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A rectangular microchip component has a length of L=(2.5×10−3 m+4.0×10−4 m)L = (2.5 \times 10^{-3}\,\text{m} + 4.0 \times 10^{-4}\,\text{m}) and a width of W=(2.5×10−3 m−4.0×10−4 m)W = (2.5 \times 10^{-3}\,\text{m} - 4.0 \times 10^{-4}\,\text{m}). Using the difference of squares identity, compute the exact surface area of the component in square millimeters (mm2\text{mm}^2).

Answer and steps

Final answer

6.09,mm\text{mm}^2

Steps

01

Convert the measurements to millimeters

To find the area using the difference of squares identity, recognize that the area of a rectangle is given by Area=L⋅W\text{Area} = L \cdot W, which matches the algebraic pattern: (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2

Since the final area is required in square millimeters (mm2\text{mm}^2), converting each component term to millimeters (mm\text{mm}) before squaring simplifies the calculations. Recall that 1 m=103 mm1\,\text{m} = 10^3\,\text{mm}:

  • Let a=2.5×10−3 ma = 2.5 \times 10^{-3}\,\text{m}: a=2.5×10−3×103 mm=2.5 mma = 2.5 \times 10^{-3} \times 10^3\,\text{mm} = 2.5\,\text{mm}
  • Let b=4.0×10−4 mb = 4.0 \times 10^{-4}\,\text{m}: b=4.0×10−4×103 mm=0.40 mmb = 4.0 \times 10^{-4} \times 10^3\,\text{mm} = 0.40\,\text{mm}

Thus, the dimensions can be expressed as: L=2.5 mm+0.40 mmL = 2.5\,\text{mm} + 0.40\,\text{mm} W=2.5 mm−0.40 mmW = 2.5\,\text{mm} - 0.40\,\text{mm}

02

Apply the difference of squares identity

Substitute a=2.5 mma = 2.5\,\text{mm} and b=0.40 mmb = 0.40\,\text{mm} into the product identity: Area=(a+b)(a−b)=a2−b2\text{Area} = (a + b)(a - b) = a^2 - b^2 Area=(2.5 mm)2−(0.40 mm)2\text{Area} = (2.5\,\text{mm})^2 - (0.40\,\text{mm})^2

03

Calculate the squares and find the final area

Compute each squared value: (2.5)2=6.25(2.5)^2 = 6.25 (0.40)2=0.16(0.40)^2 = 0.16

Subtract the results: Area=6.25 mm2−0.16 mm2=6.09 mm2\text{Area} = 6.25\,\text{mm}^2 - 0.16\,\text{mm}^2 = 6.09\,\text{mm}^2

Common Pitfall: If calculating the area in square meters first, remember that the conversion factor for area is squared: 1 m2=(103 mm)2=106 mm21\,\text{m}^2 = (10^3\,\text{mm})^2 = 10^6\,\text{mm}^2, not 103 mm210^3\,\text{mm}^2. Converting the linear dimensions first avoids power-of-ten conversion errors.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student uses the linear conversion factor 1 m = 10³ mm instead of 1 m² = 10⁶ mm² when converting square meters to square millimeters.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Area\text{Area} = (2.5 ×\times 10^{-3})^2 - (4.0 ×\times 10^{-4})^2   ⟹  \implies 6.25 ×\times 10^{-6} - 0.16 ×\times 10^{-6}   ⟹  \implies Area\text{Area} = 6.09 ×\times 10^{-6},m\text{m}^2

  2. Step 2

    Area\text{Area} = 6.09 ×\times 10^{-6} ×\times 10^3,mm\text{mm}^2 = 6.09 ×\times 10^{-3},mm\text{mm}^2

    Correct

    Area\text{Area} = 6.09 ×\times 10^{-6} ×\times (10^3)^2,mm\text{mm}^2 = 6.09,mm\text{mm}^2

    Because area is a two-dimensional quantity, converting from square meters to square millimeters requires squaring the linear conversion factor: 1 m2=(103 mm)2=106 mm21\,\text{m}^2 = (10^3\,\text{mm})^2 = 10^6\,\text{mm}^2, not 103 mm210^3\,\text{mm}^2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified and applied the difference of squares identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2.

You handled the powers of ten and decimal arithmetic accurately when computing the intermediate area in square meters.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Linear vs. Area Unit Conversion Error

You multiplied by 10310^3 instead of 10610^6 during the unit conversion step.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You applied the one-dimensional linear scale factor between meters and millimeters (1 m=103 mm1\,\text{m} = 10^3\,\text{mm}) to a two-dimensional area quantity, forgetting that area units scale with the square of the linear factor (1 m2=(103 mm)2=106 mm21\,\text{m}^2 = (10^3\,\text{mm})^2 = 10^6\,\text{mm}^2).

REASONING PATH
Your ACEpath teacher’s suggested approach
Square the unit conversion factor for area

  • If 1 unitA=k unitB1\,\text{unit}_A = k\,\text{unit}_B, then 1 unitA2=k2 unitB21\,\text{unit}_A^2 = k^2\,\text{unit}_B^2.

  • Always write out (103 mm/1 m)2(10^3\,\text{mm}/1\,\text{m})^2 with units so the conversion factor squares to 10610^6.

Alternative: Convert linear units before multiplying

  • Converting dimensions to the requested target unit before using an area identity avoids power-of-ten unit mistakes.

  • Convert a=2.5×10−3 m=2.5 mma = 2.5 \times 10^{-3}\,\text{m} = 2.5\,\text{mm} and b=4.0×10−4 m=0.40 mmb = 4.0 \times 10^{-4}\,\text{m} = 0.40\,\text{mm} first, then calculate (2.5)2−(0.40)2(2.5)^2 - (0.40)^2 directly in mm2\text{mm}^2.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Recall that 1 m=103 mm1\,\text{m} = 10^3\,\text{mm}, which means 1 m2=(103 mm)2=106 mm21\,\text{m}^2 = (10^3\,\text{mm})^2 = 10^6\,\text{mm}^2.

  2. Multiply the area in square meters by 106 mm2/m210^6\,\text{mm}^2/\text{m}^2: 6.09×10−6×106 mm2=6.09 mm26.09 \times 10^{-6} \times 10^6\,\text{mm}^2 = 6.09\,\text{mm}^2.

Final answer

6.09,mm\text{mm}^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice converting derived units (area and volume) by explicitly squaring or cubing both the numerical factor and the unit symbol.

  • When dealing with algebraic formulas involving geometric dimensions, try converting lengths to the required target unit before applying the algebraic identity.

OverviewACEpath Teacher's Diagnosis

02 The student misapplies the difference of squares identity by adding the squared terms instead of subtracting them.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Area\text{Area} = (2.5 ×\times 10^{-3})^2 + (4.0 ×\times 10^{-4})^2

    Correct

    Area\text{Area} = (2.5 ×\times 10^{-3})^2 - (4.0 ×\times 10^{-4})^2

    The product (a+b)(a−b)(a+b)(a-b) expands to the difference of squares a2−b2a^2 - b^2, which requires subtracting the second square rather than adding it.

  2. Step 2

    Area\text{Area} = 6.25 ×\times 10^{-6} + 0.16 ×\times 10^{-6}   ⟹  \implies Area\text{Area} = 6.41 ×\times 10^{-6},m\text{m}^2

    Correct

    Area\text{Area} = 6.25 ×\times 10^{-6},m\text{m}^2 - 0.16 ×\times 10^{-6},m\text{m}^2 = 6.09 ×\times 10^{-6},m\text{m}^2

    Your squaring and power-of-ten alignment were accurate, but because of the initial sign error, you computed a sum rather than a difference.

  3. Step 3

    Area\text{Area} = 6.41 ×\times 10^{-6} ×\times 10^6,mm\text{mm}^2 = 6.41,mm\text{mm}^2

    Correct

    Area\text{Area} = 6.09 ×\times 10^{-6} ×\times 10^6,mm\text{mm}^2 = 6.09,mm\text{mm}^2

    Your area conversion factor 1 m2=106 mm21\,\text{m}^2 = 10^6\,\text{mm}^2 was correctly applied, but the carried-over value led to an incorrect final answer.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified the corresponding terms a=2.5×10−3a = 2.5 \times 10^{-3} and b=4.0×10−4b = 4.0 \times 10^{-4}.

You correctly squared the terms in scientific notation and adjusted them to a common exponent.

You used the correct conversion factor 1 m2=106 mm21\,\text{m}^2 = 10^6\,\text{mm}^2 to convert square meters to square millimeters.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Difference of Squares Sign Error

You wrote a plus sign between the squared terms instead of a minus sign.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the difference of squares identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 with a sum of squares or misremembering the middle-term cancellation sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Identity Check

  • (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2

  • Check the sign between the squared terms: product of a sum and difference always yields a minus sign.

Middle-Term Cancellation

  • FOIL: a2−ab+ba−b2=a2−b2a^2 - ab + ba - b^2 = a^2 - b^2

  • Verify that +ab+ab and −ab-ab cancel out, leaving −b2-b^2 rather than +b2+b^2.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the difference of squares formula: Area=(2.5×10−3)2−(4.0×10−4)2\text{Area} = (2.5 \times 10^{-3})^2 - (4.0 \times 10^{-4})^2.

  2. Evaluate the squares: 6.25×10−6 m2−0.16×10−6 m2=6.09×10−6 m26.25 \times 10^{-6}\,\text{m}^2 - 0.16 \times 10^{-6}\,\text{m}^2 = 6.09 \times 10^{-6}\,\text{m}^2.

  3. Convert to square millimeters: 6.09×10−6 m2×106 mm2m2=6.09 mm26.09 \times 10^{-6}\,\text{m}^2 \times 10^6\,\frac{\text{mm}^2}{\text{m}^2} = 6.09\,\text{mm}^2.

Final answer

6.09,mm\text{mm}^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the special product patterns, noting specifically that (a+b)(a−b)(a+b)(a-b) always produces a difference (subtraction), not a sum.

  • When dealing with linear dimensions that convert easily to the target units, consider converting to millimeters before squaring to simplify the arithmetic.

OverviewACEpath Teacher's Diagnosis

03 The student squares the power of ten by doubling the base rather than multiplying the exponents, writing (10⁻⁴)² as 10⁻¹⁶.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Area\text{Area} = (2.5 ×\times 10^{-3})^2 - (4.0 ×\times 10^{-4})^2

  2. Step 2

    Area\text{Area} = 6.25 ×\times 10^{-6} - 16 ×\times 10^{-16} m\text{ m}^2

    Correct

    Area\text{Area} = 6.25 ×\times 10^{-6} - 16 ×\times 10^{-8} m\text{ m}^2 = 6.25 ×\times 10^{-6} - 0.16 ×\times 10^{-6} m\text{ m}^2

    You calculated (10−4)2(10^{-4})^2 as 10−1610^{-16} by squaring the exponent (−4)2=16(-4)^2 = 16, rather than applying the power-of-a-power rule (10a)b=10a⋅b(10^a)^b = 10^{a \cdot b}, which gives 10−4×2=10−810^{-4 \times 2} = 10^{-8}.

  3. Step 3

    Area≈6.25×10−6 m2\text{Area} \approx 6.25 \times 10^{-6}\text{ m}^2

    Correct

    Area=6.09×10−6 m2\text{Area} = 6.09 \times 10^{-6}\text{ m}^2

    Because 16×10−1616 \times 10^{-16} appeared negligibly small compared to 10−610^{-6}, you approximated the expression by dropping the second term. The problem asked for the exact surface area.

  4. Step 4

    Area\text{Area} = 6.25 ×\times 10^{-6} ×\times 10^6 mm\text{ mm}^2 = 6.25 mm\text{ mm}^2

    Correct

    Area\text{Area} = 6.09 ×\times 10^{-6} ×\times 10^6 mm\text{ mm}^2 = 6.09 mm\text{ mm}^2

    Your unit conversion method (1 m2=106 mm21\text{ m}^2 = 10^6\text{ mm}^2) was correct, but the final numerical value carried over the previous arithmetic error.

WHAT YOU DID WELL
Compliments from Your Teacher

You recognized and applied the difference of squares identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 correctly.

You correctly calculated (2.5×10−3)2=6.25×10−6 m2(2.5 \times 10^{-3})^2 = 6.25 \times 10^{-6}\text{ m}^2.

You used the correct area conversion factor of 1 m2=106 mm21\text{ m}^2 = 10^6\text{ mm}^2.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Power of a Power Rule Error

You evaluated (10−4)2(10^{-4})^2 as 10−1610^{-16} instead of 10−810^{-8}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the power of a power (xa)b(x^a)^b as raising the exponent to a power (xabx^{a^b}) rather than multiplying the exponents (xa⋅bx^{a \cdot b}). Since 10−1610^{-16} was ten orders of magnitude smaller than 10−610^{-6}, it led you to approximate away a significant term instead of computing the exact difference.

REASONING PATH
Your ACEpath teacher’s suggested approach
Power of a Power Rule

  • When raising a power to another exponent, multiply the exponents: (10m)n=10m⋅n(10^m)^n = 10^{m \cdot n}.

  • Always multiply: (10−4)2=10(−4)×2=10−8(10^{-4})^2 = 10^{(-4) \times 2} = 10^{-8}, never square the exponent (−4)2(-4)^2.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Correctly square (4.0×10−4)2=16×10−8 m2=0.16×10−6 m2(4.0 \times 10^{-4})^2 = 16 \times 10^{-8}\text{ m}^2 = 0.16 \times 10^{-6}\text{ m}^2.

  2. Subtract the two values with matching powers of ten: Area=(6.25−0.16)×10−6 m2=6.09×10−6 m2\text{Area} = (6.25 - 0.16) \times 10^{-6}\text{ m}^2 = 6.09 \times 10^{-6}\text{ m}^2.

  3. Convert to square millimeters: 6.09×10−6 m2×106mm2m2=6.09 mm26.09 \times 10^{-6}\text{ m}^2 \times 10^6\frac{\text{mm}^2}{\text{m}^2} = 6.09\text{ mm}^2.

Final answer

6.09,mm\text{mm}^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the power-of-a-power exponent law (xa)b=xab(x^a)^b = x^{ab} and contrast it with (xa)(xb)=xa+b(x^a)(x^b) = x^{a+b}.

  • Practice converting scientific notation terms to the same unit before squaring to keep the numbers simple.

  • Always keep exact values when an exact answer is requested, rather than rounding small-looking terms.

Lock it in

Question

How do you find the area conversion factor between two metric units if you know their linear relationship 1 m=10k mm1\,\text{m} = 10^k\,\text{mm}?

Answer

Square the entire relationship: 1 m2=(10k mm)2=102k mm21\,\text{m}^2 = (10^k\,\text{mm})^2 = 10^{2k}\,\text{mm}^2. For meters to millimeters, 1 m2=(103 mm)2=106 mm21\,\text{m}^2 = (10^3\,\text{mm})^2 = 10^6\,\text{mm}^2.

Similar question

01

A square sensor has side length s=3.0×10−2 ms = 3.0 \times 10^{-2}\,\text{m}. Compute its surface area in square millimeters (mm2\text{mm}^2).

Answer

Convert side length first: s=3.0×10−2 m×103 mm/m=30 mms = 3.0 \times 10^{-2}\,\text{m} \times 10^3\,\text{mm}/\text{m} = 30\,\text{mm}. Then area =s2=(30 mm)2=900 mm2= s^2 = (30\,\text{mm})^2 = 900\,\text{mm}^2 (or 9.0×10−4 m2×106=900 mm29.0 \times 10^{-4}\,\text{m}^2 \times 10^6 = 900\,\text{mm}^2).

02

A silicon plate has length L=(6.0×10−3 m+2.0×10−3 m)L = (6.0 \times 10^{-3}\,\text{m} + 2.0 \times 10^{-3}\,\text{m}) and width W=(6.0×10−3 m−2.0×10−3 m)W = (6.0 \times 10^{-3}\,\text{m} - 2.0 \times 10^{-3}\,\text{m}). Using the difference of squares identity, determine its area in square millimeters (mm2\text{mm}^2).

Answer

Let a=6.0×10−3 m=6.0 mma = 6.0 \times 10^{-3}\,\text{m} = 6.0\,\text{mm} and b=2.0×10−3 m=2.0 mmb = 2.0 \times 10^{-3}\,\text{m} = 2.0\,\text{mm}. Area =a2−b2=(6.0)2−(2.0)2=36−4=32 mm2= a^2 - b^2 = (6.0)^2 - (2.0)^2 = 36 - 4 = 32\,\text{mm}^2.

03

A thin optical film has a rectangular cross-section with dimensions L=(1.2×10−2 m+5.0×10−4 m)L = (1.2 \times 10^{-2}\,\text{m} + 5.0 \times 10^{-4}\,\text{m}) and W=(1.2×10−2 m−5.0×10−4 m)W = (1.2 \times 10^{-2}\,\text{m} - 5.0 \times 10^{-4}\,\text{m}). Using the difference of squares identity, find the exact area in square centimeters (cm2\text{cm}^2).

Answer

Convert each term to centimeters (1 m=102 cm1\,\text{m} = 10^2\,\text{cm}): a=1.2 cma = 1.2\,\text{cm} and b=0.05 cmb = 0.05\,\text{cm}. Applying a2−b2a^2 - b^2: (1.2)2−(0.05)2=1.44−0.0025=1.4375 cm2(1.2)^2 - (0.05)^2 = 1.44 - 0.0025 = 1.4375\,\text{cm}^2.

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