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LLemma 34B on 4.MD.A.3

Common Core Standard 4.MD.A.3

This standard covers the following:

Apply the area and perimeter formulas for rectangles in real world and mathematical problems.

🔗 Problem

Problem: A rectangle has a length of 21m and a width of 37m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 21+21+37+37 = <<21+21+37+37=126>>126m.
### Answer: 126m

Expected answer: 116

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 50 feet and a width of 20 feet. What is the perimeter of the garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the garden is 2*50 + 2*20 = <<2*50+2*20=140>>140 feet.
### Answer: 140

Expected answer: 140

Correct? Yes

🔗 Follow-up

Problem: A landscaper plans to reduce the length of the rectangular garden to 15 feet while keeping the width the same. What will the new perimeter of the garden be?

LLemma 34B:

 Reasoning: The new perimeter of the garden is 2*15 + 2*20 = <<2*15+2*20=60>>60 feet.
### Answer: 60

Expected answer: 70

Correct? No

🔗 Problem

Problem: A rectangle has a width of 7m and a length of 26m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The area of this rectangle is 7*26 = <<7*26=182>>182 m^2.
### Answer: 182

Expected answer: 182

Correct? Yes

🔗 Follow-up

Problem: The width of the rectangle stays the same at 7m, but the length extends by 9m to become 35m. What is the area of the rectangle now?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The area of the rectangle is 7*35 = <<7*35=245>>245 m^2.
### Answer: 245

Expected answer: 245

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 50m and a width of 15m. What is its perimeter?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of its length and its width.
The perimeter of the rectangle is 50+15 = <<50+15=65>>65m.
### Answer: 65

Expected answer: 130

Correct? No

🔗 Problem

Problem: A rectangular fence has a side length of 5m and another side of 19m. What is the total distance, in meters, around the fence (perimeter)?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this rectangle is 2*5 + 2*19 = <<2*5+2*19=44>>44 meters.
### Answer: 44

Expected answer: 48

Correct? No

🔗 Problem

Problem: A rectangle has sides of length 49m and 6m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 49m+6m+49m+6m = <<49+6+49+6=100>>100m.
### Answer: 100m

Expected answer: 110

Correct? No

🔗 Problem

Problem: Joshua plans to lay a wire around his rectangular garden. If the garden is 7m wide and 26m long, how much wire will Joshua need to enclose the garden completely?

LLemma 34B:

 Reasoning: The garden is 7m wide and 26m long.
The garden is 2(7)+2(26) = <<2*7+2*26=54+52=106>>106m long.
Joshua will need 106m of wire to enclose the garden completely.
### Answer: 106

Expected answer: 66

Correct? No

🔗 Problem

Problem: A rectangle has a length of 18 meters and a width of 11 meters. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 18+11+18+11 = <<18+11+18+11=58>>58 meters.
### Answer: 58

Expected answer: 58

Correct? Yes

🔗 Follow-up

Problem: Suppose the width of the rectangle changed to 44 meters, while the length remained at 18 meters. What would be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The new perimeter of the rectangle is 18+44+18+44 = <<18+44+18+44=114>>114 meters.
### Answer: 114

Expected answer: 124

Correct? No

🔗 Problem

Problem: The dimensions of a rectangle are 22 feet by 39 feet. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 22+39+22+39 = <<22+39+22+39=122>>122 feet.
### Answer: 122

Expected answer: 122

Correct? Yes

🔗 Follow-up

Problem: If the length of the rectangle is now 26 feet instead of 39 feet, while the width remains the same at 22 feet, what would then be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 26+22+26+22 = <<26+22+26+22=96>>96 feet.
### Answer: 96

Expected answer: 96

Correct? Yes

🔗 Problem

Problem: A rectangle has two sides measuring 45m and the other two sides measuring 25m each. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 45+45+25+25 = <<45+45+25+25=130>>130m.
### Answer: 130

Expected answer: 140

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 8m and a width of 5m. What is the perimeter of this garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this garden is 2*8+2*5 = <<2*8+2*5=26>>26m.
### Answer: 26m

Expected answer: 26

Correct? Yes

🔗 Follow-up

Problem: The owner of the rectangular garden decided to extend its length to 46m while keeping the width the same, i.e., 5m. What is the new perimeter of the garden now?

LLemma 34B:

 Reasoning: The new perimeter of the garden is 2*46+2*5 = <<2*46+2*5=102>>102m.
### Answer: 102m

Expected answer: 102

Correct? Yes

🔗 Problem

Problem: In a rectangle, the length is 19 units and the width is 17 units. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 19*17 = <<19*17=323>>323 square units.
### Answer: 323

Expected answer: 323

Correct? Yes

🔗 Follow-up

Problem: In the previous rectangle problem, what if the length was 22 units instead of 19 units while the width remained the same at 17 units? What would be the new area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 22*17 = <<22*17=374>>374 square units.
### Answer: 374

Expected answer: 374

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 42 meters and a width of 8 meters. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 42+8+42+8 = <<42+8+42+8=96>>96 meters.
### Answer: 96

Expected answer: 100

Correct? No

🔗 Problem

Problem: A rectangle has a width of 16m and a length of 38m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The area of this rectangle is 16*38 = <<16*38=608>>608 square meters.
### Answer: 608

Expected answer: 608

Correct? Yes

🔗 Follow-up

Problem: In the rectangle from the previous problem, suppose we replace the length of 38m with a new length of 22m. What would the new area of the rectangle be?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The new area of the rectangle is 16*22 = <<16*22=352>>352 square meters.
### Answer: 352

Expected answer: 352

Correct? Yes

🔗 Problem

Problem: Consider a rectangular plot of land. The plot has a length of 28m and a width of 17m. What is the area of the plot?

LLemma 34B:

 Reasoning: The area of the plot is 28*17 = <<28*17=476>>476 square meters.
### Answer: 476

Expected answer: 476

Correct? Yes

🔗 Follow-up

Problem: Suppose instead of the original 28m, the length of the rectangular plot of land were only 10m while keeping the width the same at 17m. What would then be the area of the rectangular plot of land?

LLemma 34B:

 Reasoning: The area of the plot is 10*17 = <<10*17=170>>170 square meters.
### Answer: 170

Expected answer: 170

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 43m and a width of 41m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 43+41+43+41 = <<43+41+43+41=168>>168m.
### Answer: 168m

Expected answer: 168

Correct? Yes

🔗 Follow-up

Problem: Suppose the rectangle from the previous problem had its length reduced to 18m while keeping the width the same. What would be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The new perimeter of the rectangle is 18+41+18+41 = <<18+41+18+41=110>>110m.
### Answer: 110m

Expected answer: 118

Correct? No

🔗 Problem

Problem: A rectangle has one side that measures 44m, and the other side that measures 17m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 44+17+44+17 = <<44+17+44+17=112>>112m.
### Answer: 112

Expected answer: 122

Correct? No

🔗 Problem

Problem: A rectangle's perimeter is twice the sum of its width and length. If the rectangle's width is 2m and its length is 20m, what is its perimeter?

LLemma 34B:

 Reasoning: The rectangle's perimeter is twice the sum of its width and length.
The rectangle's perimeter is 2*(2+20) = <<2*(2+20)=44>>44m.
### Answer: 44m

Expected answer: 44

Correct? Yes

🔗 Follow-up

Problem: If the width of the rectangle is increased from 2m to 32m, while keeping the length the same at 20m, what is the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The rectangle's perimeter is twice the sum of its width and length.
The rectangle's perimeter is 2*(32+20) = <<2*(32+20)=84>>84m.
### Answer: 84m

Expected answer: 104

Correct? No

🔗 Problem

Problem: A rectangle has a length of 4 meters and a width of 5 meters. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 4+5+4+5 = <<4+5+4+5=18>>18 meters.
### Answer: 18

Expected answer: 18

Correct? Yes

🔗 Follow-up

Problem: Consider the same rectangle from before, but now suppose its length is extended to 42 meters while the width remains 5 meters. What would then be the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 42+5+42+5 = <<42+5+42+5=94>>94 meters.
### Answer: 94

Expected answer: 94

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 45m and a width of 37m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 45+37+45+37 = <<45+37+45+37=164>>164m.
### Answer: 164

Expected answer: 164

Correct? Yes

🔗 Follow-up

Problem: If the length of the rectangle remains the same at 45m, but the width is increased to 47m, what would be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 45+47+45+47 = <<45+47+45+47=184>>184m.
### Answer: 184

Expected answer: 184

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 27m and a width of 21m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 27+21+27+21 = <<27+21+27+21=96>>96m.
### Answer: 96

Expected answer: 96

Correct? Yes

🔗 Follow-up

Problem: Suppose we extended the length of the rectangle from the previous problem to 45m, while keeping the width at 21m. What would the new perimeter of this rectangle be?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The new perimeter of this rectangle is 45+21+45+21 = <<45+21+45+21=132>>132m.
### Answer: 132

Expected answer: 132

Correct? Yes

🔗 Problem

Problem: A rectangular garden has a length of 30m and a width of 20m. What is the perimeter of the garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the garden is 2*30 + 2*20 = <<2*30+2*20=100>>100m.
### Answer: 100m

Expected answer: 100

Correct? Yes

🔗 Follow-up

Problem: Suppose the width of the rectangular garden is increased to 48m, while the length remains the same at 30m. What would be the new perimeter of the garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The new perimeter of the garden is 2*30 + 2*48 = <<2*30+2*48=140>>140m.
### Answer: 140m

Expected answer: 156

Correct? No

🔗 Problem

Problem: A rectangle has a length of 30m and a width of 27m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 30+30+27+27 = <<30+30+27+27=114>>114m.
### Answer: 114

Expected answer: 114

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 23m and a width of 11m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 23*11 = <<23*11=253>>253 square meters.
### Answer: 253

Expected answer: 253

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 3m and a width of 9m. What would be the total perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this rectangle is 2*3 + 2*9 = <<2*3+2*9=26>>26 m.
### Answer: 26 m

Expected answer: 24

Correct? No

🔗 Problem

Problem: A rectangle has a length of 31m and a width of 6m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 2(31+6) = <<2*(31+6)=2*(37)=74>>74m.
### Answer: 74

Expected answer: 74

Correct? Yes

🔗 Follow-up

Problem: Now consider, if instead of a length of 31m, the rectangle had a length of 43m but the width remained the same at 6m. What would be the new perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 2(43+6) = <<2*(43+6)=2*(49)=98>>98m.
### Answer: 98

Expected answer: 98

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 12m and a width of 33m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 12*33 = <<12*33=396>>396 m^2.
### Answer: 396 m^2

Expected answer: 396

Correct? Yes

🔗 Follow-up

Problem: Now, consider that the length of the rectangle was initially 18m instead of 12m, while the width remains the same at 33m. What would be the area of the rectangle in this scenario?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 18*33 = <<18*33=594>>594 m^2.
### Answer: 594 m^2

Expected answer: 594

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 20m and all of its sides are of equal length. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this rectangle is 2*20 + 2*20 = <<2*20+2*20=80>>80m.
### Answer: 80

Expected answer: 80

Correct? Yes

🔗 Problem

Problem: A rectangular playground has a length of 35m and a width of 9m. What is the perimeter of the playground?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the playground is 2*35+2*9 = <<2*35+2*9=88>>88m.
### Answer: 88

Expected answer: 88

Correct? Yes

🔗 Follow-up

Problem: Suppose the length of the rectangular playground was initially stated incorrectly and it's actually 26m, while the width remains 9m. What would be the correct perimeter of the playground in this case?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the playground is 2*26+2*9 = <<2*26+2*9=68>>68m.
### Answer: 68

Expected answer: 70

Correct? No

🔗 Problem

Problem: A rectangle has a length of 9m and a width of 25m. What is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 9*25 = <<9*25=225>>225 square meters.
### Answer: 225

Expected answer: 225

Correct? Yes

🔗 Follow-up

Problem: Suppose the length of the rectangle was actually 16m and the width remained the same at 9m. What would be the new area of the rectangle in this case?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 16*9 = <<16*9=144>>144 square meters.
### Answer: 144

Expected answer: 144

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 13m and a width of 43m. What is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 13*43 = <<13*43=549>>549 square meters.
### Answer: 549

Expected answer: 559

Correct? No

🔗 Problem

Problem: A rectangle has a width of 27m and a length of 47m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 27+27+47+47 = <<27+27+47+47=140>>140m.
### Answer: 140m

Expected answer: 148

Correct? No

🔗 Problem

Problem: A rectangular football field has an area of 21m * 42m, calculate the area of the football field.

LLemma 34B:

 Reasoning: The area of the football field is 21*42 = <<21*42=882>>882 m^2.
### Answer: 882 m^2

Expected answer: 882

Correct? Yes

🔗 Problem

Problem: A rectangle's area is found by multiplying its length by its width. If the width of a certain rectangle is 27m and the length is 25m, what is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of the rectangle is 27*25 = <<27*25=675>>675m^2.
### Answer: 675m^2

Expected answer: 675

Correct? Yes

🔗 Follow-up

Problem: Suppose the width of the rectangle in the previous problem was not 27m, but was instead 11m, with the length remaining at 25m. What would then be the new area of the rectangle?

LLemma 34B:

 Reasoning: The area of the rectangle is 11*25 = <<11*25=275>>275m^2.
### Answer: 275m^2

Expected answer: 275

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 45m and a width of 48m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 45+45+48+48 = <<45+45+48+48=196>>196m.
### Answer: 196

Expected answer: 186

Correct? No

🔗 Problem

Problem: A rectangle has a length of 11 meters and a width of 7 meters. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 11+11+7+7 = <<11+11+7+7=36>>36 meters.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Follow-up

Problem: Suppose the length of the rectangle was instead 10 meters while the width remained 7 meters. What would be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 10+10+7+7 = <<10+10+7+7=34>>34 meters.
### Answer: 34

Expected answer: 34

Correct? Yes

🔗 Problem

Problem: A rectangular field has a length of 30m and a width of 2m. Can you calculate the perimeter of the field?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the field is 2*30 + 2*2 = <<2*30+2*2=64>>64m.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Follow-up

Problem: As a follow-up to the previous problem, let's say that the length of the rectangular field is reduced to 20m instead of 30m, while the width remains the same. What would the new perimeter of the field be?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the field is 2*20 + 2*2 = <<2*20+2*2=44>>44m.
### Answer: 44

Expected answer: 44

Correct? Yes

🔗 Problem

Problem: A rectangle has one side that is 7m long and another side that is 8m long. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 7+8+7+8 = <<7+8+7+8=22>>22m.
### Answer: 22m

Expected answer: 30

Correct? No

🔗 Problem

Problem: A rectangle has a length of 14m and a width of 18m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 14+14+18+18 = <<14+14+18+18=64>>64m.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Follow-up

Problem: Now let's consider another rectangle. The length of this rectangle is 9m rather than 14m, while the width remains the same at 18m. What is the perimeter of this new rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 9+9+18+18 = <<9+9+18+18=54>>54m.
### Answer: 54

Expected answer: 54

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 43 units and a width of 3 units. Calculate the perimeter of this rectangle.

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 43+43+3+3 = <<43+43+3+3=89>>89 units.
### Answer: 89

Expected answer: 92

Correct? No

🔗 Problem

Problem: A rectangle has one side that measures 36 meters and another side that measures 42 meters. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 36+42 = <<36+42=78>>78 meters.
### Answer: 78

Expected answer: 156

Correct? No

🔗 Problem

Problem: A rectangle has a length of 1m and a width of 6m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 1+1+6+6 = <<1+1+6+6=14>>14m.
### Answer: 14m

Expected answer: 14

Correct? Yes

🔗 Follow-up

Problem: Now consider that the length of the rectangle has changed to 28m, while the width remains at 6m. What is the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 28+28+6+6 = <<28+28+6+6=68>>68m.
### Answer: 68m

Expected answer: 68

Correct? Yes

🔗 Problem

Problem: A rectangle has two sides measuring 45m and two sides measuring 8m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 45+45+8+8 = <<45+45+8+8=104>>104m.
### Answer: 104

Expected answer: 106

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 25 meters and a width of 37 meters. What is the perimeter of the garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the garden is 2*25 + 2*37 = <<2*25+2*37=104>>104 meters.
### Answer: 104

Expected answer: 124

Correct? No

🔗 Problem

Problem: A rectangle has length 36m and width 3m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is twice the sum of the length and the width.
The perimeter of this rectangle is 2(36+3) = <<2*(36+3)=72>>72m.
### Answer: 72m

Expected answer: 78

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 45 meters and a width of 50 meters. Calculate the perimeter of the garden.

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the garden is 2*45 + 2*50 = <<2*45+2*50=130>>130 meters.
### Answer: 130

Expected answer: 190

Correct? No

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 34m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 44+44+34+34 = <<44+44+34+34=112>>112m.
### Answer: 112

Expected answer: 156

Correct? No

🔗 Problem

Problem: A rectangle has a length of 42 m and a width of 29 m. Calculate the perimeter of the rectangle.

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 42+29+42+29 = <<42+29+42+29=132>>132 m.
### Answer: 132

Expected answer: 142

Correct? No

🔗 Problem

Problem: A rectangle has a length of 44 meters and a width of 35 meters. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 44+35+44+35 = <<44+35+44+35=158>>158 meters.
### Answer: 158

Expected answer: 158

Correct? Yes

🔗 Follow-up

Problem: Consider the same rectangle from the previous problem, but now the width has changed to 43 meters while the length remains the same at 44 meters. What is the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 44+43+44+43 = <<44+43+44+43=174>>174 meters.
### Answer: 174

Expected answer: 174

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 45m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 44+45+44+45 = <<44+45+44+45=138>>138m.
### Answer: 138

Expected answer: 178

Correct? No

🔗 Problem

Problem: A rectangle has a length of 3 meters and a width of 39 meters. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 3+3+39+39 = <<3+3+39+39=84>>84 meters.
### Answer: 84

Expected answer: 84

Correct? Yes

🔗 Follow-up

Problem: Suppose the length of the rectangle was actually 17 meters instead of 3 meters, keeping the width the same. What would be the new perimeter of the rectangle now?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 17+17+39+39 = <<17+17+39+39=102>>102 meters.
### Answer: 102

Expected answer: 112

Correct? No

🔗 Problem

Problem: A rectangular garden has sides that measure 32m and 12m. What is the perimeter of this garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this garden is 2*32 + 2*12 = <<2*32+2*12=80>>80 meters.
### Answer: 80

Expected answer: 88

Correct? No

🔗 Problem

Problem: A rectangular swimming pool has a length of 24 feet and a width of 6 feet. What is the perimeter of the swimming pool?

LLemma 34B:

 Reasoning: The perimeter of the swimming pool is 2(24+6) = <<2(24+6)=52>>52 feet.
### Answer: 52

Expected answer: 60

Correct? No

🔗 Problem

Problem: A rectangle has a length of 27m and a width of 21m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 27*21 = <<27*21=547>>547 square meters.
### Answer: 547

Expected answer: 567

Correct? No

🔗 Problem

Problem: A rectangle has twice the length of a square. If the square's side is 49m, what is the length of the rectangle?

LLemma 34B:

 Reasoning: The rectangle's length is 2*49 = <<2*49=98>>98m.
### Answer: 98

Expected answer: 98

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, we found the length of the rectangle when the square's side was 49m. Now, consider a square with a side length of 18m. What would be the length of the rectangle if it's still twice the length of the square?

LLemma 34B:

 Reasoning: The rectangle's length is 2*18 = <<2*18=36>>36m.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 50m and a width of 36m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 50+36+50+36 = <<50+36+50+36=162>>162m.
### Answer: 162m

Expected answer: 172

Correct? No

🔗 Problem

Problem: A rectangle has a length of 8m and a width of 44m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 8+8+44+44 = <<8+8+44+44=96>>96m.
### Answer: 96

Expected answer: 104

Correct? No

🔗 Problem

Problem: If a rectangle has a length of 47 feet and a width of 21 feet, what is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 47*21 = <<47*21=947>>947 square feet.
### Answer: 947

Expected answer: 987

Correct? No

🔗 Problem

Problem: A rectangle has a length of 19m and a breadth of 28m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 19+19+28+28 = <<19+19+28+28=94>>94m.
### Answer: 94m

Expected answer: 94

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, what if the length of the rectangle was 27m instead of 19m while keeping the breadth the same? What would be the new perimeter?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 27+27+28+28 = <<27+27+28+28=100>>100m.
### Answer: 100m

Expected answer: 110

Correct? No

🔗 Problem

Problem: One side of a rectangle is four times the length of the other side. If the longer side measures 39 meters, what is the area of the rectangle when the shorter side is 10 meters?

LLemma 34B:

 Reasoning: The shorter side is 10 meters. The longer side is 4 times the length of the shorter side. So the longer side is 4*10 = <<4*10=40>>40 meters.
The area of the rectangle is 40*10 = <<40*10=400>>400 square meters.
### Answer: 400

Expected answer: 390

Correct? No

🔗 Problem

Problem: A rectangle has identical sides, each measuring 19 meters. What is the total perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of a rectangle is 2*19 + 2*19 = <<2*19+2*19=76>>76 meters.
### Answer: 76

Expected answer: 76

Correct? Yes

🔗 Follow-up

Problem: The rectangle mentioned in the previous problem has been resized. The new rectangle now has a length of 19 meters but a width of 6 meters. What is the new total perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of a rectangle is 2*19 + 2*6 = <<2*19+2*6=50>>50 meters.
### Answer: 50

Expected answer: 50

Correct? Yes

🔗 Problem

Problem: The area of a rectangle is given by the equation length * width. If the length is 9m and the width is 26m, what is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of the rectangle is 9*26 = <<9*26=234>>234 square meters.
### Answer: 234

Expected answer: 234

Correct? Yes

🔗 Follow-up

Problem: The rectangle now has a new dimension with its length increased to 47m while the width is still 9m. What is the new area of the rectangle?

LLemma 34B:

 Reasoning: The new area of the rectangle is 47*9 = <<47*9=423>>423 square meters.
### Answer: 423

Expected answer: 423

Correct? Yes

🔗 Problem

Problem: A rectangle has one side measuring 19cm and another side measuring 7cm. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 19+7+19+7 = <<19+7+19+7=52>>52 cm.
### Answer: 52

Expected answer: 52

Correct? Yes

🔗 Follow-up

Problem: Suppose we extended the rectangle's length from the previous question from 19cm to 21cm. In this case, what would be the new perimeter?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 21+7+21+7 = <<21+7+21+7=56>>56 cm.
### Answer: 56

Expected answer: 56

Correct? Yes

🔗 Problem

Problem: A shutter is in the shape of a rectangle. Its length is 10m and its width is 6m. What is the total area of the shutter?

LLemma 34B:

 Reasoning: The area of a rectangle is length times width.
The area of the shutter is 10*6 = <<10*6=60>>60 m^2.
### Answer: 60 m^2

Expected answer: 60

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 24 units and a width of 15 units. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 24+24+15+15 = <<24+24+15+15=78>>78 units.
### Answer: 78

Expected answer: 78

Correct? Yes

🔗 Follow-up

Problem: Now, imagine the rectangle's width was reduced to 10 units, while the length remained 24 units. What would be the new perimeter of the rectangle under these conditions?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 24+24+10+10 = <<24+24+10+10=68>>68 units.
### Answer: 68

Expected answer: 68

Correct? Yes

🔗 Problem

Problem: If a rectangular plot of land has a length of 32m and a width of 27m, what is the area of the plot in square meters?

LLemma 34B:

 Reasoning: The area of the plot is 32*27 = <<32*27=864>>864 square meters.
### Answer: 864

Expected answer: 864

Correct? Yes

🔗 Follow-up

Problem: In the previous scenario, suppose the width of the rectangular plot of land was actually 18m rather than 27m. What would then be the area of the plot in square meters?

LLemma 34B:

 Reasoning: The area of the plot is 32*18 = <<32*18=576>>576 square meters.
### Answer: 576

Expected answer: 576

Correct? Yes

🔗 Problem

Problem: A rectangle has one side that measures 15m and another side that measures 10m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 15+10+15+10 = <<15+10+15+10=40>>40m.
### Answer: 40m

Expected answer: 50

Correct? No

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 37m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 44+37+44+37 = <<44+37+44+37=162>>162m.
### Answer: 162m

Expected answer: 162

Correct? Yes

🔗 Follow-up

Problem: What would the perimeter of the rectangle be if the length was 25m instead of 44m but the width remained the same at 37m?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 25+37+25+37 = <<25+37+25+37=104>>104m.
### Answer: 104m

Expected answer: 124

Correct? No

🔗 Problem

Problem: A rectangle has length of 14 cm and width 8 cm. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 14+8+14+8 = <<14+8+14+8=44>>44 cm.
### Answer: 44

Expected answer: 44

Correct? Yes

🔗 Follow-up

Problem: If the length of the rectangle remains 14 cm, but the width is increased to 18 cm, what would be the new perimeter of the rectangle now?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 14+18+14+18 = <<14+18+14+18=54>>54 cm.
### Answer: 54

Expected answer: 64

Correct? No

🔗 Problem

Problem: A rectangle has a length of 26m and a width of 28m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 26+26+28+28 = <<26+26+28+28=116>>116m.
### Answer: 116

Expected answer: 108

Correct? No

🔗 Problem

Problem: A rectangle has a length of 15m and a width of 37m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 15+15+37+37 = <<15+15+37+37=104>>104m.
### Answer: 104

Expected answer: 104

Correct? Yes

🔗 Follow-up

Problem: The width of the rectangle mentioned in the previous problem is now changed to 19m, while the length is still 15m. What is the new perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 15+15+19+19 = <<15+15+19+19=68>>68m.
### Answer: 68

Expected answer: 68

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 42m and a width of 40m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this rectangle is 2*42 + 2*40 = <<2*42+2*40=164>>164m.
### Answer: 164m

Expected answer: 164

Correct? Yes

🔗 Follow-up

Problem: The rectangle you were previously given now has a length of 42m and the width has been adjusted to 24m. What is the new perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of this rectangle is 2*42 + 2*24 = <<2*42+2*24=120>>120m.
### Answer: 120m

Expected answer: 132

Correct? No

🔗 Problem

Problem: A rectangular garden has one side that measures 21 feet and another side that measures 30 feet. What is the perimeter of the garden?

LLemma 34B:

 Reasoning: The perimeter of the garden is 21+21+30+30 = <<21+21+30+30=102>>102 feet.
### Answer: 102

Expected answer: 102

Correct? Yes

🔗 Follow-up

Problem: Now, imagine that the side of the garden that initially measured 30 feet was actually only 5 feet long. What would be the new perimeter of the garden?

LLemma 34B:

 Reasoning: The new perimeter of the garden is 21+21+5+5 = <<21+21+5+5=52>>52 feet.
### Answer: 52

Expected answer: 52

Correct? Yes

🔗 Problem

Problem: A certain playground is in the shape of a rectangle. The area of the playground is given as the product of 17 and 44. How many square meters is the playground?

LLemma 34B:

 Reasoning: The area of the playground is 17*44 = <<17*44=776>>776 square meters.
### Answer: 776

Expected answer: 748

Correct? No

🔗 Problem

Problem: Rectangle ABCD has a width of 47m and a length of 50m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of rectangle ABCD is 47+47+50+50 = <<47+47+50+50=194>>194m.
### Answer: 194

Expected answer: 194

Correct? Yes

🔗 Follow-up

Problem: Now consider that the rectangle ABCD's dimensions have changed slightly. The width is still 47m, but the length has adjusted to 48m. What is the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of rectangle ABCD is 47+47+48+48 = <<47+47+48+48=190>>190m.
### Answer: 190

Expected answer: 190

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 11m and a width of 31m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 11+11+31+31 = <<11+11+31+31=84>>84m.
### Answer: 84m

Expected answer: 84

Correct? Yes

🔗 Follow-up

Problem: Suppose we later find out that the actual length of the rectangle is 36m, not 11m as we originally thought, while the width remains the same at 31m. What would then be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 36+36+31+31 = <<36+36+31+31=134>>134m.
### Answer: 134m

Expected answer: 134

Correct? Yes

🔗 Problem

Problem: A rectangular garden has a length of 39m and a width of 19m. What is the area of this garden?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this garden is 39*19 = <<39*19=721>>721 square meters.
### Answer: 721

Expected answer: 741

Correct? No

🔗 Problem

Problem: A carpenter wants to install a wooden floor in a rectangular hallway. The wooden plank has a width of 29 cm and the carpenter has cut it to a length of 11 cm. How many square centimeters of area does one cut wooden plank cover?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The area of one cut wooden plank is 29*11 = <<29*11=319>>319 square centimeters.
### Answer: 319

Expected answer: 319

Correct? Yes

🔗 Follow-up

Problem: If instead the carpenter decided to cut the same 29 cm wide wooden plank into shorter lengths of 5 cm each, how many square centimeters of area does one of these shorter cut wooden planks cover?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its width and length.
The area of one cut wooden plank is 29*5 = <<29*5=145>>145 square centimeters.
### Answer: 145

Expected answer: 145

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 25m and a width of 11m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 25+25+11+11 = <<25+25+11+11=72>>72m.
### Answer: 72m

Expected answer: 72

Correct? Yes

🔗 Follow-up

Problem: Now, suppose the rectangle's width is changed from 11m to 17m. What will be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 25+25+17+17 = <<25+25+17+17=84>>84m.
### Answer: 84m

Expected answer: 84

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 49m and a breadth of 9m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 49+9+49+9 = <<49+9+49+9=106>>106m.
### Answer: 106

Expected answer: 116

Correct? No

🔗 Problem

Problem: A rectangle has a length of 16m and a width of 7m. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 16*7 = <<16*7=112>>112 square meters.
### Answer: 112

Expected answer: 112

Correct? Yes

🔗 Problem

Problem: The perimeter of a rectangular playground is found by the formula 2*(Length + Breadth). If the length of the playground is 14m and the breadth is 2m, what is the perimeter of the playground?

LLemma 34B:

 Reasoning: The perimeter of the playground is 2*(14+2) = <<2*(14+2)=34>>34m.
### Answer: 34

Expected answer: 32

Correct? No

🔗 Problem

Problem: A square has a side length of 16 units. What is the perimeter of the square?

LLemma 34B:

 Reasoning: The perimeter of a square is 4 times the length of a side.
The perimeter of the square is 4*16 = <<4*16=64>>64 units.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, the square's side length was 16 units. Now consider a rectangle with a length of 22 units and a width of 16 units. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is twice the length plus twice the width.
The perimeter of the rectangle is 2*22+2*16 = <<2*22+2*16=68>>68 units.
### Answer: 68

Expected answer: 76

Correct? No

🔗 Problem

Problem: A rectangular garden has a width of 7 meters and a length of 21 meters. What is the perimeter of this garden?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*width + 2*length.
The perimeter of this garden is 2*7 + 2*21 = <<2*7+2*21=54>>54 meters.
### Answer: 54

Expected answer: 56

Correct? No

🔗 Problem

Problem: A rectangle has a length of 20 meters and a width of 5 meters. What is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 20*5 = <<20*5=100>>100 square meters.
### Answer: 100

Expected answer: 100

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, we calculated the area of a rectangle with a length of 20 meters and a width of 5 meters. This time, instead of having the width as 5 meters, we have a new rectangle where the width is 17 meters. With the length still being 20 meters, what is the area of this new rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 20*17 = <<20*17=340>>340 square meters.
### Answer: 340

Expected answer: 340

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 37m and a width of 43m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 37+37+43+43 = <<37+37+43+43=156>>156m.
### Answer: 156

Expected answer: 160

Correct? No

🔗 Problem

Problem: A rectangle has a length of 9m. If the width of the rectangle is 1m, what is the area?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 9*1 = <<9*1=9>>9m^2.
### Answer: 9m^2

Expected answer: 9

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, we considered a rectangle with a length of 9m and a width of 1m. Now, consider if the width of the rectangle was actually 40m instead of 1m. What would be the new area of the rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 9*40 = <<9*40=360>>360m^2.
### Answer: 360m^2

Expected answer: 360

Correct? Yes

🔗 Problem

Problem: A rectangle's area is the product of 22m and 23m. What is the area of the rectangle?

LLemma 34B:

 Reasoning: The area of the rectangle is 22m*23m = <<22*23=466>>466.
### Answer: 466

Expected answer: 506

Correct? No

🔗 Problem

Problem: A rectangle has one side that measures 14 cm and another side that measures 21 cm. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 14+21+14+21 = <<14+21+14+21=60>>60 cm.
### Answer: 60

Expected answer: 70

Correct? No

🔗 Problem

Problem: In Clara's backyard, there is a rectangular swimming pool. The length of the pool is 23 meters, while the width is 12 meters. What is the area of the swimming pool in Clara's backyard?

LLemma 34B:

 Reasoning: The area of the swimming pool is 23*12 = <<23*12=276>>276 square meters.
### Answer: 276

Expected answer: 276

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 39 meters and a width of 8 meters. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 39+39+8+8 = <<39+39+8+8=84>>84 meters.
### Answer: 84

Expected answer: 94

Correct? No

🔗 Problem

Problem: A rectangle has a width of 27m and a length of 40m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 27+27+40+40 = <<27+27+40+40=134>>134m.
### Answer: 134

Expected answer: 134

Correct? Yes

🔗 Follow-up

Problem: What if the length of the rectangle was 44m instead of 40m while keeping the width still at 27m? What would the new perimeter of the rectangle be?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 27+27+44+44 = <<27+27+44+44=138>>138m.
### Answer: 138

Expected answer: 142

Correct? No

🔗 Problem

Problem: A rectangle's length is 35m and its width is 9m. What is the perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the rectangle is 2*35+2*9 = <<2*35+2*9=88>>88m.
### Answer: 88m

Expected answer: 88

Correct? Yes

🔗 Follow-up

Problem: Suppose we resized the rectangle and shortened its length from 35m to 19m, while keeping the width the same. What would the new perimeter of the rectangle be?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2*length + 2*width.
The perimeter of the rectangle is 2*19+2*9 = <<2*19+2*9=46>>46m.
### Answer: 46m

Expected answer: 56

Correct? No

🔗 Problem

Problem: A rectangle has a length of 38m and a width of 11m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 38+38+11+11 = <<38+38+11+11=88>>88m.
### Answer: 88

Expected answer: 98

Correct? No

🔗 Problem

Problem: One side of a rectangle is 28m long, and the other side is 29m long. What is the area of this rectangle?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 28*29 = <<28*29=812>>812 square meters.
### Answer: 812

Expected answer: 812

Correct? Yes

🔗 Follow-up

Problem: Suppose the side of the rectangle previously mentioned gets extended to 35m instead of 29m, while the other side remains 28m. What would be the new area of the rectangle in this case?

LLemma 34B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 28*35 = <<28*35=980>>980 square meters.
### Answer: 980

Expected answer: 980

Correct? Yes

🔗 Problem

Problem: A square has a side length of 9m. What is the total length of fencing needed to enclose the square's perimeter?

LLemma 34B:

 Reasoning: The perimeter of a square is 4 times the length of a side.
The perimeter of the square is 4*9 = <<4*9=36>>36m.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 38m and a width of 29m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 38+38+29+29 = <<38+38+29+29=134>>134m.
### Answer: 134

Expected answer: 134

Correct? Yes

🔗 Follow-up

Problem: Now consider that the rectangle from the previous question actually had a length of 6m instead of 38m. What would be the new perimeter of the rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of this rectangle is 6+6+29+29 = <<6+6+29+29=64>>64m.
### Answer: 64

Expected answer: 70

Correct? No

🔗 Problem

Problem: A rectangle has one side of length 45m and another side of length 47m. Find the perimeter of the rectangle.

LLemma 34B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 45+47 = <<45+47=92>>92m.
### Answer: 92

Expected answer: 184

Correct? No

🔗 Problem

Problem: Given a rectangle where the length is 24m and the width is 20m, what is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is 2(length + width).
The perimeter of this rectangle is 2(24+20) = <<2(24+20)=96>>96m.
### Answer: 96m

Expected answer: 88

Correct? No

🔗 Problem

Problem: The length and width of a rectangle are both 32m. What is the perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is twice the sum of its length and width.
The perimeter of this rectangle is 2(32+32) = <<2(32+32)=128>>128m.
### Answer: 128

Expected answer: 128

Correct? Yes

🔗 Follow-up

Problem: The length of the rectangle from the previous problem remains the same, but the width of the rectangle is now adjusted to 13m. What is the new perimeter of this rectangle?

LLemma 34B:

 Reasoning: The perimeter of a rectangle is twice the sum of its length and width.
The perimeter of this rectangle is 2(32+13) = <<2(32+13)=80>>80m.
### Answer: 80

Expected answer: 90

Correct? No