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

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==Solution==
 
==Solution==
 
Assume that the area of each square is <math>1</math>.  Then, the area of the bolded region in the top left square is <math>\dfrac{1}{4}</math>.  The area of the top right bolded region is <math>\dfrac{1}{8}</math>.  The area of the bottom left bolded region is <math>\dfrac{3}{8}</math>.  And the area of the bottom right bolded region is <math>\dfrac{1}{4}</math>.  Add the four fractions:  <math>\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1</math>.  The four squares together have an area of <math>4</math>, so the percentage bolded is <math>\dfrac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}</math>.
 
Assume that the area of each square is <math>1</math>.  Then, the area of the bolded region in the top left square is <math>\dfrac{1}{4}</math>.  The area of the top right bolded region is <math>\dfrac{1}{8}</math>.  The area of the bottom left bolded region is <math>\dfrac{3}{8}</math>.  And the area of the bottom right bolded region is <math>\dfrac{1}{4}</math>.  Add the four fractions:  <math>\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1</math>.  The four squares together have an area of <math>4</math>, so the percentage bolded is <math>\dfrac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}</math>.
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/j3QSD5eDpzU?t=261
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/HrwrL26vd2o
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=6|num-a=8}}
 
{{AMC8 box|year=2011|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:04, 18 November 2024

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] $\textbf{(A) }12\frac{1}{2} \qquad\textbf{(B) }20 \qquad\textbf{(C) }25 \qquad\textbf{(D) }33\frac{1}{3} \qquad\textbf{(E) }37\frac{1}{2}$

Solution

Assume that the area of each square is $1$. Then, the area of the bolded region in the top left square is $\dfrac{1}{4}$. The area of the top right bolded region is $\dfrac{1}{8}$. The area of the bottom left bolded region is $\dfrac{3}{8}$. And the area of the bottom right bolded region is $\dfrac{1}{4}$. Add the four fractions: $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1$. The four squares together have an area of $4$, so the percentage bolded is $\dfrac{1}{4} \cdot 100 = \boxed{\textbf{(C)}\ 25}$.

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=261

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/HrwrL26vd2o

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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