}}{2}a$.
• Area: $\text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{h^2}{\sqrt{3}}$.
• Inradius ($r$): $r = \frac{h}{3} = \frac{a}{2\sqrt{3}}$.
• Circumradius ($R$): $R = \frac{2h}{3} = \frac{a}{\sqrt{3}} = 2r$.
• Area Ratio of Circles: $\text{Area of Incircle} : \text{Area of Circumcircle} = \pi r^2 : \pi R^2 = 1 : 4$.

3. Right-Angled Triangle

• For perpendicular legs $a, b$ and hypotenuse $c = \sqrt{a^2 + b^2}$:

$$\text{Area} = \frac{1}{2}ab$$

• Altitude to Hypotenuse ($p$):

$$p = \frac{a \times b}{c}$$

---

Quadrilaterals: Exhaustive Formulas & Pathway Rules

1. Rectangle (Length $l$, Breadth $b$)

• $\text{Area} = l \times b$, $\text{Perimeter} = 2(l + b)$, $\text{Diagonal } d = \sqrt{l^2 + b^2}$.
• Pathway Formulas (Width $w$): All lengths are positive. An inner path needs $2w\le\min(l,b)$; central cross-paths need $w\le\min(l,b)$.
  1. Uniform Outer Path: Runs outside a rectangle $l \times b$:

$$\text{Area of Outer Path} = (l + 2w)(b + 2w) - lb = 2w(l + b + 2w)$$

1. Uniform Inner Path: Runs inside a rectangle $l \times b$:

$$\text{Area of Inner Path} = lb - (l - 2w)(b - 2w) = 2w(l + b - 2w)$$

1. Central Cross Paths: Two pathways of width $w$ running parallel to sides and intersecting at center:

$$\text{Area of Cross Paths} = w(l + b - w)$$

+─────────────────────────────────────────────────────────────+
|                    RECTANGULAR PATHWAY PROFILES             |
+─────────────────────────────────────────────────────────────+
| OUTER PATH (Width w):        INNER PATH (Width w):          |
| ┌───────────────────────┐    ┌─────────────────────────┐    |
| │      Pathway (w)      │    │  Outer Boundary (l x b) │    |
| │   ┌───────────────┐   │    │   ┌─────────────────┐   │    |
| │   │ Plot (l x b)  │   │    │   │  Inner Unpaved  │   │    |
| │   └───────────────┘   │    │   │ (l-2w) x (b-2w) │   │    |
| └───────────────────────┘    │   └─────────────────┘   │    |
| Area = 2w(l + b + 2w)        └─────────────────────────┘    |
|                              Area = 2w(l + b - 2w)          |
+─────────────────────────────────────────────────────────────+

2. Square (Side $a$)

• $\text{Area} = a^2 = \frac{d^2}{2}$.
• $\text{Perimeter} = 4a$, $\text{Diagonal } d = a\sqrt{2}$.
• Inradius $r = a/2$; Circumradius $R = a/\sqrt{2}$.

3. Parallelogram (Sides $a, b$ and Height $h$)

• $\text{Area} = \text{base} \times \text{height} = a \times h = ab \sin \theta$.
• Diagonal Identity: The sum of the squares of the diagonals equals the sum of the squares of all four sides:

$$d_1^2 + d_2^2 = 2(a^2 + b^2)$$

4. Rhombus (Side $a$, Diagonals $d_1, d_2$)

• All four sides are equal; diagonals bisect each other at right angles ($90^\circ$).
• Area:

$$\text{Area} = \frac{1}{2} d_1 d_2$$

• Side-Diagonal Fundamental Identity:

$$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \iff d_1^2 + d_2^2 = 4a^2$$ $$a = \frac{1}{2}\sqrt{d_1^2 + d_2^2}, \quad \text{Perimeter} = 4a = 2\sqrt{d_1^2 + d_2^2}$$

5. Trapezium / Trapezoid (Parallel Sides $a$ and $b$, Height $h$)

• Area:

$$\text{Area} = \frac{1}{2}(a + b)h$$

• Median Rule: The line segment connecting the midpoints of the non-parallel sides has length $m = \frac{a+b}{2}$. Thus:

$$\text{Area} = \text{Median} \times h$$

---

Circles, Sectors, Segments & Annulus Systems

1. Fundamental Circle & Semicircle Metrics

• Full Circle: $\text{Area} = \pi r^2$, $\text{Circumference} = 2\pi r = \pi d$.
• Semicircle: $\text{Area} = \frac{1}{2}\pi r^2$.
  • Perimeter of Semicircle: Unlike circumference, the perimeter includes the straight diameter boundary:

$$\text{Perimeter} = \pi r + 2r = r(\pi + 2) \approx r\left(\frac{22}{7} + 2\right) = \frac{36}{7}r \quad (\text{when } \pi \approx 22/7)$$

• Quadrant (Quarter Circle): $\text{Area} = \frac{1}{4}\pi r^2$.
  • $\text{Perimeter} = \frac{1}{2}\pi r + 2r = r\left(\frac{\pi}{2} + 2\right) \approx r\left(\frac{11}{7} + 2\right) = \frac{25}{7}r$.

2. Sectors and Segments of a Circle

For a sector subtending central angle $\theta$ (in degrees), use the minor-segment formula below for $0<\theta\le180^\circ$:

• Arc Length ($l$):

$$l = \frac{\theta}{360^\circ} \times 2\pi r$$

• Sector Area:

$$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} l r$$

• Area of Segment: The region bounded by a chord and arc:

$$\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of } \Delta = \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2 \sin \theta$$

3. Annulus / Circular Ring

The area bounded between two concentric circles of outer radius $R$ and inner radius $r$ (track width $w = R - r$):

$$\text{Area} = \pi(R^2 - r^2) = \pi(R + r)(R - r)$$ $$\text{Area} = \pi(R^2 - r^2) = \pi x^2 = \frac{\pi}{4}(\text{Chord Length})^2$$

4. Classical Shaded Region Problems

• Four Equal Circles at Square Vertices: Place four radius-$r$ circles with centers at the vertices of a square of side $2r$. Adjacent circles touch; this explicitly stipulated square arrangement gives: The enclosed central shaded region is:

$$\text{Shaded Area} = \text{Area of Square} - 4 \times (\text{Quadrant Area}) = (2r)^2 - \pi r^2 = (4 - \pi)r^2 \approx \frac{6}{7}r^2$$

---

Regular Polygons

For any regular polygon with $n$ equal sides and side length $a$:

---

Worked Examples

Worked Example 1: Rhombus Land Parcel & Paving Cost

Problem: A railway goods siding plot is laid out in the shape of a rhombus. The perimeter of the plot is $104\text{ m}$, and one of its diagonals measures $48\text{ m}$. Calculate: (a) the length of the second diagonal, (b) the total surface area of the plot, and (c) the cost of leveling the plot at ₹15 per $\text{m}^2$.

• Step 1: Compute Side Length of Rhombus:

$$\text{Perimeter} = 4a = 104\text{ m} \implies a = \frac{104}{4} = 26\text{ m}$$

• Step 2: Determine Half-Diagonal Lengths: Given diagonal $d_1 = 48\text{ m}$, half-diagonal is $d_1 / 2 = 24\text{ m}$. Since diagonals of a rhombus bisect each other perpendicularly:

$$a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2$$ $$26^2 = 24^2 + \left(\frac{d_2}{2}\right)^2 \implies 676 = 576 + \left(\frac{d_2}{2}\right)^2$$ $$\left(\frac{d_2}{2}\right)^2 = 676 - 576 = 100 \implies \frac{d_2}{2} = 10\text{ m}$$

• Step 3: Compute Full Second Diagonal Length:

$$d_2 = 2 \times 10 = 20\text{ m}$$

• Step 4: Compute Total Area:

$$\text{Area} = \frac{1}{2} d_1 d_2 = \frac{1}{2} \times 48 \times 20 = 480\text{ m}^2$$

• Step 5: Compute Total Cost:

$$\text{Cost} = 480\text{ m}^2 \times \text{INR } 15/\text{m}^2 = \text{INR } 7,200$$

---

Worked Example 2: Railway Track Formation Cross-Section (Trapezium)

Problem: A broad-gauge railway embankment has a trapezoidal cross-section. The top formation width is $12\text{ m}$, the base width is $18\text{ m}$, and the vertical height of the embankment is $4\text{ m}$. Determine: (a) the cross-sectional area of the embankment, and (b) the total volume of earthwork needed to construct a $500\text{ m}$ section of this track embankment.

• Step 1: Compute Cross-Sectional Area:

$$\text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(12 + 18) \times 4 = \frac{1}{2} \times 30 \times 4 = 60\text{ m}^2$$

• Step 2: Compute Earthwork Volume over $500\text{ m}$ Track Length:

$$\text{Volume} = \text{Cross-Sectional Area} \times \text{Length} = 60\text{ m}^2 \times 500\text{ m} = 30,000\text{ m}^3$$

---

Worked Example 3: Annulus Turntable Inspection Pathway

Problem: A circular locomotive turntable pit of radius $r = 14\text{ m}$ is surrounded by an outer concrete inspection walkway of uniform width $w = 3.5\text{ m}$. Using $\pi = 22/7$, calculate: (a) the outer radius $R$, (b) the area of the inspection walkway, and (c) the total expenditure of paving the pathway at ₹80 per $\text{m}^2$.

• Step 1: Determine Outer Radius:

$$R = r + w = 14 + 3.5 = 17.5\text{ m} = \frac{35}{2}\text{ m}$$

• Step 2: Express Area via Difference of Squares:

$$\text{Area} = \pi(R^2 - r^2) = \pi(R + r)(R - r)$$ $$R + r = 17.5 + 14 = 31.5\text{ m} = \frac{63}{2}\text{ m}, \quad R - r = 3.5\text{ m} = \frac{7}{2}\text{ m}$$

• Step 3: Calculate Walkway Area:

$$\text{Area} = \frac{22}{7} \times \frac{63}{2} \times \frac{7}{2} = 11 \times \frac{63}{2} = \frac{693}{2} = 346.5\text{ m}^2$$

• Step 4: Calculate Total Paving Cost:

$$\text{Cost} = 346.5 \times 80 = \text{INR } 27,720$$

---

Calculation Traps & Speed Heuristics

1. The Semicircle Perimeter Trap ($P = \pi r$ vs. $\pi r + 2r$): A common perimeter error is taking the perimeter of a semicircle as $\pi r$. That is only the curved arc! The full perimeter includes the straight diameter: $P = \pi r + 2r = \frac{36}{7}r$. When an exercise stipulates $\pi=22/7$ and gives $P = 108\text{ m}$, immediately write $\frac{36}{7}r = 108 \implies r = 21\text{ m}$.
2. The Diagonal-Area Relation in Squares ($a^2$ vs. $d^2/2$): Never waste time finding side $a$ from diagonal $d$ by computing $a = d/\sqrt{2}$. Directly square the diagonal and divide by 2: $\text{Area} = d^2/2$. If diagonal is $12\text{ cm}$, area is $144/2 = 72\text{ cm}^2$.
3. The Polygon Interior-Exterior Shortcut: When a problem states "each interior angle of a regular polygon is $150^\circ$", avoid solving $\frac{(n-2) \times 180}{n} = 150$. Immediately compute the exterior angle: $\text{Ext Angle} = 180^\circ - 150^\circ = 30^\circ$. Then the number of sides is $n = \frac{360^\circ}{30^\circ} = 12$ sides.