Plane Geometry

LessonAbout 1.5 h1 of 2 in Plane and Solid Geometry

Plane Geometry

Key Ideas

  • Plane geometry examines two-dimensional flat figures.
  • Triangles and quadrilaterals form foundational plane polygons.
  • Polygons sum interior angles based on side count.
  • Circle sectors and segments measure partial circular areas.

Plane Geometry is the branch of geometry dealing with two-dimensional shapes, angles, polygons, and circular figures constructed on a flat surface.

Angles and Intersecting Lines

  • Complementary Angles: two angles summing to \(90^\circ\).
  • Supplementary Angles: two angles summing to \(180^\circ\).
  • Explementary Angles: two angles summing to \(360^\circ\).
  • Vertical Angles: equal opposite angles from intersecting lines.

Angle Supplement and Complement Relation

\[ (180^\circ - \theta) = k(90^\circ - \theta) \]

Example 1: A certain angle has a supplement equal to 5 times its complement. What is the angle?

Solution: Let \(\theta\) be the angle. Express supplement and complement in equation form:

\[ 180^\circ - \theta = 5(90^\circ - \theta) \]
\[ 180^\circ - \theta = 450^\circ - 5\theta \]
\[ 4\theta = 270^\circ \]
\[ \theta = 67.5^\circ \]

Answer: 67.5°

Triangles and Circles

  • Sum of Interior Angles: interior angles total \(180^\circ\).
  • Median: line from vertex to midpoint of opposite side.
  • Incircle Radius: circle tangent to all three triangle sides.
  • Circumcircle Radius: circle passing through all three vertices.
  • Escribed Circle Radius: circle tangent to one side and extended sides.

Heron’s Area Formula

\[ A_T = \sqrt{s(s - a)(s - b)(s - c)} \]

Inscribed Circle Radius Formula

\[ r = \dfrac{A_T}{s} \]

Circumscribed Circle Radius Formula

\[ R = \dfrac{abc}{4A_T} \]

Escribed Circle Radius Formula

\[ r_a = \dfrac{A_T}{s - a} \]

Example 2: The sides of a triangle are 8 cm, 10 cm, and 14 cm. Determine the radius of the inscribed circle and the circumscribed circle.

Solution: Compute semi-perimeter \(s\) and area \(A_T\) using Heron’s formula:

\[ s = \dfrac{8 + 10 + 14}{2} = 16\text{ cm} \]
\[ A_T = \sqrt{16(16 - 8)(16 - 10)(16 - 14)} = \sqrt{1536} = 39.19\text{ cm}^2 \]

Inscribed circle radius \(r\):

\[ r = \dfrac{39.19}{16} = 2.45\text{ cm} \]

Circumscribed circle radius \(R\):

\[ R = \dfrac{8 \times 10 \times 14}{4 \times 39.19} = 7.14\text{ cm} \]

Answer: Incircle radius = 2.45 cm, Circumcircle radius = 7.14 cm

Example 3: The sides of a triangular lot are 130 m, 180 m, and 190 m. The lot is divided by a median drawn from the vertex opposite the longest side to its midpoint. Find the length of the dividing line.

Solution: Let \(a = 130\text{ m}\), \(b = 180\text{ m}\), \(c = 190\text{ m}\). Find angle \(B\) opposite side \(b\) using Cosine Law:

\[ 180^2 = 130^2 + 190^2 - 2(130)(190) \cos B \]
\[ 32400 = 16900 + 36100 - 49400 \cos B \implies \cos B = 0.4170 \]

Apply Cosine Law to the sub-triangle formed by side \(a\), half of \(c\) (\(95\text{ m}\)), and median \(m\):

\[ m^2 = 130^2 + 95^2 - 2(130)(95)(0.4170) \]
\[ m^2 = 16900 + 9025 - 10299.9 = 15625.1 \implies m = 125\text{ m} \]

Answer: 125 m

Quadrilaterals

  • Trapezoid Area: half product of height and parallel base sum.
  • Rhombus Area: half product of diagonals or \(a^2 \sin \theta\).
  • Parallelogram Area: base times altitude or \(ab \sin \theta\).
  • General Quadrilateral Area: half diagonal product times sine of included angle.
  • Cyclic Quadrilateral: vertices on a circle; opposite angles sum to \(180^\circ\).

Trapezoid Area Formula

\[ A = \dfrac{B + b}{2} h \]

Trapezoid Equal Area Dividing Line

\[ y = \sqrt{\dfrac{a^2 + b^2}{2}} \]

Cyclic Quadrilateral Area (Brahmagupta’s Formula)

\[ A = \sqrt{(s - a)(s - b)(s - c)(s - d)} \]

Ptolemy’s Theorem

\[ d_1 d_2 = ac + bd \]

General Quadrilateral Area (Bretschneider’s Formula)

\[ A = \sqrt{(s - a)(s - b)(s - c)(s - d) - abcd \cos^2 \theta} \]

Example 4: Find the area of a quadrilateral with sides \(a = 10\text{ cm}\), \(b = 5\text{ cm}\), \(c = 14.14\text{ cm}\), and \(d = 15\text{ cm}\), if the sum of opposite angles equals \(225^\circ\).

Solution: Calculate semi-perimeter \(s\) and angle parameter \(\theta\):

\[ s = \dfrac{10 + 5 + 14.14 + 15}{2} = 22.07\text{ cm} \]
\[ \theta = \dfrac{225^\circ}{2} = 112.5^\circ \implies \cos^2(112.5^\circ) = 0.1464 \]

Apply Bretschneider’s formula:

\[ (s - a)(s - b)(s - c)(s - d) = (12.07)(17.07)(7.93)(7.07) = 11550 \]
\[ abcd \cos^2 \theta = (10 \times 5 \times 14.14 \times 15) \times 0.1464 = 1552 \]
\[ A = \sqrt{11550 - 1552} = \sqrt{9998} = 100\text{ cm}^2 \]

Answer: 100 cm²

Example 5: Parallel sides of a trapezoidal lot measure 160 m and 240 m. Find the length of a dividing line parallel to the two bases that splits the lot into two equal areas.

Solution: Apply the equal area dividing line formula for bases \(a = 160\) and \(b = 240\):

\[ y = \sqrt{\dfrac{160^2 + 240^2}{2}} = \sqrt{\dfrac{25600 + 57600}{2}} = \sqrt{41600} = 203.96\text{ m} \]

Answer: 203.96 m

Polygons

  • Convex Polygon: every interior angle is less than \(180^\circ\).
  • Sum of Interior Angles: total interior degrees equal \((n - 2) \times 180^\circ\).
  • Sum of Exterior Angles: sum of exterior angles always equals \(360^\circ\).
  • Number of Diagonals: non-adjacent vertex line count.
  • Apothem: radius of inscribed circle inside a regular polygon.

Sum of Interior Angles Formula

\[ S = (n - 2) \times 180^\circ \]

Interior Angle of a Regular Polygon

\[ \theta = \dfrac{(n - 2) \times 180^\circ}{n} \]

Number of Diagonals Formula

\[ D = \dfrac{n}{2}(n - 3) \]

Area of a Regular Polygon

\[ A = \dfrac{1}{2} n b r = \dfrac{n b^2}{4 \tan\left(\dfrac{180^\circ}{n}\right)} \]

Example 6: How many sides has a polygon if the sum of its interior angles is \(1080^\circ\), and how many diagonals does it have?

Solution: Solve for side count \(n\) using interior angle sum formula:

\[ (n - 2) \times 180^\circ = 1080^\circ \implies n - 2 = 6 \implies n = 8\text{ sides (Octagon)} \]

Compute diagonal count \(D\) for \(n = 8\):

\[ D = \dfrac{8}{2}(8 - 3) = 4 \times 5 = 20 \]

Answer: 8 sides, 20 diagonals

Example 7: Find the area of a regular hexagon inscribed in a circle of radius \(R = 10\text{ cm}\).

Solution: A regular hexagon divides into 6 central equilateral triangles with side \(r = R = 10\text{ cm}\):

\[ A = 6 \times \left(\dfrac{1}{2} R^2 \sin 60^\circ\right) = 3 \times (10)^2 \times \dfrac{\sqrt{3}}{2} = 259.8\text{ cm}^2 \]

Answer: 259.8 cm²

Example 8: Find the sum of the interior angles of the 5 vertices of a five-pointed star inscribed in a circle.

Solution: Each star tip angle subtends one-fifth of the circle circumference minus twice adjacent arcs, yielding tip angle \(\alpha = 36^\circ\). Total sum of 5 tip angles:

\[ S = 5 \times 36^\circ = 180^\circ \]

Answer: 180°

Circles, Sectors, and Segments

  • Circumference: perimeter length equal to \(2\pi r\) or \(\pi d\).
  • Sector Area: pie-shaped area bounded by two radii and arc.
  • Segment Area: region bounded by a chord and subtended arc.
  • Annulus Area: ring area bounded between two concentric circles.

Arc Length Formula

\[ C = r \theta_{\text{rad}} = \dfrac{\pi r \theta^\circ}{180^\circ} \]

Sector Area Formula

\[ A_{\text{sector}} = \dfrac{1}{2} r^2 \theta_{\text{rad}} = \dfrac{\pi r^2 \theta^\circ}{360^\circ} \]

Segment Area Formula

\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \dfrac{1}{2} r^2 (\theta_{\text{rad}} - \sin \theta) \]

Annulus Area Formula

\[ A = \pi (R^2 - r^2) \]

Example 9: A circle with radius \(6\text{ cm}\) has half its area removed by cutting off a border of uniform width \(x\). Find the width \(x\).

Solution: Total area \(A = \pi(6)^2 = 36\pi\). Remaining inner area \(A_{\text{inner}} = 18\pi\). Solve for inner radius \(r\):

\[ \pi r^2 = 18\pi \implies r = \sqrt{18} = 4.24\text{ cm} \]

Border width \(x\):

\[ x = R - r = 6 - 4.24 = 1.76\text{ cm} \]

Answer: 1.76 cm

Example 10: A pool is shaped in two partially overlapping identical circles of radius \(9\text{ m}\), each passing through the center of the other. Find the total surface area of the pool.

Solution: Center distance \(d = r = 9\text{ m}\). Central angle subtended by overlapping chord is \(\theta = 120^\circ = \dfrac{2\pi}{3}\text{ rad}\).

Area of single overlapping segment \(A_{\text{segment}}\):

\[ A_{\text{segment}} = \dfrac{1}{2}(9)^2 \left(\dfrac{2\pi}{3} - \sin 120^\circ\right) = 40.5 \times (2.0944 - 0.8660) = 49.75\text{ m}^2 \]

Total pool area equals combined circle areas minus twice overlapping segment area:

\[ A_{\text{total}} = 2\pi(9)^2 - 2(49.75) = 162\pi - 99.5 = 508.94 - 99.5 = 409.44\text{ m}^2 \]

Answer: 409.44 m²

Formula Summary

Formula Name Formula
Angle Supplement and Complement Relation \((180^\circ - \theta) = k(90^\circ - \theta)\)
Heron’s Area Formula \(A_T = \sqrt{s(s - a)(s - b)(s - c)}\)
Inscribed Circle Radius Formula \(r = \dfrac{A_T}{s}\)
Circumscribed Circle Radius Formula \(R = \dfrac{abc}{4A_T}\)
Escribed Circle Radius Formula \(r_a = \dfrac{A_T}{s - a}\)
Trapezoid Area Formula \(A = \dfrac{B + b}{2} h\)
Trapezoid Equal Area Dividing Line \(y = \sqrt{\dfrac{a^2 + b^2}{2}}\)
Cyclic Quadrilateral Area (Brahmagupta’s Formula) \(A = \sqrt{(s - a)(s - b)(s - c)(s - d)}\)
Ptolemy’s Theorem \(d_1 d_2 = ac + bd\)
General Quadrilateral Area (Bretschneider’s Formula) \(A = \sqrt{(s - a)(s - b)(s - c)(s - d) - abcd \cos^2 \theta}\)
Sum of Interior Angles Formula \(S = (n - 2) \times 180^\circ\)
Interior Angle of a Regular Polygon \(\theta = \dfrac{(n - 2) \times 180^\circ}{n}\)
Number of Diagonals Formula \(D = \dfrac{n}{2}(n - 3)\)
Area of a Regular Polygon \(A = \dfrac{1}{2} n b r = \dfrac{n b^2}{4 \tan\left(\dfrac{180^\circ}{n}\right)}\)
Arc Length Formula \(C = r \theta_{\text{rad}} = \dfrac{\pi r \theta^\circ}{180^\circ}\)
Sector Area Formula \(A_{\text{sector}} = \dfrac{1}{2} r^2 \theta_{\text{rad}} = \dfrac{\pi r^2 \theta^\circ}{360^\circ}\)
Segment Area Formula \(A_{\text{segment}} = \dfrac{1}{2} r^2 (\theta_{\text{rad}} - \sin \theta)\)
Annulus Area Formula \(A = \pi (R^2 - r^2)\)