Solve for X: Rectangle with Area 72 and Dimensions 2X by 4X

Rectangle Area Problems with Algebraic Variables

The area of a rectangle is equal to 72.

AC = 2X

AB = 4X

Calculate X.

AAABBBDDDCCC4X722X

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1

Understand the problem

The area of a rectangle is equal to 72.

AC = 2X

AB = 4X

Calculate X.

AAABBBDDDCCC4X722X

2

Step-by-step solution

The area of a rectangle is equal to its length multiplied by its width.

Let's begin by inserting the known data into the formula as follows :

72=2x×4x 72=2x\times4x

72=8x2 72=8x^2

Let's proceed to simplify both sides of the equation by the (HCF) the highest common factor, in this case 8 :

9=x2 9=x^2

Finally let's remove the square root:

x=3 x=3

3

Final Answer

3

Key Points to Remember

Essential concepts to master this topic
  • Area Formula: Rectangle area equals length times width: A = l × w
  • Technique: Substitute variables: 72 = 2x × 4x = 8x²
  • Check: Verify x = 3: Area = 2(3) × 4(3) = 6 × 12 = 72 ✓

Common Mistakes

Avoid these frequent errors
  • Adding dimensions instead of multiplying
    Don't calculate 2x + 4x = 6x and set equal to 72! Addition gives perimeter, not area. Area requires multiplication. Always multiply length times width: 2x × 4x = 8x².

Practice Quiz

Test your knowledge with interactive questions

Look at the rectangle ABCD below.

Side AB is 6 cm long and side BC is 4 cm long.

What is the area of the rectangle?
666444AAABBBCCCDDD

FAQ

Everything you need to know about this question

Why do I multiply 2x by 4x instead of adding them?

+

Area formula requires multiplication! Length × Width = Area. Adding dimensions gives you perimeter, which is the distance around the rectangle, not the space inside it.

How do I solve x² = 9 to get x = 3?

+

Take the square root of both sides: x2=9 \sqrt{x^2} = \sqrt{9} . Since we're looking for length, we only need the positive solution: x = 3.

What if I get x² = negative number?

+

That's impossible for a real rectangle! Lengths must be positive numbers. Check your setup - you might have mixed up the area formula or made an arithmetic error.

Can the dimensions be different expressions like 3x and 5x?

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Absolutely! The method stays the same: multiply the expressions together. For 3x × 5x = 15x², then solve 15x2=given area 15x^2 = \text{given area} .

How do I check if my final dimensions make sense?

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Calculate both dimensions using your x-value, then multiply them. The result should equal the given area. Also check that both dimensions are positive numbers!

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