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2005 AMC 10B Problem 8

2005 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 2 of 5 · about 2 min · position P1-10

An $8$ -foot by $10$ -foot bathroom floor is tiled with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

A)
$80-20\pi$
B)
$60-10\pi$
C)
$80-10\pi$
D)
$60+10\pi$
E)
$80+10\pi$

Answer

Key insight

The four quarter circles on one tile assemble into one full circle of radius 1/2, so each tile has 1 - pi/4 shaded, and there are 80 tiles.

Solution

Look at a single tile. The four white pieces are quarter circles of the same radius $\frac{1}{2}$ , so together they have the area of one whole circle:
$\pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}.$

The tile has area $1$ , so the shaded part of one tile is $1 - \frac{\pi}{4}$ .

The floor holds $8 \cdot 10 = 80$ tiles, giving a total shaded area of
$80\left(1 - \frac{\pi}{4}\right) = 80 - 20\pi.$

The answer is $\boxed{\textbf{(A)}\ 80-20\pi}$ .

Why this works

Repeated patterns should be analyzed on one cell and then multiplied. Within the cell, congruent sectors are reassembled into a whole circle so no sector formula is needed. "Shaded equals total minus unshaded" is the standard route when the white region is the one with a simple shape.

The trap

Using radius 1 instead of 1/2 for the quarter circles, which gives 80 - 80 pi.

Common mistakes

Using radius 1 instead of 1/2 for the quarter circles, which gives 80 - 80 pi.
Computing the white area ( $20\pi$ ) or miscounting the tiles (using $8 + 10$ or $18$ instead of $80$ ).

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)