# 2012 AMC 8 Problems/Problem 5

## Problem

In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters? $[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("X",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]$ $\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$

## Solution

The figure is the same height on both sides, so the sum of the lengths contributing to the height on the left side will equal the sum of the lengths contributing to the height on the left side. $1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$

Thus, the answer is $\boxed{\textbf{(E)}\ 5}$.

## Video Solution

https://youtu.be/m4g-Nmot-c8 ~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 