10.2 Quadrilaterals, Polygons, Circles & Coordinate Geometry

Quadrilaterals, Polygons, Circles & Coordinate Geometry

This section expands upon fundamental geometry to explore multi-sided plane figures (polygons and quadrilaterals), circular geometry (arcs, sectors, tangents, and inscribed angles), and coordinate geometry on the Cartesian plane. Mastering these topics is essential for solving quantitative items on the NTS GAT General.

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1. Properties and Angle Theorems of Quadrilaterals & Polygons

A polygon is a closed two-dimensional figure bounded by straight line segments joining non-collinear vertices. A polygon is convex if all interior angles measure less than $180^\circ$, and regular if all sides are equal in length (equilateral) and all interior angles are equal in measure (equiangular).

Classification and Properties of Quadrilaterals (4-Sided Polygons)

Quadrilaterals form a core sub-category of polygons tested heavily on the NTS GAT:

General Polygon Formulas for $n$-Sided Polygons

For any convex polygon with $n$ sides ($n \ge 3$):

1. Sum of Interior Angles ($S$): Derived by dividing the polygon into $(n - 2)$ non-overlapping triangles from one vertex: $S = (n - 2) \times 180^\circ$
2. Each Interior Angle of a Regular $n$-gon ($I$): $I = \frac{(n - 2) \times 180^\circ}{n}$
3. Sum of Exterior Angles ($E_{\text{total}}$): The sum of exterior angles (one per vertex) for any convex polygon is strictly constant: $E_{\text{total}} = 360^\circ$
4. Each Exterior Angle of a Regular $n$-gon ($E$): $E = \frac{360^\circ}{n}$ Note: At any given vertex, $\text{Interior Angle } I + \text{Exterior Angle } E = 180^\circ$.
5. Number of Diagonals ($D$): The total number of line segments connecting non-adjacent vertices: $D = \frac{n(n - 3)}{2}$

Worked NTS GAT Example 1

Problem: Each interior angle of a regular polygon measures $144^\circ$. How many sides does this polygon possess, and what is its total number of diagonals?

Solution:

1. Find Exterior Angle ($E$): $E = 180^\circ - 144^\circ = 36^\circ$.
2. Find Number of Sides ($n$): Using $E = \frac{360^\circ}{n} \implies 36^\circ = \frac{360^\circ}{n} \implies n = \frac{360}{36} = 10$ sides (Decagon).
3. Alternatively, apply interior angle formula: $\frac{(n-2) \times 180}{n} = 144 \implies 180n - 360 = 144n \implies 36n = 360 \implies n = 10$.
4. Calculate Number of Diagonals ($D$): $D = \frac{n(n - 3)}{2} = \frac{10(10 - 3)}{2} = \frac{10 \times 7}{2} = 35 \text{ diagonals}$
5. Answer: The polygon has 10 sides and 35 diagonals.

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2. Circle Geometry: Arcs, Sectors, and Chords

A circle is the set of all points in a plane situated at a fixed distance (radius $r$) from a central point $O$.

Core Metric Formulas

• Radius ($r$): Distance from center to outer boundary.
• Diameter ($d$): Longest chord passing through center: $d = 2r$.
• Circumference ($C$): Total perimeter boundary of circle: $C = 2\pi r = \pi d$.
• Area ($A$): Flat surface region enclosed: $A = \pi r^2$.

Arcs and Circular Sectors

An arc is a curved continuous section of a circle's circumference. A sector is a pie-shaped region bounded by two radii and an arc. For a central angle $\theta$ expressed in degrees:

$\text{Arc Length } (L) = \frac{\theta}{360^\circ} \times 2\pi r$

$\text{Sector Area } (K) = \frac{\theta}{360^\circ} \times \pi r^2$

Important Circle Theorems

1. Inscribed Angle Theorem: An angle inscribed in a circle (vertex on boundary) is equal to exactly half of the central angle subtending the same intercepted arc. $\angle \text{inscribed} = \frac{1}{2} \theta_{\text{central}}$
2. Angle in a Semicircle: Any angle inscribed within a semicircle (subtended by the diameter) is strictly a right angle ($90^\circ$).
3. Tangent-Radius Perpendicularity: A tangent line touching a circle at point $P$ is strictly perpendicular ($90^\circ$) to the radius $OP$.
4. Perpendicular Bisector of a Chord: A line drawn from the center perpendicular to a chord bisects the chord into two equal segments.

Worked NTS GAT Example 2

Problem: A circular sector has a central angle of $60^\circ$ and a radius of $12\text{ cm}$. Find the exact arc length and sector area in terms of $\pi$.

Solution:

1. Calculate Arc Length ($L$): $L = \frac{60^\circ}{360^\circ} \times 2\pi(12) = \frac{1}{6} \times 24\pi = 4\pi\text{ cm}$
2. Calculate Sector Area ($K$): $K = \frac{60^\circ}{360^\circ} \times \pi(12^2) = \frac{1}{6} \times 144\pi = 24\pi\text{ cm}^2$
3. Answer: Arc length is $4\pi\text{ cm}$ and Sector area is $24\pi\text{ cm}^2$.

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3. Cartesian Coordinate Geometry

Coordinate geometry unifies algebra and spatial geometry on the two-dimensional Cartesian plane defined by perpendicular horizontal ($x$-axis) and vertical ($y$-axis) axes intersecting at origin $(0,0)$.

Fundamental Coordinate Formulas

For any two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$:

1. Distance Formula: Derived directly from the Pythagorean theorem: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
2. Midpoint Formula: Calculates coordinates of the middle point: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
3. Slope Formula ($m$): Quantifies steepness and direction: $m = \frac{\text{Rise}}{\text{Run}} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2)$

Parallel and Perpendicular Line Rules

• Parallel Lines ($L_1 \parallel L_2$): Slopes are equal ($m_1 = m_2$).
• Perpendicular Lines ($L_1 \perp L_2$): Slopes are negative reciprocals ($m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1}$).
• Horizontal Lines: Slope $m = 0$; equation $y = k$.
• Vertical Lines: Slope is undefined; equation $x = h$.

Equations of Lines

• Slope-Intercept Form: $y = mx + c$ (where $m$ is slope, $c$ is $y$-intercept).
• Point-Slope Form: $y - y_1 = m(x - x_1)$.
• Standard Form: $Ax + By = C$.

Worked NTS GAT Example 3

Problem: Line $L_1$ passes through points $(2, 5)$ and $(6, 13)$. Line $L_2$ is perpendicular to $L_1$ and passes through $(4, 1)$. Find the equation of line $L_2$ in slope-intercept form.

Solution:

1. Find slope of $L_1$ ($m_1$): $m_1 = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$
2. Find perpendicular slope of $L_2$ ($m_2$): $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$
3. Write equation of $L_2$ using point-slope form with $(4, 1)$: $y - 1 = -\frac{1}{2}(x - 4)$
4. Convert to slope-intercept form ($y = mx + c$): $y - 1 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 3$
5. Answer: The equation of line $L_2$ is $y = -\frac{1}{2}x + 3$.