Lesson goal: Index Card Geometry and Area (AMC 10A Problem 10)

Previous: Distance formula with variables | Home | Next: Changing a variable in one line

In the 2022 AMC 10A competition, Problem #10 uses the distance formula and the Pythagorean theorem to solve for the area of a rectangle without ever finding its individual side lengths:

1 1 1 1 8 4√2

Contest Problem Statement:
"Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters. Daniel then cuts out equal squares of side 1 cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters. What is the area of the original index card?"

Setting Up the Coordinate Geometry

Let the length of the card be $L$ and the width be $W$.
  1. The Diagonal Formula: By the Pythagorean theorem:
    $$L^2 + W^2 = 8^2 = 64$$
  2. The Cutout Distance: Place the card on the Cartesian plane with vertices at $(0,0), (L,0), (L,W),$ and $(0,W)$.
    Removing $1 \times 1\text{ cm}$ squares at opposite corners leaves inner vertices at:
    $$(1, W - 1) \quad \text{and} \quad (L - 1, 1)$$
    The horizontal distance between these vertices is $(L - 1) - 1 = L - 2$.
    The vertical distance between them is $(W - 1) - 1 = W - 2$.
    By the distance formula:
    $$(L - 2)^2 + (W - 2)^2 = (4\sqrt{2})^2 = 16 \times 2 = 32$$

Algebraic Reduction

Expanding the distance equation:
$$L^2 - 4L + 4 + W^2 - 4W + 4 = 32$$
$$(L^2 + W^2) - 4(L + W) + 8 = 32$$
Substitute $L^2 + W^2 = 64$:
$$64 - 4(L + W) + 8 = 32$$
$$72 - 4(L + W) = 32 \implies 4(L + W) = 40 \implies L + W = 10$$
Now square $L + W = 10$:
$$(L + W)^2 = 10^2 = 100$$
$$L^2 + 2LW + W^2 = 100$$
Substitute $L^2 + W^2 = 64$ again:
$$64 + 2LW = 100 \implies 2LW = 36 \implies \text{Area} = LW = 18$$
We found the area of the card is exactly 18 without ever needing to calculate $L$ or $W$ individually!
diag_sq = 8^2
inner_sq = (4 * math.sqrt(2))^2
sum_LW = (diag_sq + 8 - inner_sq) / 4
-- Card Area = L * W = ((L+W)^2 - diag^2) / 2: Replace ???? with (sum_LW^2 - diag_sq) / 2 area = (sum_LW^2 - diag_sq) / 2
Move the mouse over a dotted box for more information.

By relating the geometric distance formulas to algebraic identities, we solve for the area 18, matching choice (E) 18.

Now you try. Replace ???? with (L_plus_W^2 - diag_sq) / 2 and click Run to find the exact area (18 cm$^2$)! Then check Example 1 to test other cutout sizes (such as 0.5 cm or 2 cm cutouts)!

Type your code here:


See your results here: