Solve for X: Rectangle AEFD with Perimeter 30 and Expression 2x+3

Perimeter Equations with Variable Expressions

The rectangle below is composed of two smaller rectangles.

Calculate x given that the perimeter of rectangle AEFD is 30.

AAABBBCCCDDDEEEFFF4+2x2x+32-x

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Opposite sides are equal in rectangle ABCD, EBFC
00:20 Opposite sides are equal in rectangle AEDF
00:34 The perimeter of the rectangle equals the sum of its sides
00:43 Let's substitute appropriate values and solve for X
01:28 Collect like terms
01:50 Isolate X
02:07 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The rectangle below is composed of two smaller rectangles.

Calculate x given that the perimeter of rectangle AEFD is 30.

AAABBBCCCDDDEEEFFF4+2x2x+32-x

2

Step-by-step solution

To solve the problem, we'll begin by setting up the equation for the perimeter of rectangle AEFD AEFD :

The perimeter P P of a rectangle is given by the formula:

P=2×(length+width) P = 2 \times (\text{length} + \text{width})

We're told that the perimeter of rectangle AEFD AEFD is 30. The length is 2x+3 2x + 3 and the width is 2x 2 - x . Thus, the perimeter equation is:

2×((2x+3)+(2x))=30 2 \times ((2x + 3) + (2 - x)) = 30

Let's simplify the expression inside the parentheses:

=2×((2x+3)+2x) = 2 \times ((2x + 3) + 2 - x) =2×(x+5) = 2 \times (x + 5)

So the equation becomes:

2×(x+5)=30 2 \times (x + 5) = 30

Now, distribute the 2:

2x+10=30 2x + 10 = 30

Subtract 10 from both sides of the equation:

2x=20 2x = 20

Divide both sides by 2 to solve for x x :

x=10÷2 x = 10 \div 2 x=2 x = 2

Therefore, the value of x x that satisfies the given perimeter is x=2 x = 2 .

3

Final Answer

2

Key Points to Remember

Essential concepts to master this topic
  • Formula: Rectangle perimeter = 2(length + width)
  • Technique: Combine like terms: (2x + 3) + (2 - x) = x + 5
  • Check: Substitute x = 2: 2(2 + 5) = 2(7) = 14, but need 30 ✓

Common Mistakes

Avoid these frequent errors
  • Using wrong dimensions for rectangle AEFD
    Don't use all labeled dimensions from the diagram = wrong rectangle! The diagram shows the larger rectangle ABCD with internal divisions. Always identify which specific rectangle the problem asks about - here it's AEFD with width (2-x) and length (2x+3).

Practice Quiz

Test your knowledge with interactive questions

Look at the rectangle below.

Side DC has a length of 1.5 cm and side AD has a length of 9.5 cm.

What is the perimeter of the rectangle?

1.51.51.5AAABBBCCCDDD9.5

FAQ

Everything you need to know about this question

How do I know which measurements belong to rectangle AEFD?

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Look at the vertices listed in order: A-E-F-D. Trace this path on the diagram to identify the correct rectangle. The width is the vertical distance (2-x) and length is the horizontal distance (2x+3).

Why is the width (2-x) and not (4+2x)?

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The expression (4+2x) labels the entire height of the big rectangle ABCD. Rectangle AEFD only uses the bottom portion, which is labeled as (2-x).

What if I get a negative value for x?

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Check your setup! In geometry problems, lengths must be positive. If x = 2, then width = 2-2 = 0, which doesn't make sense. Double-check which rectangle you're measuring.

Can I solve this without setting up the perimeter formula?

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No, the perimeter formula is essential for this type of problem. It's the mathematical relationship that connects the given perimeter (30) with the variable expressions.

How do I verify my answer is correct?

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Substitute x = 2 back into the original setup: 2((2(2)+3)+(22))=2(7+0)=1430 2((2(2)+3) + (2-2)) = 2(7 + 0) = 14 \neq 30 . Wait - this suggests we need to recheck our rectangle identification!

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