Appendix A — Answer Key, Chapter 13: Area and Perimeter — Triangles and Parallelograms

SOL 6.MG.2 · Covers textbook Chapter 13 and the companion workbook. Item numbers match the textbook; where the workbook repeats the same problems, this key serves both, and workbook-only items are listed at the end. Reasoning answers show an acceptable response, not the only wording. Every area is reported in square units and every perimeter in linear units; answers without units should be treated as incomplete.

Lesson 13.1 — Perimeter of Triangles and Parallelograms

Guided practice

1. $P = 5 + 8 + 11 = 24$ in
2. $P = 2(10 + 4) = 28$ cm
3. $P = 3(9) = 27$ ft
4. $P = 4(13) = 52$ m
5. Perimeter is the distance around the outside, so it adds only the lengths of the sides you would walk along. The height is a measurement across the inside of the figure and is not a side at all.

Independent practice

6. a) $6 + 6 + 9 = 21$ cm b) $12.5 + 9 + 10.5 = 32$ m c) $15 + 20 + 25 = 60$ in
7. a) $2(7 + 5) = 24$ ft b) $2(20 + 11) = 62$ cm c) $2(6.5 + 3.5) = 20$ m
8. $2(15 + s) = 48$, so $15 + s = 24$ and $s = 9$ in. Check: $2(15 + 9) = 48$ in.
9. $9 + 12 = 21$, and $31 - 21 = 10$ cm. Check: $9 + 12 + 10 = 31$ cm.
10. $P = 2(9 + 5) = 28$ m. The height of 4 m was not used, because the height is not a side.
11. $P = 18 + 24 + 30 = 72$ ft. Fenced length: $72 - 3 = 69$ ft. Cost: $69 \times 4 = \$276$.
12. No. The first has $P = 2(10 + 6) = 32$ cm and the second has $P = 2(10 + 8) = 36$ cm. Sharing a base is not enough — perimeter depends on both side lengths, so the 2 cm difference in the adjacent side produces a 4 cm difference in perimeter.

Exit ticket 13.1

1. $7 + 10 + 13 = 30$ m
2. $2(11 + 6) = 34$ in
3. $2(12 + s) = 40$, so $12 + s = 20$ and $s = 8$ ft. Check: $2(12 + 8) = 40$ ft.
4. Perimeter adds side lengths only, and the height is not a side. It is the perpendicular distance across the inside of the figure, so it never appears in a perimeter calculation.

Lesson 13.2 — Developing the Area Formula for Parallelograms

Guided practice

1. $A = 7 \times 3 = 21$ cm²
2. $A = 15 \times 8 = 120$ in²
3. $A = 9 \times 4 = 36$ m². The adjacent side of 6 m was not used, because area needs the perpendicular height.
4. $8h = 40$, so $h = 5$ ft. Check: $8 \times 5 = 40$ ft².
5. Cutting and sliding moves a piece but does not add or remove any surface. The two pieces still cover exactly the same amount of space, so the area is unchanged — and the resulting rectangle has the same base and the same height as the original parallelogram.

Independent practice

6. a) $12 \times 5 = 60$ cm² b) $6.5 \times 4 = 26$ m² c) $20 \times 11 = 220$ in² d) $9 \times 9 = 81$ ft²
7. a) $8h = 72$, so $h = 9$ cm b) $6b = 90$, so $b = 15$ m
8. $A = 5 \times 3 = 15$ square units. Cutting along the height removes a right triangle from one end; sliding it to the other end produces a 5-by-3 rectangle whose 15 unit squares can all be counted whole, with no partial squares left over.
9. The areas are the same: both are $10 \times 4 = 40$ cm², because area depends only on the base and the height. The perimeters are different, because perimeter depends on the slanted side, and those lengths differ.
10. The height, the slanted side, and a piece of the base form a right triangle in which the slanted side is the longest edge. The height is the straight-across, perpendicular path between the two parallel sides, and no slanted path between them can be shorter than that.
11. $A = 9 \times 18 = 162$ ft². $162 \div 50 = 3.24$, so $3$ bottles are not enough and $4$ whole bottles are needed.
12. The 5 ft measurement is the slanted side, not the perpendicular height, and the slanted side is always longer than the height — so $10 \times 5$ must overstate the area before any computation is done. The correct area is $A = 10 \times 3 = 30$ ft².

Exit ticket 13.2

1. $A = 14 \times 6 = 84$ in²
2. $7h = 63$, so $h = 9$ m. Check: $7 \times 9 = 63$ m².
3. The 5 cm dashed segment is the height, because it meets the base at a right angle. The 8 cm measurement is a side.
4. Cut a right triangle off one end along the height and slide it to the other end. The result is a rectangle with the same base and the same height, and moving a piece does not change the amount of surface covered. Since the rectangle's area is $b \times h$, the parallelogram's area is $bh$ as well.

Lesson 13.3 — Developing the Area Formula for Triangles

Guided practice

1. $A = \tfrac{1}{2}(12)(5) = 30$ cm²
2. $A = \tfrac{1}{2}(9)(4) = 18$ in²
3. $A = \tfrac{1}{2}(6)(8) = 24$ m². The hypotenuse of 10 m was not used.
4. $\tfrac{1}{2}(10)h = 30$, so $5h = 30$ and $h = 6$ ft. Check: $\tfrac{1}{2}(10)(6) = 30$ ft².
5. Two congruent copies of a triangle fit together to form a parallelogram with the same base and height. The parallelogram covers $bh$, and the triangle is one of its two equal halves, so it covers $\tfrac{1}{2}bh$.

Independent practice

6. a) $\tfrac{1}{2}(16)(5) = 40$ cm² b) $\tfrac{1}{2}(7)(6) = 21$ m² c) $\tfrac{1}{2}(15)(8) = 60$ in² d) $\tfrac{1}{2}(9)(4) = 18$ ft²
7. a) $2 \times 48 = 96$ and $96 \div 8 = 12$ cm b) $2 \times 35 = 70$ and $70 \div 10 = 7$ m
8. $A = \tfrac{1}{2}(6)(9) = 27$ in². The height falling outside the triangle does not change the formula.
9. Triangle: $\tfrac{1}{2}(10)(6) = 30$ cm². Parallelogram: $10 \times 6 = 60$ cm². The triangle's area is exactly half the parallelogram's.
10. $A = \tfrac{1}{2}(5)(12) = 30$ ft² and $P = 5 + 12 + 13 = 30$ ft. The numbers match by coincidence. The area counts square units of surface and is written ft², while the perimeter counts a distance along the edges and is written ft — the two quantities measure different things and are never interchangeable.
11. $A = \tfrac{1}{2}(8)(15) = 60$ ft². Cost: $60 \times 9 = \$540$.
12. The student used the hypotenuse as the height. In a right triangle the two legs are perpendicular to each other, so the legs are the base and the height; the hypotenuse is the longest side and is never the height. The correct area is $\tfrac{1}{2}(5)(12) = 30$ ft².

Exit ticket 13.3

1. $A = \tfrac{1}{2}(14)(6) = 42$ m²
2. $2 \times 54 = 108$ and $108 \div 12 = 9$ in. Check: $\tfrac{1}{2}(12)(9) = 54$ in².
3. $A = \tfrac{1}{2}(3)(8) = 12$ cm²
4. Two congruent copies of the triangle join to form a parallelogram with that same base and height. The triangle is one of the two equal halves, so its area is half of $bh$.

Lesson 13.4 — Area and Perimeter Problems in Context

Guided practice

1. $A = \tfrac{1}{2}(10)(6) = 30$ ft²; $30 \div 10 = 3$ bags
2. $A = 8 \times 5 = 40$ ft²
3. $P = 2(24 + 15) = 78$ in of trim
4. $A = \tfrac{1}{2}(18)(12) = 108$ in²
5. Perimeter. Edging goes around the outside border of the bed, so the gardener needs a distance around, not an amount of surface.

Independent practice

6. $A = 14 \times 9 = 126$ m²; $P = 2(14 + 10) = 48$ m
7. One banner: $\tfrac{1}{2}(4)(3) = 6$ ft². Five banners: $5 \times 6 = 30$ ft².
8. $2 \times 84 = 168$ and $168 \div 12 = 14$ ft. Check: $\tfrac{1}{2}(12)(14) = 84$ ft².
9. $A = 60 \times 40 = 2{,}400$ yd²; fencing $P = 2(60 + 45) = 210$ yd
10. Triangle A: $\tfrac{1}{2}(9)(8) = 36$ in². Triangle B: $\tfrac{1}{2}(12)(6) = 36$ in². The two areas are equal, even though the triangles have different shapes.
11. $A = 12 \times 8 = 96$ ft². $96 \div 80 = 1.2$, so $2$ whole cans are needed and the cost is $2 \times 22 = \$44$. Rounding up is required because one can covers only 80 ft², leaving 16 ft² unpainted, and paint is sold only in whole cans.
12. Fabric is an area: $A = 5 \times 3 = 15$ ft². Ribbon is a perimeter: $P = 2(5 + 3.5) = 17$ ft. The height is used for the fabric because area is the surface between the base and the opposite side, measured straight across. It is not used for the ribbon because the ribbon runs along the edges, and the height is not an edge.

Exit ticket 13.4

1. $A = \tfrac{1}{2}(6)(4) = 12$ ft²
2. $P = 2(11 + 7) = 36$ m
3. $12h = 84$, so $h = 7$ in. Check: $12 \times 7 = 84$ in².
4. Ask whether the question is about the inside of the shape or its outline. Covering, filling, painting, or buying fabric is area, reported in square units. Fencing, framing, trimming, or walking around is perimeter, reported in linear units.

Chapter 13 Review

Part A — Developing the area formulas (6.MG.2a)

1. Draw the height of the parallelogram, which cuts a right triangle off one end. Cut along it and slide that triangle to the opposite end, where it fits against the matching slanted edge. The result is a rectangle. The base stays the same, the height stays the same, and the area stays the same, because a piece was moved rather than added or removed. Since the rectangle's area is $b \times h$, the parallelogram's is $bh$.
2. Make a congruent copy of the triangle, rotate it a half turn, and join it to the original along a side. The two triangles exactly fill a parallelogram with the same base and height as the triangle. That parallelogram's area is $bh$, and the triangle is one of two congruent halves, so its area is $\tfrac{1}{2}bh$.
3. $A = 7 \times 4 = 28$ square units. Cutting along the height and sliding the right triangle to the other end makes a 7-by-4 rectangle, so all 28 squares can be counted whole and no partial squares remain.
4. The height is the 6 cm dashed segment, because it meets the base at a right angle. $A = 10 \times 6 = 60$ cm². The 8 cm slanted side is not used.

Part B — Perimeter (6.MG.2b)

5. $P = 13 + 14 + 15 = 42$ in
6. $P = 2(16 + 9) = 50$ cm
7. $2(20 + s) = 54$, so $20 + s = 27$ and $s = 7$ ft. Check: $2(20 + 7) = 54$ ft.
8. $12 + 15 = 27$, and $40 - 27 = 13$ m. Check: $12 + 15 + 13 = 40$ m.

Part C — Area (6.MG.2b)

9. $A = 18 \times 7 = 126$ m²
10. $A = \tfrac{1}{2}(20)(9) = 90$ cm²
11. $8b = 96$, so $b = 12$ in. Check: $12 \times 8 = 96$ in².
12. $2 \times 45 = 90$ and $90 \div 15 = 6$ ft. Check: $\tfrac{1}{2}(15)(6) = 45$ ft².

Part D — Problems in context (6.MG.2b)

13. $A = \tfrac{1}{2}(12)(7) = 42$ ft². $42 \div 15 = 2.8$, so $3$ whole bags are needed.
14. Decking: $A = 20 \times 12 = 240$ ft². Railing: $P = 2(20 + 13) = 66$ ft. The height of 12 ft is used for the area and the side of 13 ft for the perimeter.
15. $A = \tfrac{1}{2}(9)(16) = 72$ ft². Cost: $72 \times 12 = \$864$.
16. Parallelogram: $10 \times 6 = 60$ m². Triangle: $\tfrac{1}{2}(15)(8) = 60$ m². The two areas are equal, so neither sign has more surface to paint.

Part E — Reasoning

17. Area measures the surface between the base and the side opposite it, and that distance must be measured straight across at a right angle to the base. The slanted side is a longer, tilted path between the same two parallel lines, so using it produces a product larger than the actual surface covered.
18. Yes. Two parallelograms with base 10 cm and height 4 cm both have area 40 cm², but if one has an adjacent side of 5 cm and the other has 7 cm, their perimeters are $2(10 + 5) = 30$ cm and $2(10 + 7) = 34$ cm. Base and height fix the area but do not fix the slanted side, and the slanted side is what perimeter depends on.
19. The triangle's area is exactly half the parallelogram's. Two congruent copies of the triangle fit together to fill that parallelogram, so the parallelogram covers $bh$ and each triangle covers $\tfrac{1}{2}bh$.
20. Area counts how many unit squares fit inside a figure, and each of those units is a square, so the label is square units. Perimeter is a single distance measured along a path, so its label is a plain linear unit. Reporting one with the other's units would describe the wrong kind of quantity.

Workbook-only items

Page 2, perimeter table.

Page 3, missing sides. 1. $9$ in; check $2(15 + 9) = 48$ in. 2. $10$ cm; check $9 + 12 + 10 = 31$ cm. 3. $8$ ft; check $2(12 + 8) = 40$ ft.

Page 3, cross out. Perimeter $= 2(9 + 5) = 28$ m. Unused measurement: the height, 4 m. It is not a side of the figure, so it is never added into a perimeter.

Page 5, rectangle table. $6 \times 4 = 24$ square units; $9 \times 3 = 27$ square units; $12 \times 5 = 60$ square units.

Page 5, rectangle formula. $A = b \times h$ (base times height).

Page 6, cut and slide. You get a rectangle. The base does not change, the height does not change, and the area does not change. So $A = bh$. Moving a piece leaves the area unchanged because no surface is added or removed — the same amount of material is simply arranged differently.

Page 7, areas.

Page 7, missing measurements. Height $= 72 \div 8 = 9$ cm. Base $= 90 \div 6 = 15$ m. Height $= 63 \div 7 = 9$ m.

Page 7, explain. The height is the perpendicular, straight-across distance between the two parallel sides. The slanted side is a tilted path between those same two lines and forms the longest edge of a right triangle whose leg is the height, so it must be longer.

Page 9, two triangles. The two triangles make a parallelogram. Its area is $bh$, and one triangle is half of it. So $A = \tfrac{1}{2}bh$.

Page 9, triangle area table.

Page 10, right triangle table.

Page 10, obtuse triangle. $A = \tfrac{1}{2}(6)(9) = 27$ in².

Page 10, missing measurements. Base $= (2 \times 48) \div 8 = 12$ cm. Height $= (2 \times 35) \div 10 = 7$ m. Height $= (2 \times 30) \div 10 = 6$ ft.

Page 10, error hunt. The student used the hypotenuse, 13 ft, as the height. In a right triangle the two legs are perpendicular, so they are the base and the height, and the hypotenuse is never the height. Correct area: $\tfrac{1}{2}(5)(12) = 30$ ft².

Page 12, area or perimeter. Sod for a lawn: area. Fencing for a dog run: perimeter. Ribbon around a banner: perimeter. Paint for a wall panel: area. Trim around a sign: perimeter. Pavers for a patio: area.

Page 12, rounding rule. $96 \div 80 = 1.2$ cans, so you must buy $2$ whole cans. One can covers only 80 ft², which leaves 16 ft² unpainted, and paint is sold in whole cans — so any leftover fraction forces you up to the next whole can.

Page 13, problems in context.

1. Area $= \tfrac{1}{2}(10)(6) = 30$ ft²; bags $= 30 \div 10 = 3$
2. Area $= 14 \times 9 = 126$ m²; edging $= 2(14 + 10) = 48$ m
3. Area $= \tfrac{1}{2}(8)(15) = 60$ ft²; cost $= 60 \times 9 = \$540$
4. Height $= (2 \times 84) \div 12 = 14$ ft; check $\tfrac{1}{2}(12)(14) = 84$ ft²
5. Area $= 60 \times 40 = 2{,}400$ yd²; fencing $= 2(60 + 45) = 210$ yd
6. Perimeter $= 18 + 24 + 30 = 72$ ft; length fenced $= 72 - 3 = 69$ ft; cost $= 69 \times 4 = \$276$

Page 14, which uses less fabric. Area A $= \tfrac{1}{2}(9)(8) = 36$ in². Area B $= \tfrac{1}{2}(12)(6) = 36$ in². Neither uses less — the two designs use the same amount of fabric.

Page 14, same base and height. Triangle $= \tfrac{1}{2}(10)(6) = 30$ cm². Parallelogram $= 10 \times 6 = 60$ cm². The triangle's area is exactly half the parallelogram's.

Page 14, same base and height, different perimeters. Areas: $40$ cm² and $40$ cm². Perimeters: $2(10 + 5) = 30$ cm and $2(10 + 7) = 34$ cm. Equal areas do not force equal perimeters: area depends on the base and height, while perimeter depends on the base and the slanted side.

Page 14, fabric and ribbon. Fabric $= 5 \times 3 = 15$ ft². Ribbon $= 2(5 + 3.5) = 17$ ft. The height is used for the fabric because area measures the surface straight across between the parallel sides. It is not used for the ribbon because the ribbon runs along the edges, and the height is not an edge.

Pages 16–18, Chapter 13 review. Same answers as the textbook Chapter 13 Review above.

Page 18, formula summary.