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⬅ Back to Plane Geometry
📐 Key Concepts: Quadrilaterals & Polygons

Rectangle: Area = length × width; Perimeter = 2(l + w)

Trapezoid: Area = ½(b₁ + b₂)h; for isosceles trapezoid, legs are equal and diagonals are equal.

Cyclic Quadrilateral (Brahmagupta's Formula): $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s = \frac{a+b+c+d}{2}$

General Quadrilateral: Split into two triangles by a diagonal; sum areas using the Law of Cosines and ½ab sin C.

Regular Polygon with n sides:

Problem 1: Rectangular Garden (Worded Problem)

The length of a rectangular garden is 9 m longer than its width. If the area of the garden is 78.6 square meters, find its dimensions.

Plane Geometry – Quadrilaterals – Problem 1 – Diagram Plane Geometry – Quadrilaterals – Problem 1 – Diagram Plane Geometry – Quadrilaterals – Problem 1 – Diagram
Solution Solution Solution Solution

Let $w$ = width, $l = w + 9$. Area equation:

$$w(w + 9) = 78.6 \Rightarrow w^2 + 9w - 78.6 = 0$$

Using the quadratic formula:

$$w = \frac{-9 + \sqrt{81 + 4(78.6)}}{2} = \frac{-9 + \sqrt{395.4}}{2} = \frac{-9 + 19.88}{2} \approx 5.44 \text{ m}$$
$$l = 5.44 + 9 = 14.44 \text{ m}$$

Dimensions: $\boxed{w \approx 5.44 \text{ m},\; l \approx 14.44 \text{ m}}$

Problem 2: Cyclic Quadrilateral (Brahmagupta's Formula)

The sides of a cyclic quadrilateral measure 8 cm, 9 cm, 12 cm, and 7 cm, respectively.
a. Find the area of the cyclic quadrilateral.
b. Find the area of the circumscribing circle.

Plane Geometry – Quadrilaterals – Problem 2 – Diagram
Solution

Part a — Area using Brahmagupta's Formula (sides $a=8, b=9, c=12, d=7$):

$$s = \frac{8+9+12+7}{2} = 18$$
$$A = \sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{10 \times 9 \times 6 \times 11} = \sqrt{5940} \approx \boxed{77.1 \text{ cm}^2}$$

Part b — Circumscribing circle (using the cyclic quadrilateral circumradius formula):

$$R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A}$$
$$= \frac{\sqrt{(72+84)(96+63)(56+108)}}{4(77.1)} = \frac{\sqrt{156 \times 159 \times 164}}{308.4} = \frac{\sqrt{4{,}067{,}856}}{308.4} \approx \frac{2017}{308.4} \approx 6.54 \text{ cm}$$
$$A_{\text{circle}} = \pi R^2 = \pi(6.54)^2 \approx \boxed{134.3 \text{ cm}^2}$$

Problem 3: General Quadrilateral Building Site

A building site is in the form of a quadrilateral with the following sides and angles given. AB = 52.4 m, BC = 28.5 m, CD = 34.6 m, angle B = 72°, and angle D = 75°. Determine the area of the site.

Plane Geometry – Quadrilaterals – Problem 3 – Diagram Plane Geometry – Quadrilaterals – Problem 3 – Diagram Plane Geometry – Quadrilaterals – Problem 3 – Diagram
Solution Solution Solution Solution

Split quadrilateral ABCD with diagonal AC. Triangle ABC (AB=52.4, BC=28.5, B=72°):

$$A_1 = \tfrac{1}{2}(52.4)(28.5)\sin72° = \tfrac{1}{2}(1493.4)(0.9511) \approx 710.0 \text{ m}^2$$
$$AC^2 = 52.4^2 + 28.5^2 - 2(52.4)(28.5)\cos72° = 2633.0 \Rightarrow AC \approx 51.3 \text{ m}$$

Triangle ACD (AC=51.3, CD=34.6, angle D=75° in triangle ACD). Find angle CAD via Law of Sines:

$$\sin(\angle CAD) = \frac{CD\sin D}{AC} = \frac{34.6\sin75°}{51.3} = \frac{33.4}{51.3} \approx 0.651 \Rightarrow \angle CAD \approx 40.6°$$
$$\angle ACD = 180° - 75° - 40.6° = 64.4°$$
$$A_2 = \tfrac{1}{2}(AC)(CD)\sin(\angle ACD) = \tfrac{1}{2}(51.3)(34.6)\sin64.4° \approx 800 \text{ m}^2$$
$$\text{Total Area} = A_1 + A_2 = 710 + 800 = \boxed{1510 \text{ m}^2}$$

Problem 4: Hexagon (Polygon General Formulas)

A regular hexagon whose sides measure 30 units each is inscribed in a circle. Determine the following:
a. The number of diagonals
b. The interior angle on each side
c. The radius of the circumscribing circle
d. The length of the apothem
e. The area of the polygon

Plane Geometry – Quadrilaterals – Problem 4 – Diagram Plane Geometry – Quadrilaterals – Problem 4 – Diagram Plane Geometry – Quadrilaterals – Problem 4 – Diagram
Solution Solution Solution Solution

For a regular hexagon with $n = 6$, side $s = 30$:

a. Number of diagonals:

$$\frac{n(n-3)}{2} = \frac{6(3)}{2} = \boxed{9}$$

b. Interior angle:

$$\frac{(n-2) \times 180°}{n} = \frac{4 \times 180°}{6} = \boxed{120°}$$

c. Circumradius:

$$R = \frac{s}{2\sin(\pi/n)} = \frac{30}{2\sin30°} = \frac{30}{1} = \boxed{30 \text{ units}}$$

d. Apothem:

$$a = \frac{s}{2\tan(\pi/n)} = \frac{30}{2\tan30°} = \frac{30}{2(0.5774)} = \frac{30}{1.1547} = 15\sqrt{3} \approx \boxed{25.98 \text{ units}}$$

e. Area:

$$A = \frac{1}{2} \times \text{Perimeter} \times a = \frac{1}{2}(180)(15\sqrt{3}) = \frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(900) = 1350\sqrt{3} \approx \boxed{2338 \text{ sq units}}$$

Problem 5: Cyclic Quadrilateral (Side = Diameter)

A cyclic quadrilateral ABCD is inside a circle with side AD coinciding with the diameter of the circle. AB = 8 cm, BC = 10 cm, CD = 12 cm. Find the area of the circumscribing circle.

Plane Geometry – Quadrilaterals – Problem 5 – Diagram Plane Geometry – Quadrilaterals – Problem 5 – Diagram Plane Geometry – Quadrilaterals – Problem 5 – Diagram
Solution Solution Solution Solution

Since AD is the diameter, angles ABD and ACD are both 90° (inscribed-angle theorem). Let $D = AD$ (diameter).

In right triangle ABD: $BD = \sqrt{D^2 - 64}$.   In right triangle ACD: $AC = \sqrt{D^2 - 144}$.

Apply Ptolemy's theorem ($AC \cdot BD = AB \cdot CD + BC \cdot AD$):

$$\sqrt{(D^2-144)(D^2-64)} = 8(12) + 10D = 96 + 10D$$

Squaring both sides and simplifying:

$$D^4 - 208D^2 + 9216 = 9216 + 1920D + 100D^2$$
$$D^3 - 308D - 1920 = 0 \Rightarrow D \approx 20.09 \text{ cm} \Rightarrow r = \frac{D}{2} \approx 10.04 \text{ cm}$$
$$A_{\text{circle}} = \pi r^2 \approx \pi(10.04)^2 \approx \boxed{317 \text{ cm}^2}$$

Problem 6: Isosceles Trapezoid

An isosceles trapezoid has parallel bases of 10 cm and 16 cm. The non-parallel legs are each 5 cm. Find:
a. The height of the trapezoid
b. The area of the trapezoid
c. The length of each diagonal

Plane Geometry – Quadrilaterals – Problem 6 – Diagram Plane Geometry – Quadrilaterals – Problem 6 – Diagram Plane Geometry – Quadrilaterals – Problem 6 – Diagram
Solution Solution Solution Solution

Bases: $b_1 = 10$ cm, $b_2 = 16$ cm; leg $= 5$ cm. The overhang on each side $= (16-10)/2 = 3$ cm.

a. Height (right triangle: hyp = 5, base = 3):

$$h = \sqrt{5^2 - 3^2} = \sqrt{16} = \boxed{4 \text{ cm}}$$

b. Area:

$$A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(10 + 16)(4) = \boxed{52 \text{ cm}^2}$$

c. Diagonal length (e.g., from corner of $b_2$ to far corner of $b_1$, horizontal span $= 3 + 10 = 13$ cm):

$$d = \sqrt{13^2 + 4^2} = \sqrt{169 + 16} = \sqrt{185} \approx \boxed{13.6 \text{ cm}}$$

Problem 7: Regular Pentagon Inscribed in a Circle

A regular pentagon is inscribed in a circle of radius 10 cm. Find:
a. The side length of the pentagon
b. The perimeter of the pentagon
c. The area of the pentagon
d. The area of the region inside the circle but outside the pentagon

Plane Geometry – Quadrilaterals – Problem 7 – Diagram Plane Geometry – Quadrilaterals – Problem 7 – Diagram Plane Geometry – Quadrilaterals – Problem 7 – Diagram
Solution Solution Solution Solution

For a regular pentagon ($n = 5$) inscribed in a circle of $R = 10$ cm:

a. Side length:

$$s = 2R\sin\!\left(\frac{\pi}{5}\right) = 2(10)\sin36° = 20(0.5878) \approx \boxed{11.76 \text{ cm}}$$

b. Perimeter:

$$P = 5s = 5(11.76) \approx \boxed{58.78 \text{ cm}}$$

c. Area (using apothem $a = R\cos(\pi/5) = 10\cos36° \approx 8.09$ cm):

$$A_{\text{pentagon}} = \tfrac{1}{2}Pa = \tfrac{1}{2}(58.78)(8.09) \approx \boxed{237.8 \text{ cm}^2}$$

d. Area inside circle but outside pentagon:

$$A_{\text{circle}} - A_{\text{pentagon}} = \pi(10)^2 - 237.8 = 314.2 - 237.8 \approx \boxed{76.4 \text{ cm}^2}$$