Plane Geometry

Angles, triangles, quadrilaterals, circles, area, similar triangles and 3-D solids.

Worked Examples

<details> <summary><b>Example 1: Same-side interior angles</b></summary>

Two parallel lines are cut by a transversal. Two same-side interior angles measure $(2x + 10)°$ and $(3x + 20)°$. Find both angles.

1. Same-side interior angles are one small and one big, so they are supplementary.
2. $(2x + 10) + (3x + 20) = 180 \implies 5x + 30 = 180 \implies x = 30$.
3. The angles are $2(30) + 10 = 70°$ and $3(30) + 20 = 110°$. Check: $70 + 110 = 180$.

Trap avoided: setting them equal would give $x = -10$, a negative angle, which tells you the setup was wrong.

</details> <details> <summary><b>Example 2: Complement and supplement in one equation</b></summary>

The complement of an angle is one-third of its supplement. Find the angle.

1. Translate: complement $= 90 - x$, supplement $= 180 - x$.
2. $90 - x = \frac{1}{3}(180 - x)$. Multiply by 3: $270 - 3x = 180 - x$.
3. $90 = 2x \implies x = 45°$.
4. Check: complement 45°, supplement 135°, and $135 \div 3 = 45$. ✓

</details> <details> <summary><b>Example 3: Sides of a regular polygon from one angle</b></summary>

Each interior angle of a regular polygon is 150°. How many sides does it have?

1. Exterior angle $= 180 - 150 = 30°$.
2. Exterior angles sum to 360°, so $n = 360 \div 30 = 12$.
3. Check: $(12 - 2) \times 180 = 1{,}800$, and $1{,}800 \div 12 = 150$. ✓

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Worked Examples

<details> <summary><b>Example 1: Isosceles triangle and an exterior angle</b></summary>

In isosceles triangle ABC, AB = AC and the vertex angle A measures 40°. Side BC is extended past C to point D. Find angle ACD.

1. Base angles B and C are equal: $(180 - 40) \div 2 = 70°$ each.
2. Angle ACD is the exterior angle at C, so it equals the two remote angles: $40 + 70 = 110°$.
3. Check with the straight line: $180 - 70 = 110°$. ✓

</details> <details> <summary><b>Example 2: Area of an equilateral triangle</b></summary>

Find the height and area of an equilateral triangle with side 10.

1. The altitude makes a 30-60-90 triangle with hypotenuse 10 and short leg 5.
2. Height = long leg $= 5\sqrt{3}$.
3. Area $= \frac{1}{2}(10)(5\sqrt{3}) = 25\sqrt{3}$. Formula check: $\frac{\sqrt{3}}{4}(100) = 25\sqrt{3}$. ✓

</details> <details> <summary><b>Example 3: A ladder problem with a triple</b></summary>

A 13-foot ladder reaches 12 feet up a vertical wall. How far is its foot from the wall?

1. The ladder is the hypotenuse (it is opposite the right angle between wall and ground).
2. Spot the 5-12-13 triple, or compute $\sqrt{169 - 144} = \sqrt{25} = 5$ feet.

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Worked Examples

<details> <summary><b>Example 1: Isosceles trapezoid height and area</b></summary>

An isosceles trapezoid has bases 8 and 20 and legs of 10. Find its area.

1. Overhang on each side: $(20 - 8) \div 2 = 6$.
2. Each end is a right triangle with hypotenuse 10 and leg 6, so $h = \sqrt{100 - 36} = 8$.
3. Area $= \frac{1}{2}(8 + 20)(8) = 112$.

Trap avoided: using the slanted leg 10 as the height gives 140.

</details> <details> <summary><b>Example 2: Rectangle from its diagonal</b></summary>

A rectangle's length is 2 more than its width, and its diagonal is 10. Find the perimeter.

1. The diagonal is a hypotenuse: $w^2 + (w + 2)^2 = 100$.
2. $2w^2 + 4w - 96 = 0 \implies w^2 + 2w - 48 = 0 \implies (w + 8)(w - 6) = 0$, so $w = 6$.
3. Length 8, perimeter $2(6 + 8) = 28$. (A 6-8-10 triangle, the 3-4-5 triple doubled.)

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Worked Examples

<details> <summary><b>Example 1: Arc length and sector area together</b></summary>

A circle has radius 9. Find the arc length and sector area for a central angle of 80°.

1. Fraction of the circle: $\frac{80}{360} = \frac{2}{9}$.
2. Arc $= \frac{2}{9}(18\pi) = 4\pi$.
3. Sector area $= \frac{2}{9}(81\pi) = 18\pi$.

Trap avoided: arc uses the circumference ($2\pi r$); sector area uses the area ($\pi r^2$).

</details> <details> <summary><b>Example 2: Square inscribed in a circle</b></summary>

A square is inscribed in a circle of radius 5. Find the area of the region inside the circle but outside the square.

1. The square's diagonal is a diameter: 10.
2. Side $= \frac{10}{\sqrt{2}} = 5\sqrt{2}$, so square area $= 50$ (or $\frac{d^2}{2} = \frac{100}{2}$).
3. Circle area $= 25\pi$. Shaded region $= 25\pi - 50$.

</details> <details> <summary><b>Example 3: Triangle with a diameter side</b></summary>

Triangle ABC is inscribed in a circle, and AB is a diameter of length 20. If AC = 12, find BC.

1. Angle C is inscribed in a semicircle, so it is 90°, and AB is the hypotenuse.
2. $BC = \sqrt{400 - 144} = \sqrt{256} = 16$ (the 3-4-5 triple times 4).

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Worked Examples

<details> <summary><b>Example 1: Area and perimeter of a rectangle with a semicircle</b></summary>

A figure is a 20-by-10 rectangle with a semicircle attached outward along one 10-unit side. Find its area and perimeter.

1. Semicircle radius $= 5$. Area $= 200 + \frac{1}{2}\pi(25) = 200 + 12.5\pi$.
2. Perimeter: the covered 10-unit side is gone. Outside edges: $20 + 10 + 20 = 50$, plus the arc $\frac{1}{2}(2\pi \cdot 5) = 5\pi$.
3. Perimeter $= 50 + 5\pi$.

</details> <details> <summary><b>Example 2: A picture frame</b></summary>

An 8-by-10 inch photo has a 2-inch frame on all sides. Find the frame's area.

1. Outer dimensions: $8 + 4 = 12$ by $10 + 4 = 14$, area 168.
2. Frame $= 168 - 80 = 88$ square inches.

Trap avoided: adding only 2 to each dimension (10 by 12) gives $120 - 80 = 40$.

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Worked Examples

<details> <summary><b>Example 1: A shadow problem</b></summary>

A 6-foot person casts a 4-foot shadow at the same time a tree casts a 30-foot shadow. How tall is the tree?

1. Person and tree each make a right angle with the ground, and the sun's angle is the same, so the triangles are similar (AA).
2. $\frac{\text{height}}{\text{shadow}}$: $\frac{6}{4} = \frac{h}{30}$.
3. $h = 45$ feet.

</details> <details> <summary><b>Example 2: Cylinder volume and surface area</b></summary>

A cylinder has radius 3 and height 10. Find its volume and total surface area.

1. Volume $= \pi(3^2)(10) = 90\pi$.
2. Two circular ends: $2\pi(9) = 18\pi$. Curved side (unrolls to a rectangle with width $2\pi r$ and height $h$): $2\pi(3)(10) = 60\pi$.
3. Surface area $= 78\pi$.

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Worked Examples

<details> <summary><b>Example 1: Rectangle inscribed in a circle</b></summary>

A 6-by-8 rectangle is inscribed in a circle. Find the area of the region inside the circle but outside the rectangle.

1. Link: the rectangle's diagonal is a diameter. Diagonal $= \sqrt{36 + 64} = 10$, so $r = 5$.
2. Circle area $= 25\pi$; rectangle area $= 48$.
3. Shaded region $= 25\pi - 48$.

</details> <details> <summary><b>Example 2: Angle algebra with an exterior angle</b></summary>

In triangle ABC, angle A $= (x + 15)°$, angle B $= (2x)°$, and the exterior angle at C is $(4x - 25)°$. Find angle C.

1. Exterior angle = sum of remote angles: $4x - 25 = (x + 15) + 2x \implies x = 40$.
2. Angle A $= 55°$, angle B $= 80°$, exterior angle at C $= 135°$.
3. Angle C $= 180 - 135 = 45°$. Check: $55 + 80 + 45 = 180$. ✓

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