Solve for Rectangle Dimensions: Area = 12 cm² and Perimeter = 14 cm

Quadratic Equations with Geometry Applications

The perimeter of a rectangle is 14 cm.

The area of the rectangle is 12 cm².

What are the lengths of its sides?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Determine the sides of the rectangle
00:03 Draw the rectangle
00:06 Mark the sides with X and Y
00:12 The perimeter of the rectangle equals the sum of its sides
00:18 Substitute in the perimeter value and proceed to solve for the sum of sides
00:26 This is the sum of the sides
00:29 Now apply the formula to calculate the rectangle's area (side multiplied by side)
00:33 Express X in terms of Y
00:43 Substitute in our X and solve for Y
00:46 Open the parentheses properly - multiply each term
00:57 Arrange the equation so that the right side equals 0
01:06 Determine the possible solutions for Y
01:10 These are the possible solutions for Y
01:17 Substitute them in to the expression to find X
01:28 Note that these are the options for our sides
01:36 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The perimeter of a rectangle is 14 cm.

The area of the rectangle is 12 cm².

What are the lengths of its sides?

2

Step-by-step solution

Since in a rectangle each pair of opposite sides are equal to each other, let's call each pair of sides X and Y

Now let's set up a formula to calculate the perimeter of the rectangle:

2x+2y=14 2x+2y=14

Let's divide both sides by 2:

x+y=7 x+y=7

From this formula, we'll calculate X:

x=7y x=7-y

We know that the area of the rectangle equals length times width:

x×y=12 x\times y=12

We know that X equals 7 minus Y, let's substitute this in the formula:

(7y)×y=12 (7-y)\times y=12

7yy2=12 7y-y^2=12

y27y+12=0 y^2-7y+12=0

(y3)×(y4)=0 (y-3)\times(y-4)=0

From this we can claim that:

y=3,y=4 y=3,y=4

Let's go back to the formula we found earlier:

x=7y x=7-y

Let's substitute y equals 3 and we get:

x=73=4 x=7-3=4

Now let's substitute y equals 4 and we get:

x=74=3 x=7-4=3

Therefore, the lengths of the rectangle's sides are 4 and 3

3

Final Answer

3, 4

Key Points to Remember

Essential concepts to master this topic
  • Rectangle Properties: Opposite sides equal, perimeter = 2(length + width)
  • Substitution Method: Express x = 7 - y, then substitute into xy = 12
  • Verification: Check both conditions: 2(3) + 2(4) = 14 and 3 × 4 = 12 ✓

Common Mistakes

Avoid these frequent errors
  • Setting up incorrect perimeter formula
    Don't write perimeter as x + y = 14! This ignores that rectangles have four sides, not two. This gives wrong dimensions like (7,0). Always use perimeter = 2x + 2y for rectangles.

Practice Quiz

Test your knowledge with interactive questions

Calculate the perimeter of the rectangle below.

181818222

FAQ

Everything you need to know about this question

Why do we get two solutions (3,4) and (4,3) - which one is correct?

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Both are correct! In a rectangle, it doesn't matter which dimension you call length and which you call width. A 3×4 rectangle and a 4×3 rectangle are the same shape.

How do I know when to use substitution vs other methods?

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Use substitution when one equation is easier to solve for a variable. Here, x+y=7 x + y = 7 quickly gives us x=7y x = 7 - y , making substitution the best choice.

What if I get a negative solution for the quadratic?

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Always check if your solutions make physical sense! Rectangle sides must be positive, so reject any negative values. Only keep solutions where both dimensions are greater than zero.

Can I solve this problem without using quadratic equations?

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Not easily! You could try guess-and-check with factor pairs of 12, but the systematic algebraic approach guarantees you find all solutions and understand why they work.

Why do we rearrange to standard form y² - 7y + 12 = 0?

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Standard form makes it easier to factor or use the quadratic formula. When we see y27y+12=0 y^2 - 7y + 12 = 0 , we can quickly find factors that multiply to 12 and add to -7.

What if the quadratic doesn't factor nicely?

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Use the quadratic formula! y=7±49482 y = \frac{7 \pm \sqrt{49-48}}{2} works for any quadratic equation, even when factoring is difficult.

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