Polygons: area and perimeter

1. Area of a parallelogram: A = bh (on the formula sheet).
2. The base is b = 9 cm.
3. The height is the dashed segment that meets the base at a right angle: h = 5 cm.
4. The slanted side (6 cm) is not the height, so it is not used for area.
5. A = 9 \times 5 = 45.

The area is 45 cm².

Check: slide the left triangle over to the right side and the parallelogram becomes a 9 by 5 rectangle: 9 \times 5 = 45 ✓.

Tip: For parallelograms, area = base × perpendicular height. The slanted side is always longer than the height, so using it makes the area too big.

1. A hexagon has 6 sides.
2. "Regular" means all 6 sides are equal: each is 8 in.
3. Perimeter = n \times s = 6 \times 8 = 48.

The perimeter is 48 in.

Check: add them one by one: 8 + 8 + 8 + 8 + 8 + 8 = 48 ✓. The perimeter is a length, so the unit is plain inches, not square inches.

Tip: Memorize the side counts: pentagon 5, hexagon 6, octagon 8. Perimeter of a regular polygon = (number of sides) × (side length) — no squaring.

1. Area of a parallelogram: A = bh (on the formula sheet).
2. The base is b = 14 cm.
3. The height must be perpendicular to the base: h = 7 cm. The 9 cm slanted side is not the height.
4. Compute: A = 14 \times 7 = 98.

The area is 98 cm².

Check: cut the triangle off one end and slide it to the other end — you get a 14 \times 7 rectangle, area 98 ✓.

Tip: In a parallelogram, the slanted side is always longer than the height. Using it (14 \times 9 = 126) gives too big an area.

1. Trapezoid area: A = \tfrac{1}{2}h(b_1 + b_2).
2. Bases: b_1 = 12, b_2 = 18; height h = 10.
3. Add the bases: 12 + 18 = 30.
4. A = \tfrac{1}{2}(10)(30) = \tfrac{1}{2}(300) = 150.

The deck has an area of 150 ft².

Check: average base = 15 ft; 15 \times 10 = 150 ✓. Also, 150 is between 12 \times 10 = 120 and 18 \times 10 = 180 ✓.

Tip: The "distance between the parallel sides" is the GED's way of naming the height. A quick reasonableness check: a trapezoid's area always lies between (short base × height) and (long base × height).

1. An octagon has 8 sides.
2. It is regular, so every side is 12.4 in.
3. Perimeter = 8 \times 12.4.
4. 8 \times 12 = 96 and 8 \times 0.4 = 3.2; 96 + 3.2 = 99.2.

The strip is 99.2 in long.

Check: estimate 8 \times 12 = 96; the answer is a little more ✓. Dividing back, 99.2 \div 8 = 12.4 in per side ✓.

Tip: "Oct-" means 8 (an *oct*opus has 8 arms). Going around the edge of a shape is always a perimeter question.

1. The fencing goes around the garden, so the perimeter is P = 65 m.
2. A pentagon has 5 equal sides (regular): P = 5s.
3. 65 = 5s, so s = 65 \div 5 = 13.

Each side is 13 m.

Check: 5 \times 13 = 65 ✓. Each side (13 m) is a reasonable fraction of the total — about one fifth of 65.

Tip: Working backwards from a regular polygon's perimeter: divide by the number of sides. "Penta-" means 5 (the Pentagon building has 5 sides).

1. The area of a trapezoid is A = \tfrac{1}{2}h(b_1 + b_2) (on the formula sheet).
2. Here b_1 = 5, b_2 = 17 and h = 8.
3. Substitute: A = \tfrac{1}{2}(5 + 17)(8).
4. Value: \tfrac{1}{2}(22)(8) = 11 \times 8 = 88\text{ cm}^2.

Correct expression: \tfrac{1}{2}(5 + 17)(8).

Check: the average of the bases is (5 + 17) \div 2 = 11; an 11 \times 8 rectangle has the same area, 88 ✓.

Tip: Add the two bases (do not multiply them), then take half, then multiply by the height. The order of multiplying does not matter, but the addition inside the parentheses must come first.

1. A regular hexagon has 6 equal sides.
2. Perimeter = 6s, so 87 = 6s.
3. Divide: s = 87 \div 6 = 14.5.

Each side is 14.5 inches.

Check: 6 \times 14.5 = 87 ✓ (since 6 \times 14 = 84 and 6 \times 0.5 = 3). Estimate: 90 \div 6 = 15, so a side a little under 15 in makes sense ✓.

Tip: Working backwards from the perimeter of a regular polygon: divide by the number of sides. Hexagon → 6, not 5 or 8.

1. Draw lines from the center to each corner: this makes 5 identical triangles.
2. Each triangle has base 10 cm (one side) and height 6.9 cm (the apothem).
3. One triangle: \tfrac{1}{2}(10)(6.9) = 34.5\text{ cm}^2.
4. Five triangles: 5 \times 34.5 = 172.5\text{ cm}^2.

Same thing with one formula: A = \tfrac{1}{2} \times \text{apothem} \times \text{perimeter} = \tfrac{1}{2}(6.9)(50) = 172.5.

The area is 172.5 cm².

Check: a circle with radius about 7 would have area about 3.14 \times 49 \approx 154; the pentagon reaches a bit beyond the apothem at its corners, so a value a little above that is reasonable ✓.

Tip: This formula is not on the GED formula sheet — but you never need it: split a regular polygon into triangles from the center and use \tfrac{1}{2}bh.

1. Base: the bottom side runs from x = -4 to x = 2 at y = -2, so b = 2 - (-4) = 6 units.
2. Height: the top side is at y = 3, so h = 3 - (-2) = 5 units.
3. Area: A = bh = 6 \times 5 = 30.

The area is 30 square units.

Check: the slanted side goes 2 right and 5 up, so it is longer than 5; using it would give more than 30. The height must be the straight vertical distance, 5 ✓.

Tip: On a grid, choose a horizontal side as the base; then the height is just the difference in y-values. No need to measure slanted sides.

1. Use A = \tfrac{1}{2}h(b_1 + b_2): 165 = \tfrac{1}{2}h(11 + 19).
2. Add the bases: 165 = \tfrac{1}{2}h(30).
3. Simplify: \tfrac{1}{2} \times 30 = 15, so 165 = 15h.
4. Divide: h = 165 \div 15 = 11.

The height is 11 ft.

Check: \tfrac{1}{2}(11)(11 + 19) = \tfrac{1}{2}(11)(30) = 165 ✓.

Tip: Combine the known numbers first (\tfrac{1}{2} \times 30 = 15), leaving a one-step equation. Forgetting the \tfrac{1}{2} gives 165 \div 30 = 5.5.

1. Missing horizontal side: the top pieces must equal the bottom, 20 m. The top is 12 m, so the step is 20 - 12 = 8 m.
2. Missing vertical side: the right-side pieces must equal the left, 14 m. The lower right is 8 m, so the step is 14 - 8 = 6 m.
3. Add all six sides: 20 + 8 + 8 + 6 + 12 + 14 = 68 m.

The perimeter is 68 m.

Check: for any outline with only right-angle corners like this one, the perimeter equals that of the surrounding rectangle: 2(20) + 2(14) = 68 ✓.

Tip: Find every missing side before adding. Adding only the labeled sides (20 + 8 + 12 + 14 = 54) is the classic trap.

1. One triangle has base b = 10 in (a side) and height h = 8.66 in (the apothem).
2. Area of one triangle: \tfrac{1}{2}(10)(8.66) = \tfrac{1}{2}(86.6) = 43.3 in².
3. The hexagon is 6 such triangles: 6 \times 43.3 = 259.8 in².

The area is 259.8 in².

Check: with A = \tfrac{1}{2}aP: P = 6 \times 10 = 60, so A = \tfrac{1}{2}(8.66)(60) = 259.8 ✓.

Tip: You don't need a special hexagon formula. Split any regular polygon into triangles from the center, find one triangle with \tfrac{1}{2}bh, and multiply by the number of sides.

1. Area of a parallelogram: A = bh = 60 \times 35 = 2{,}100\text{ ft}^2. (The 40 ft slanted side is not the height.)
2. Cost: 2{,}100 \times \$0.35 = \$735.

The sod costs $735.

Check: 2{,}000 \times 0.35 = 700 and 100 \times 0.35 = 35, so 700 + 35 = 735 ✓.

Tip: In cost problems, first decide what you are buying for: sod/paint/flooring → area; fence/trim → perimeter. Then use the correct height (perpendicular, not slanted).

1. Lines from the center to the six corners split the hexagon into 6 identical triangles.
2. Each triangle: base 6 ft (one side), height 5.2 ft (the apothem).
3. One triangle: \tfrac{1}{2}(6)(5.2) = 15.6\text{ ft}^2.
4. Six triangles: 6 \times 15.6 = 93.6\text{ ft}^2.

Same result: \tfrac{1}{2} \times \text{apothem} \times \text{perimeter} = \tfrac{1}{2}(5.2)(36) = 93.6.

The area is about 93.6 ft².

Check: a circle of radius 6 (through the corners) would have area 3.14 \times 36 \approx 113; the hexagon fits inside that circle, so it must be smaller — 93.6 < 113 ✓.

Tip: Regular polygon area = n triangles × \tfrac{1}{2} × side × apothem. This is not on the formula sheet, but the triangle method always works.

1. Parallelogram area: A = bh, using the perpendicular height (80 ft), not the slanted side.
2. A = 120 \times 80 = 9{,}600 ft².
3. Buckets: 9{,}600 \div 2{,}500 = 3.84.
4. Buckets are sold whole, and 3 buckets cover only 7,500 ft², so round up to 4.
5. Cost: 4 \times 89 = 356.

The sealant costs $356.

Check: 4 buckets cover 4 \times 2{,}500 = 10{,}000 \ge 9{,}600 ft² ✓; 3 buckets (7{,}500) would not be enough ✓.

Tip: In "buy enough" problems, round up to a whole unit before multiplying by the price. Paying for 3.84 buckets isn't possible, and buying 3 leaves part of the lot bare.

1. Let the height be h. Then the base is 2h.
2. Area: A = bh = (2h)(h) = 2h^2.
3. Set equal to 98: 2h^2 = 98, so h^2 = 49.
4. h = \sqrt{49} = 7 in.
5. Base: 2h = 14 in.

The base is 14 inches.

Check: 14 \times 7 = 98 ✓, and 14 is twice 7 ✓. Estimate: a square with area 98 would have sides near 10; a base longer than 10 and a height shorter than 10 fits a figure twice as long as it is tall ✓.

Tip: Do not stop at h^2 = 49 or h = 7 — re-read what the question asks. Here it wants the base.

1. One paver: 6 triangles, each \tfrac{1}{2}(6)(5.2) = 15.6\text{ in}^2, so 6 \times 15.6 = 93.6\text{ in}^2.
2. Patio in square inches: 65 \times 144 = 9{,}360\text{ in}^2.
3. Pavers: 9{,}360 \div 93.6 = 100.

100 pavers are needed.

Check: 100 \times 93.6 = 9{,}360\text{ in}^2, and 9{,}360 \div 144 = 65\text{ ft}^2 ✓.

Tip: Square units convert with the square of the length factor: 1\text{ ft}^2 = 12^2 = 144\text{ in}^2, not 12. Get both areas in the same unit before dividing.

1. Base AB: both points have y = 1, so AB = 2 - (-3) = 5 units.
2. Side DC is parallel to AB and passes through C(4, -3), so it lies on y = -3. (In fact D = (-1, -3).)
3. Height: the vertical distance between y = 1 and y = -3 is 1 - (-3) = 4 units.
4. Area: A = bh = 5 \times 4 = 20.

The area is 20 square units.

Check: DC must also be 5 long: from x = -1 to x = 4 is 5 ✓. Slanted side BC goes 2 right and 4 down, so it is longer than 4 — not the height ✓.

Tip: You do not need the fourth vertex to find the area: a horizontal base and the vertical distance to the opposite side are enough.

1. Perimeter: a regular octagon has 8 sides, so P = 8 \times 15 = 120 in.
2. Convert to feet: 120 \div 12 = 10 ft.
3. Cost: 10 \times \$1.25 = \$12.50.

The trim costs $12.50.

Check: each side is 15\text{ in} = 1.25 ft, so 8 \times 1.25 = 10 ft, and 10 \times \$1.25 = \$12.50 ✓.

Tip: When the price is "per foot" but the sides are in inches, convert before you multiply by the price. Pricing 120 inches at $1.25 each would cost $150 — a sign that units were skipped.