Worksheet featuring quadratic area word problems for solving real-world geometry applications.

Let’s solve each of the four quadratic area word problems step by step.

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Problem 1:

> A plot of land for sale has a width of *x* feet, and a length that is 8 feet less than its width. A farmer will only purchase the land if it measures 240 square feet. What value of *x* will allow the farmer to purchase the land?

Step 1: Define variables
- Width = *x* feet
- Length = *x – 8* feet

Step 2: Write area equation
Area = width × length
→ *x(x – 8) = 240*

Step 3: Expand and rearrange into standard quadratic form
*x² – 8x = 240*
→ *x² – 8x – 240 = 0*

Step 4: Factor the quadratic
We need two numbers that multiply to -240 and add to -8.
Try: -20 and +12 → (-20)(12) = -240, -20 + 12 = -8 ✔

So:
*(x – 20)(x + 12) = 0*

Step 5: Solve for x
*x – 20 = 0* → *x = 20*
*x + 12 = 0* → *x = -12* (discard — width can’t be negative)

✔ Answer: x = 20 feet

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Problem 2:

> The area of a rectangular field is 2000 sq.cm and its perimeter is 180 cm. Find the length and width of the field.

Step 1: Let variables
Let length = *l*, width = *w*

Step 2: Write equations
Area: *l × w = 2000*
Perimeter: *2(l + w) = 180* → *l + w = 90*

Step 3: Express one variable in terms of the other
From perimeter: *l = 90 – w*

Plug into area equation:
*(90 – w) × w = 2000*
→ *90w – w² = 2000*

Step 4: Rearrange into standard quadratic form
*–w² + 90w – 2000 = 0*
Multiply both sides by -1:
*w² – 90w + 2000 = 0*

Step 5: Factor
Find two numbers that multiply to 2000 and add to -90.
Try: -40 and -50 → (-40)(-50)=2000, -40 + (-50) = -90 ✔

So:
*(w – 40)(w – 50) = 0*

Step 6: Solve
*w = 40* or *w = 50*

If *w = 40*, then *l = 90 – 40 = 50*
If *w = 50*, then *l = 90 – 50 = 40*

Either way, dimensions are 40 cm and 50 cm.

✔ Answer: Length = 50 cm, Width = 40 cm (or vice versa)

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Problem 3:

> A rectangle whose area is 192 square feet has a width that is 4 feet less than the length. Find the dimensions of the rectangle.

Step 1: Define variables
Let length = *l*
Then width = *l – 4*

Step 2: Area equation
*l(l – 4) = 192*

Step 3: Expand and rearrange
*l² – 4l = 192*
→ *l² – 4l – 192 = 0*

Step 4: Factor
Need two numbers that multiply to -192 and add to -4.
Try: -16 and +12 → (-16)(12) = -192, -16 + 12 = -4 ✔

So:
*(l – 16)(l + 12) = 0*

Step 5: Solve
*l – 16 = 0* → *l = 16*
*l + 12 = 0* → *l = -12* (discard — length can’t be negative)

Width = *l – 4 = 16 – 4 = 12*

✔ Answer: Length = 16 ft, Width = 12 ft

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Problem 4:

> A photograph 8cm × 11cm will be framed as shown in the diagram. The total area of the frame and photograph is 180 cm². Find the width of the frame, marked *x*.

Step 1: Understand the diagram
The frame adds *x* cm on all sides. So:

- Total width = 11 + 2x (x on left, x on right)
- Total height = 8 + 2x (x on top, x on bottom)

Step 2: Write area equation
Total area = (11 + 2x)(8 + 2x) = 180

Step 3: Expand the expression
= 11×8 + 11×2x + 2x×8 + 2x×2x
= 88 + 22x + 16x + 4x²
= 4x² + 38x + 88

Set equal to 180:
*4x² + 38x + 88 = 180*

Step 4: Subtract 180 from both sides
*4x² + 38x – 92 = 0*

Step 5: Simplify by dividing all terms by 2
*2x² + 19x – 46 = 0*

Step 6: Use quadratic formula
*a = 2, b = 19, c = -46*

Discriminant:
*D = b² – 4ac = 19² – 4(2)(-46) = 361 + 368 = 729*

√729 = 27

Solutions:
*x = [–19 ± 27] / (2×2)*

First solution:
*x = (–19 + 27)/4 = 8/4 = 2*

Second solution:
*x = (–19 – 27)/4 = –46/4 = –11.5* (discard — width can’t be negative)

✔ Answer: x = 2 cm

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## ✔ Final Answers:

1. x = 20 feet
2. Length = 50 cm, Width = 40 cm
3. Length = 16 ft, Width = 12 ft
4. x = 2 cm

All solved using quadratic equations derived from area formulas and given constraints.