Difference between revisions of "2011 AMC 8 Problems/Problem 7"
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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); | 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> | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | ||
+ | <math> \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} </math> | ||
==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>\ | + | 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 | ||
==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 23:30, 3 January 2023
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?
Solution
Assume that the area of each square is . Then, the area of the bolded region in the top left square is . The area of the top right bolded region is . The area of the bottom left bolded region is . And the area of the bottom right bolded region is . Add the four fractions: . The four squares together have an area of , so the percentage bolded is .
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=261
~ pi_is_3.14
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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