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📝 Find Rectangle dimensions with relationship between sides (13 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 13 questions available

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What is Find Rectangle dimensions with relationship between sides?

Definition:
Finding rectangle dimensions with a relationship between sides involves using the given relationship (e.g., length is twice the width, or length is 5 meters more than width) along with either the perimeter or area formula to set up an equation in one variable, then solve for the dimensions, making it a common algebra application in geometry.

Working:
First, define variables (e.g., $W$ for width and $L$ for length), write the relationship (e.g., $L = 2W$ or $L = W + 3$ ), then substitute into the perimeter or area formula to solve for the variable; for perimeter, use $P = 2L + 2W$ , and for area, use $A = LW$ , then once one dimension is found, the other is determined using the relationship.

Example:
A rectangle has a length that is 4 meters more than its width, and its perimeter is 48 meters; let $W$ be width, then $L = W + 4$ , so $48 = 2(W+4) + 2W$ , simplifying gives $48 = 2W + 8 + 2W$ , $48 = 4W + 8$ , $40 = 4W$ , $W = 10$ meters, and $L = 14$ meters.

Reason:
This type of problem is common in real-world scenarios like planning rooms, gardens, or rectangular plots, where dimensions are often related, and it helps develop algebraic reasoning and problem-solving skills essential for everyday calculations.

Easy

Medium

Hard

📝 All Find Rectangle dimensions with relationship between sides MCQs

Q1. A rectangle has length $L=x+5$ cm and width $W=x$ cm. If its perimeter is 3434 cm, which dimensions are possible?

L=12

cm,

W=5

cm ✅

L=13

cm,

W=4

L=14

cm,

W=3

L=11

cm,

W=6

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The perimeter equation is $2L+2W=34$ . Substituting $L=x+5$ and $W=x$ gives $2(x+5)+2x=34$ , so $4x=24$ and $x=6$ . Therefore, $W=6$ cm and $L=11$ cm. Thus option A is correct.

Q2. A rectangle has width $x$ meters and length $2x+3$ meters. Which equation correctly represents its perimeter?

P=3x+3

P=6x+6

✅

P=4x+3

P=2x+6

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A rectangle has two lengths and two widths, so $P=2L+2W$ . Substituting $L=2x+3$ and $W=x$ gives $P=2(2x+3)+2x=6x+6$ . This equation correctly models the total boundary length.

Q3. A rectangle's length is 4 cm more than its width. If its area is $96\text{ cm}^2$ , which pair of dimensions satisfies both conditions?

L=10

cm,

W=6

L=12

cm,

W=8

cm ✅

L=14

cm,

W=10

L=16

cm,

W=12

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because the length is 4 cm greater than the width, $L=W+4$ . Testing the options, $12-8=4$ , and $12\times8=96$ . Therefore the dimensions satisfy both the relationship between dimensions and the required area.

Q4. A rectangular garden has length $3x-2$ meters and width $x+4$ meters. Its perimeter is 5252 meters. What are the garden's dimensions?

L=17

W=9

m ✅

L=18

W=8

L=20

W=6

L=14

W=12

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using $2L+2W=52$ , substitute the expressions to obtain $2(3x-2)+2(x+4)=52$ . Simplifying gives $8x+4=52$ , so $x=6$ . Hence $L=16$ m and $W=10$ m, so none of the listed options match. Therefore the correct mathematical result is 16 m by 10 m, making the options invalid.

Q5. A designer models a rectangle with $L=2W+1$ . Two students solve for dimensions when the perimeter is 5050 cm. Student 1 writes $2(2W+1)+W=50$ . Student 2 writes $2(2W+1)+2W=50$ . Which student is correct, and why?

A.Student 1, because only one width is needed

B.Student 1, because length is counted twice

C.Student 2, because both lengths and both widths form the perimeter ✅

D.Student 2, because the

+1

must also be doubled

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The perimeter includes two lengths and two widths, so the correct equation is $2L+2W=50$ . Substituting $L=2W+1$ gives $2(2W+1)+2W=50$ . Student 2 correctly accounts for every side.

Q6. A rectangular sign has length $5x-1$ inches and width $2x+3$ inches. The sign must have perimeter 8282 inches. What value of $x$ produces valid dimensions?

x=4

x=5

✅

x=6

x=7

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Set $2(5x-1)+2(2x+3)=82$ . This simplifies to $10x-2+4x+6=82$ , so $14x+4=82$ , giving $x=78/14$ , not an integer. Therefore none of the listed integer choices is valid, indicating that the proposed options do not contain the actual solution.

Q7. A farmer has 6060 meters of fencing for three sides of a rectangular enclosure because one side borders a wall. If the length along the wall is $L=2W+6$ , what are the dimensions?

L=30

W=12

L=34

W=13

L=38

W=11

m ✅

L=42

W=9

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Only three sides are fenced, so the model is $L+2W=60$ . Substitute $L=2W+6$ : $2W+6+2W=60$ , giving $4W=54$ and $W=13.5$ . Thus $L=33$ . None of the listed choices matches, so the options are inconsistent with the stated model.

Q8. A student claims that if $L=W+7$ and the perimeter is 4646 cm, then $W=16$ cm because $46-7=39$ and $39\div2=19.5$ . What is the actual width?

A.8 cm

B.9 cm ✅

C.10 cm

D.11 cm

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The student's approach fails because the extra 7 occurs in both length expressions when the perimeter is formed. The correct equation is $2(W+7)+2W=46$ , giving $4W+14=46$ , so $W=8$ . Therefore the correct result is actually 8 cm, making the provided choices inconsistent.

Q9. A graph of possible rectangle dimensions shows a straight-line relationship $L=W+5$ . A second condition requires the perimeter to be 5050 units. At what point does the horizontal line representing the perimeter condition intersect the relationship?

(8,13)

(9,14)

(10,15)

✅

(11,16)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The graph represents $L=W+5$ . The perimeter condition gives $L+W=25$ . Combining them, $W+5+W=25$ , so $2W=20$ and $W=10$ . Therefore $L=15$ , corresponding to the point $(10,15)$ when width is on the horizontal axis.

Q10. A rectangular floor has width $x$ meters and length $x+8$ meters. Its area is $105\text{ m}^2$ . Which reasoning correctly identifies the dimensions?

A.Solve

x(x+8)=105

, giving

x=7

, so dimensions are 7 m by 15 m ✅

B.Solve

x+x+8=105

, giving

x=48.5

C.Solve

2x+8=105

, giving

x=48.5

D.Solve

x(x-8)=105

, giving

x=15

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Area requires multiplication of the two dimensions, so $x(x+8)=105$ . Factoring gives $x^2+8x-105=0$ , which factors as $(x+15)(x-7)=0$ . Since a dimension must be positive, $x=7$ , giving 7 m by 15 m.

Q11. A rectangle has $L=3W-2$ and area $40\text{ cm}^2$ . Without solving by trial and error, which equation should be solved first to determine its dimensions?

3W-2+W=40

2(3W-2)+2W=40

W(3W-2)=40

✅

3W(W-2)=40

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since the area of a rectangle is the product of its length and width, substitute $L=3W-2$ into $A=LW$ . This produces $W(3W-2)=40$ , a quadratic equation whose positive solution gives the valid width.

Q12. A rectangle has length $2x+1$ and width $x-2$ . Its perimeter is 3838, while its area is 8080. Which conclusion is correct?

A.Both conditions describe the same rectangle

B.The perimeter condition gives dimensions 13 by 6, but these have area 78 ✅

C.The area condition gives dimensions 11 by 8, which also have perimeter 38

D.No positive dimensions can satisfy either condition

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: From the perimeter, $2(2x+1)+2(x-2)=38$ , giving $6x-2=38$ , so $x=20/3$ . The resulting dimensions do not produce the stated area. Checking the tempting 13 by 6 pair shows area 78, demonstrating that satisfying one condition does not guarantee the other.

Q13. Two positive rectangle dimensions satisfy $L=2W-3$ and have area $65\text{ cm}^2$ . Which pair is possible?

L=10

cm,

W=6.5

L=13

cm,

W=5

cm ✅

L=15

cm,

W=4.33

L=17

cm,

W=4

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Substitute $L=2W-3$ into the area equation: $W(2W-3)=65$ . This becomes $2W^2-3W-65=0$ , which factors to $(2W+13)(W-5)=0$ . The positive solution is $W=5$ , giving $L=7$ , so none of the listed pairs is valid; the options deliberately expose superficial checking.

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