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| Line 38: |
Line 38: |
| | dot("$J$", J, SE); | | dot("$J$", J, SE); |
| | </asy> | | </asy> |
| − |
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| − | ==Video Solution for Problems 21-25=
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| − | https://youtu.be/-mi3qziCuec
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| − |
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| − | ==Video Solution==
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| − | https://youtu.be/sbBjMvq5GG4 ~savannahsolver
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| − |
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| − | Easy Solution1-<br>
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| − | let side of 1 square be= 2 (doesn't matter if we are calculating ratios)<br>
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| − | Then, total area= 3x2^2=12<br>
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| − | Shaded area=Total area - Unshaded area<br>
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| − | Unshaded area = [AJFEDA]<br>
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| − | We can now extend AD to the centre point of FG. This can be called y.<br>
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| − | △AJY[{(2+2) x 2+(2/2)}/2]=6 + [DEFY]=2x1=2<br>
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| − | Unshaded area=6+2=8<br>
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| − | Shaded area = 12-8=4 <br>
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| − | Hence, ratio= 4/12 = 1/3<br>
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| − | ~aarnavSOni
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| − |
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| − | ==Easiest Solution==
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| − |
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| − | We can see that the Pentagon is made of two congruent shapes. We can fit one triangle into the gap in the upper square. Therefore, the answer is just <math>\frac{1}{3}\implies\boxed{C}</math>
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| − |
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| − | ==Solution 1==
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| − | <asy>
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| − | pair A,B,C,D,E,F,G,H,I,J,X;
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| − | A = (0.5,2);
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| − | B = (1.5,2);
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| − | C = (1.5,1);
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| − | D = (0.5,1);
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| − | E = (0,1);
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| − | F = (0,0);
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| − | G = (1,0);
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| − | H = (1,1);
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| − | I = (2,1);
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| − | J = (2,0);
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| − | X= extension(I,J,A,B);
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| − | dot(X,red);
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| − | draw(I--X--B,red);
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| − | draw(A--B);
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| − | draw(C--B);
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| − | draw(D--A);
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| − | draw(F--E);
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| − | draw(I--J);
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| − | draw(J--F);
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| − | draw(G--H);
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| − | draw(A--J);
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| − | filldraw(A--B--C--I--J--cycle,grey);
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| − | draw(E--I);
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| − | dot("$A$", A, NW);
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| − | dot("$B$", B, NE);
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| − | dot("$C$", C, NE);
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| − | dot("$D$", D, NW);
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| − | dot("$E$", E, NW);
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| − | dot("$F$", F, SW);
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| − | dot("$G$", G, S);
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| − | dot("$H$", H, N);
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| − | dot("$I$", I, NE);
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| − | label("$X$", X,SE);
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| − | dot("$J$", J, SE);</asy>
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| − |
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| − |
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| − | First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
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| − |
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| − | ==Solution 2==
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| − |
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| − | <asy>
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| − | pair A,B,C,D,E,F,G,H,I,J,X;
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| − | A = (0.5,2);
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| − | B = (1.5,2);
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| − | C = (1.5,1);
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| − | D = (0.5,1);
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| − | E = (0,1);
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| − | F = (0,0);
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| − | G = (1,0);
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| − | H = (1,1);
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| − | I = (2,1);
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| − | J = (2,0);
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| − | X= (1.25,1);
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| − | draw(A--B);
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| − | draw(C--B);
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| − | draw(D--A);
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| − | draw(F--E);
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| − | draw(I--J);
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| − | draw(J--F);
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| − | draw(G--H);
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| − | draw(A--J);
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| − | filldraw(A--B--C--I--J--cycle,grey);
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| − | draw(E--I);
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| − | dot(X,red);
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| − | label("$A$", A, NW);
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| − | label("$B$", B, NE);
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| − | label("$C$", C, NE);
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| − | label("$D$", D, NW);
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| − | label("$E$", E, NW);
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| − | label("$F$", F, SW);
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| − | label("$G$", G, S);
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| − | label("$H$", H, N);
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| − | label("$I$", I, NE);
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| − | label("$X$", X,SW,red);
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| − | label("$J$", J, SE);</asy>
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| − |
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| − | Let the side length of each square be <math>1</math>.
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| − |
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| − | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>.
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| − |
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| − | Since <math>[ABCD]=[GHIJ]</math>, <math>AD=IJ</math>. Since <math>\angle IXJ</math> and <math>\angle AXD</math> are vertical angles, they are congruent. We also have <math>\angle JIH\cong\angle ADC</math> by definition.
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| − |
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| − | So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>.
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| − |
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| − | Since <math>C</math> and <math>D</math> are midpoints of sides, <math>DH=CJ=\dfrac{1}{2}</math>. This combined with <math>DX=JX</math> yields <math>HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}</math>.
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| − |
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| − | The area of trapezoid <math>ABCX</math> is <math>\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}</math>.
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| − |
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| − | The area of triangle <math>JIX</math> is <math>\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}</math>.
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| − |
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| − | So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>.
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| − |
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| − | The area of the <math>3</math> squares is <math>1\times 3=3</math>.
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| − |
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| − | Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
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| − |
| |
| − | ==Solution 3==
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| − |
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| − | <asy>
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| − | pair A,B,C,D,E,F,G,H,I,J,K;
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| − | A = (0.5,2);
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| − | B = (1.5,2);
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| − | C = (1.5,1);
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| − | D = (0.5,1);
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| − | E = (0,1);
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| − | F = (0,0);
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| − | G = (1,0);
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| − | H = (1,1);
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| − | I = (2,1);
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| − | J = (2,0);
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| − | K= (1.25,1);
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| − | draw(A--B);
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| − | draw(C--B);
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| − | draw(D--A);
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| − | draw(F--E);
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| − | draw(I--J);
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| − | draw(J--F);
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| − | draw(G--H);
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| − | draw(A--J);
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| − | filldraw(A--B--C--I--J--cycle,grey);
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| − | draw(E--I);
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| − | dot(K,red);
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| − | label("$A$", A, NW);
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| − | label("$B$", B, NE);
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| − | label("$C$", C, NE);
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| − | label("$D$", D, NW);
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| − | label("$E$", E, NW);
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| − | label("$F$", F, SW);
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| − | label("$G$", G, S);
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| − | label("$H$", H, N);
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| − | label("$I$", I, NE);
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| − | label("$K$", K,SW,red);
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| − | label("$J$", J, SE);</asy>
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| − |
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| − | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>.
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| − |
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| − | Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.
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| − |
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| − | Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.
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| − |
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| − | Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.
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| − |
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| − | Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.
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| − |
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| − | Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.
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| − |
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| − | Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math>
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| − |
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| − | So the shaded area now completely covers the square <math>ABCD</math>
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| − |
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| − | Set the area of a square as <math>x</math>
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| − |
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| − | Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
| |
| − |
| |
| − | ==Video Solution by OmegaLearn==
| |
| − | https://youtu.be/j3QSD5eDpzU?t=346
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| − |
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| − |
| |
| − | ==See Also==
| |
| − | {{AMC8 box|year=2013|num-b=23|num-a=25}}
| |
| − | {{MAA Notice}}
| |
, 
, and 
are equal in area. Points 
and 
are the midpoints of sides 
and 
, respectively. What is the ratio of the area of the shaded pentagon 
to the sum of the areas of the three squares?
![[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); [/asy]](http://latex.artofproblemsolving.com/1/6/9/169139bda69e81f8342f246f95cc3f636c1f9abe.png)
