Difference between revisions of "2011 AMC 8 Problems/Problem 7"

Problem

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded? $[asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Solution

Assume that the area of each square is one. Then, the area of the shade region in the top left square is 1/4. The area of the top right shaded region is 1/8. The area of the bottom left shaded region is 3/8. And the area of the bottom right shaded region is 1/4. Add the four fractions: 1/4 + 1/8 + 3/8 + 1/4 = 1. The four squares together have an area of 4, so the percent is 1/4 = 25%. Therefore the solution is $\boxed{\textbf{(C)}\ 25}$