2011 AMC 8 Problems/Problem 19: Difference between revisions

Latest revision as of 14:52, 2 February 2025

Problem

How many rectangles are in this figure?

$[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]$

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution 1 (Splitting It Up)

The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$

$3+5+3=\boxed{\textbf{(D)}\ 11}$

Solution 2 (Count by Reduction)

We can remove the 3 big blocks of rectangles one by one. 7 (left) + 3 (bottom) + 1 = 11 are removed in total.

~aliciawu

Video Solution by WhyMath

https://youtu.be/IEeJsGh3ltk

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 22: / Line 22: @@
 <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math>
-==Solution==
+==Solution 1 (Splitting It Up)==
 The figure can be divided into <math>7</math> sections. The number of rectangles with just one section is <math>3.</math> The number of rectangles with two sections is <math>5.</math> There are none with only three sections. The number of rectangles with four sections is <math>3.</math>

Page

Toolbox

Search