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📝 Find rectangle dimensions from perimeter or area (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

←Previous Topic:Area of a rectangle formula (A = LW)📝 Find rectangle dimensions from perimeter or areaFind Rectangle dimensions with relationship between sides:Next Topic→

What is Find rectangle dimensions from perimeter or area?

Definition:
Finding rectangle dimensions from perimeter or area involves solving for the length and width when given the perimeter or area, using the formulas $P = 2L + 2W$ and $A = LW$ , and often requires setting up and solving a system of equations if both perimeter and area are given, or using one formula with an additional relationship between length and width.

Working:
If only perimeter is given, and no other relation, there are infinite solutions, but if a relationship is provided (e.g., length is 3 times width), substitute into the perimeter equation and solve; if area and perimeter are both known, use the area to express one variable in terms of the other, then substitute into the perimeter equation to form a quadratic equation $2L^2 - P L + 2A = 0$ , and solve for $L$ using the quadratic formula $L = \frac{P \pm \sqrt{P^2 - 16A}}{4}$ .

Example:
A rectangle has a perimeter of 30 cm and an area of 54 cm $^2$ ; using $P = 2L + 2W$ and $A = LW$ , substitute $W = \frac{54}{L}$ into $30 = 2L + 2(\frac{54}{L})$ , simplify to $15 = L + \frac{54}{L}$ , multiply by $L$ : $L^2 - 15L + 54 = 0$ , factor to $(L-6)(L-9) = 0$ , so length = 9 cm and width = 6 cm.

Reason:
This skill is important for practical applications like designing rectangular spaces, optimizing materials, and solving geometry problems, and it reinforces algebraic techniques in a geometric context, enhancing problem-solving abilities.

Easy

Medium

Hard

📝 All Find rectangle dimensions from perimeter or area MCQs

Q1. A rectangular garden has a perimeter of 5454 m. If its length is 1717 m, what is its width?

A.9 m

B.10 m ✅

C.11 m

D.12 m

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a rectangle, the perimeter is $2L+2W$ . Substituting $P=54$ and $L=17$ gives $54=34+2W$ . Thus $20=2W$ , so $W=10$ m. The key reasoning step is isolating the unknown width rather than dividing the perimeter directly by two.

Q2. A rectangle has an area of $96\text{ m}^2$ and a width of 88 m. Which equation correctly represents the calculation needed to find its length?

L=96+8

L=96-8

L=96\div8

✅

L=8\div96

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Area is obtained by multiplying length and width, so $A=LW$ . To isolate the unknown length, divide the area by the known width. Therefore $L=96\div8=12$ m. Adding or subtracting the dimensions would not preserve the area relationship.

Q3. Two rectangles have the same perimeter of 4040 m. Rectangle A has length 1212 m, while Rectangle B has length 1515 m. Which statement correctly compares their widths?

A.A is 1 m wider than B

B.A is 2 m wider than B ✅

C.B is 2 m wider than A

D.Both widths are equal

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using $P=2L+2W$ , Rectangle A has $40=24+2W$ , giving $W=8$ m. Rectangle B has $40=30+2W$ , giving $W=5$ m. Therefore A is 3 m wider, not 2 m. This demonstrates how increasing length while holding perimeter fixed decreases width.

Q4. A farmer has 7272 m of fencing for a rectangular enclosure. He wants the length to be 66 m greater than the width. Which dimensions satisfy both conditions?

A.Length 15 m, width 21 m

B.Length 21 m, width 15 m ✅

C.Length 18 m, width 12 m

D.Length 24 m, width 12 m

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Let the width be $W$ , so the length is $W+6$ . The perimeter equation becomes $72=2(W+6)+2W$ . Simplifying gives $72=4W+12$ , so $W=15$ m and $L=21$ m. Both the difference and perimeter conditions are satisfied.

Q5. A designer claims that a rectangle with area $120\text{ cm}^2$ and width 1010 cm must have length 1111 cm because $120-10=110$ , then adjusting for units gives 11. What is the correct length?

A.10 cm

B.12 cm ✅

C.20 cm

D.110 cm

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The designer incorrectly subtracts dimensions from area. Since $A=LW$ , the length must be found by division: $L=A\div W=120\div10=12$ cm. Area measures square units, while length measures linear units, so subtraction cannot produce the required dimension.

Q6. A rectangular playground has perimeter 100100 m. A student calculates its length as $100\div2=50$ m and concludes that the width must be 50 m as well. What is the flaw?

A.Perimeter should be multiplied by 2

B.The perimeter only determines

L+W=50

, not each dimension individually ✅

C.Length must always be smaller than width

D.The width must be found by subtracting 50 from the area

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Dividing 100 by 2 gives $L+W=50$ , not $L=50$ and $W=50$ . A perimeter alone does not uniquely determine both dimensions. Additional information, such as one dimension or their difference, is required to find the missing dimension.

Q7. A rectangular plot has area $240\text{ m}^2$ . Its width is 44 m less than its length. A student tests 2020 m by 1212 m and rejects it because the dimensions differ by 88 m. What should the student conclude?

A.The pair is correct because its area is

240\text{ m}^2

B.The pair is correct because all rectangles have equal dimensions

C.The pair is incorrect because it does not satisfy the stated difference ✅

D.The pair is incorrect because rectangles cannot have a 20-m length

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Although $20\times12=240$ , the dimensions differ by 8 m, not 4 m. Therefore the pair satisfies the area condition but violates the width-length relationship. Checking every stated condition is essential in a multi-constraint modeling problem.

Q8. A graph shows possible rectangles with constant perimeter 4040 m, represented by $W=20-L$ . A point on the graph has $L=14$ . What width does the point represent?

A.4 m

B.6 m ✅

C.14 m

D.20 m

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The graph equation $W=20-L$ describes all rectangles having perimeter 40 m. Substituting $L=14$ gives $W=20-14=6$ m. The graph therefore represents a trade-off: as length increases, width decreases while the total perimeter remains fixed.

Q9. A rectangular banner has area $300\text{ cm}^2$ . Its length is 55 cm greater than its width. Which approach is most appropriate for finding the dimensions?

A.Divide 300 by 5 immediately

B.Set

W=x

L=x+5

, then use

LW=300

✅

C.Set both dimensions equal because the area is fixed

D.Subtract 5 from 300 to obtain the width

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because the two dimensions are related by a difference of 5, one variable can represent the width and the other can be written as $x+5$ . The area condition becomes $x(x+5)=300$ , allowing the dimensions to be determined systematically.

Q10. A rectangular swimming pool has perimeter 8686 m. The designer knows the length is twice the width plus 11 m. What are the dimensions?

A.Length 29 m, width 14 m ✅

B.Length 28 m, width 15 m

C.Length 30 m, width 13 m

D.Length 27 m, width 16 m

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Let the width be $W$ , making the length $2W+1$ . The perimeter equation is $86=2(2W+1)+2W=6W+2$ . Thus $84=6W$ , giving $W=14$ m and $L=29$ m. Checking gives $2(29)+2(14)=86$ m.

Q11. A rectangle has perimeter 6464 m. One student finds $L+W=32$ , while another claims $L=W=32$ . Which evaluation is correct?

A.Both are correct because the perimeter is 64

B.Only the first is correct because the dimensions must add to 32 ✅

C.Only the second is correct because rectangles have equal sides

D.Neither is correct because

L+W=64

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: From $64=2L+2W$ , dividing by 2 gives $L+W=32$ . This relation does not mean each dimension equals 32. Many pairs, such as 20 and 12, satisfy the condition. More information is needed to determine unique dimensions.

Q12. A rectangular field has area $192\text{ m}^2$ . Its length and width are positive whole numbers, and the length is greater than the width. Which pair gives the smallest possible perimeter?

A.24 m by 8 m

B.16 m by 12 m ✅

C.32 m by 6 m

D.48 m by 4 m

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Each pair has area $192\text{ m}^2$ , but their perimeters differ. The $24\times8$ rectangle has perimeter 64 m, $16\times12$ has perimeter 56 m, $32\times6$ has 76 m, and $48\times4$ has 104 m. Thus 16 m by 12 m gives the smallest perimeter among the choices.

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