LiAuto-DSR/GeoGramBench huggingface dataset:
In this diagram, not drawn to scale, Figures $I$ and $III$ are equilateral triangular regions with respective areas of $32\sqrt{3}$ and $8\sqrt{3}$ square inches. Figure $II$ is a square region with area $32$ square inches. Let the length of segment $AD$ be decreased by $12\tfrac{1}{2}$ % of itself, while the lengths of $AB$ and $CD$ remain unchanged. The decrease in the area of the square is:

[asy] draw((0,0)--(10,20*sqrt(3)/2)--(20,0)--cycle,black+linewidth(.75)); draw((20,0)--(20,12)--(32,12)--(32,0)--cycle,black+linewidth(.75)); draw((32,0)--(37,10*sqrt(3)/2)--(42,0)--cycle,black+linewidth(.75)); MP("I",(10,0),N);MP("II",(26,0),N);MP("III",(37,0),N); MP("A",(0,0),S);MP("B",(20,0),S);MP("C",(32,0),S);MP("D",(42,0),S); [/asy]

Answer: $24$
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the $A$ is:

[asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]

Answer: $7$
In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoint of $BF$. The length of $EC$ divided by the length of $BE$ is

[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (4,0); E = (0,0); F = (1.14,2.14); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); [/asy]

Answer: $3$
Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE, AD$, and $BC$ at points $M, P$, and $Q$ respectively. The length of $MQ$ divided by the length of $PM$ is

[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); [/asy]

Answer: $\frac{19}{5}$
In the adjoining figure $ABCD$ is a square and $CMN$ is an equilateral triangle. If the area of $ABCD$ is $4$ square inch, then the area of $CMN$ in square inches is

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label("A", (0,0), S); label("B", (1,0), S); label("C", (1,1), N); label("D", (0,1), N); label("M", (0,.76), W); label("N", (.82,0), S);[/asy]

Answer: $8\sqrt{3}-12$
In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=8$ and $T'Q=18$ then $r$ is

[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(O--T2); label("$r$",midpoint(O--T2),NW); draw(T--P); label("8",midpoint(T--P),N); draw(T1--Q); label("18",midpoint(T1--Q),S); draw(P--Q);[/asy]

Answer: $12$
In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $8$, then the area of the triangle $QPO$ is equal to

[asy] size((400)); draw((0,0)--(5,0)--(6,3)--(1,3)--cycle); draw((6,3)--(-5,0)--(10,0)--(1,3)); label("A", (0,0), S); label("B", (5,0), S); label("C", (6,3), NE); label("D", (1,3), NW); label("P", (10,0), E); label("Q", (-5,0), W); label("M", (.5,1.5), NW); label("N", (5.65, 1.5), NE); label("O", (3.4,1.75));[/asy]

Answer: $9$
In the adjoining figure triangle $ABC$ is such that $AB = 4$ and $AC = 8$. IF $M$ is the midpoint of $BC$ and $AM = 3$, what is the length of $BM$?

[asy] draw((-4,0)--(4,0)--(-1,4)--cycle); draw((-1, 4)--(0, 0.00001)); label("B", (-4,0), S); label("C", (4,0), S); label("A", (-1, 4), N); label("M", (0, 0.0001), S); [/asy]

Answer: $\sqrt{31}$
In $\triangle ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC, AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{GF}{EG}$ equals

[asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy]

Answer: $\frac{2}{3}$
In the adjoining figure, circle $K$ has diameter $AB$; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$; and circle $M$ tangent to circle $K$, to circle $L$ and $AB$. The area of circle $L$ divided by the area of circle $M$ is

[asy] size(150); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); label("$A$", A, W); label("$K$", K, S); label("$B$", B, E); label("$L$", L); label("$M$", M); [/asy]

Answer: $4$
Each of the three circles in the adjoining figure is externally tangent to the other two, and each side of the triangle is tangent to two of the circles. If each circle has radius two, then the perimeter of the triangle is

[asy] size(120); real t = 2/sqrt(3); real x = 1 + sqrt(3); pair A = t*dir(90), D = x*A; pair B = t*dir(210), E = x*B; pair C = t*dir(330), F = x*C; draw(D--E--F--cycle); draw(Circle(A, 1)); draw(Circle(B, 1)); draw(Circle(C, 1)); [/asy]

Answer: $12+12\sqrt{3}$
In $\triangle ABC, AB = 10, AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The radius of circle P is

[asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); [/asy]

Answer: $2.4$
If rectangle ABCD has area 120 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is

[asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); [/asy]

Answer: $30$
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$. What is the measure of $\measuredangle DEB$ in degrees?

[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); [/asy]

Answer: $45$
The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length two. Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$.

[asy] size(150); import patterns; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; add("hatch",hatch()); draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; aux=midpoint(AA--BB); draw(BB--DD); P=midpoint(AA--aux); aux=midpoint(CC--DD); Q=midpoint(CC--aux); draw(AA--CC,dashed); dot(P); dot(Q); fill(DD--BB--CC--cycle,pattern("hatch")); label("$A$",AA,W); label("$B$",BB,S); label("$C$",CC,E); label("$D$",DD,N); label("$P$",P,S); label("$Q$",Q,E); [/asy]

Answer: $\sqrt{2}$
Circles with centers $A, B$, and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals

[asy] import cse5; pathpen=black; pointpen=black; dotfactor=3; pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); [/asy]

Answer: $1+\sqrt{3(r^2-1)}$
In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $2$ and $\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals

[asy] size(200); import cse5; pathpen=black; anglefontpen=black; pointpen=black; anglepen=black; dotfactor=3; pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE; EE=(B+C)/2; D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle); D(MP("$E$",EE,N)--MP("$D$",D,S)); D(D);D(EE); MA("80^\circ",8,D,EE,C,0.1); MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C); draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow()); MA("100^\circ",8,A,B,C,0.1,0); MA("60^\circ",8,C,A,B,0.1,0); [/asy]

Answer: $\frac{\sqrt{3}}{2}$
In the adjoining figure, CDE is an equilateral triangle and ABCD and DEFG are squares. The measure of $\angle ADE$ is

[asy] size(100);defaultpen(linewidth(0.7)+fontsize(10)); pair D=(0,0), C=D+dir(230), E=D+dir(310), F=E+dir(40), G=D+dir(40), A=D+dir(140), B=C+dir(140); draw(E--D--G--F--E--C--D--A--B--C); pair point=(0,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(-15)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G));[/asy]

Answer: $150$
If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PC$ divided by the length of $BQ$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(Q,1)); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$D$", D, S); label("$P$", P, S); label("$Q$", Q, SE); label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy]

Answer: $\frac{2\sqrt{3}}{3}$
Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 6, 8, 24, and 26, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(A)); label("$B$", B, dir(B)); label("$C$", C, dir(C)); label("$D$", D, dir(D)); label("$6$", A--B, dir(A--B)*dir(-90)); label("$8$", B--C, dir(B--C)*dir(-90)); label("$24$", C--D, dir(C--D)*dir(-90)); label("$26$", D--A, dir(D--A)*dir(-90));[/asy]

Answer: $144$
In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$, and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$. The area of quadrilateral $FDCE$ divided by the area of triangle $BDF$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy]

Answer: $5$
In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, and $BN\perp AN$. If sides $AB$ and $AC$ have lengths $28$ and $38$, respectively, then find $MN$.

[asy] size(150); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(42), C=intersectionpoint(B--(30,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(30)); label(rotate(angle(dir(A--C)))*"$38$", A--C, dir(A--C)*dir(90)); label(rotate(angle(dir(A--B)))*"$28$", A--B, dir(A--B)*dir(90)); [/asy]

Answer: $5$
In $\triangle ABC$ in the adjoining figure, $AD$ and $AE$ trisect $\angle BAC$. The lengths of $BD$, $DE$ and $EC$ are $4$, $6$, and $12$, respectively. The length of the shortest side of $\triangle ABC$ is

[asy] defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,S); label("$4$",midpoint(B--D),N); label("$6$",midpoint(D--E),NW); label("$12$",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]

Answer: $4\sqrt{10}$
In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=6, BC=12$, and $AC=9$, then the perimeter of $\triangle AMN$ is

[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0)); draw(B--M--O--B--C--O--N--C^^N--A--M); label("$A$", A, dir(90)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$M$", M, dir(90)*dir(B--A)); label("$N$", N, dir(90)*dir(A--C)); label("$O$", O, dir(90));[/asy]

Answer: $15$
In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC=2$. The length of $CN$ is

[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M); draw(M--C--A--B--C^^B--N); pair point=P; markscalefactor=0.01; draw(rightanglemark(B,C,N)); draw(rightanglemark(C,P,B)); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(C--A)*dir(90)); label("$2$", B--C, NW); [/asy]

Answer: $\sqrt{2}$
In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$. If the radius of the largest circle is $18$ and that of the smallest one is $8$, then the radius of the second circle is

[asy] size(250);defaultpen(linewidth(0.7)); real alpha=5.797939254, x=71.191836; int i; for(i=0; i<5; i=i+1) { real r=8*(sqrt(6)/2)^i; draw(Circle((x+r)*dir(alpha), r)); x=x+2r; } real x=71.191836+40+20*sqrt(6), r=18; pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2); pair A1=300*dir(origin--A), B1=300*dir(origin--B); draw(B1--origin--A1); pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X, Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y, Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z; clip(X--Y--Y1--X1--cycle); label("$L_2$", Z, S); label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]

Answer: $4\sqrt{6}$
Triangle $\triangle ABC$ in the figure has area $20$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$, are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B); draw(A--B--C--A--E--F--D); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$2$", (2,0), S); label("$3$", (7,0), S);[/asy]

Answer: $12$
In an arcade game, the "monster" is the shaded sector of a circle of radius $2$ cm, as shown in the figure. The missing piece (the mouth) has central angle $\usepackage{gensymb} 60\degree$. What is the perimeter of the monster in cm?

[asy] size(100); defaultpen(linewidth(0.7)); filldraw(Arc(origin,1,30,330)--dir(330)--origin--dir(30)--cycle, yellow, black); label("2", (sqrt(3)/4, 1/4), NW); label("$60^\circ$", (1,0));[/asy]

Answer: $\frac{10}{3}\pi+4$
Pegs are put in a board $2$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is

[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]

Answer: $24$
Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $2$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$. The area of $ABCD$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D); real r=degrees(B); pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90); markscalefactor=0.02; draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B)^^rightanglemark(N1,E,B)); pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); label("A", A, NE); label("B", B, NE); label("C", C, dir(0)); label("D", D, dir(180)); label("$L$", (x/2,0), SW); label("$L^\prime$", C, SW); label("2", D--F, NW); label("2", F--E, SE); label("2", E--B, SE); clip(W--X--Y--Z--cycle);[/asy]

Answer: $12\sqrt{2}$
In $\triangle ABC$, we have $\angle C = 3\angle A$, $a = 27$ and $c = 48$. What is $b$?

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(12,0), C=(9,5); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]

Answer: $35$
Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches. How high is the table?

[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]

Answer: $30$
A triangular corner with side lengths $DB=EB=2$ is cut from equilateral triangle ABC of side length $6$. The perimeter of the remaining quadrilateral is

[asy] draw((0,0)--(2,0)--(2.5,.87)--(1.5,2.6)--cycle, linewidth(1)); draw((2,0)--(3,0)--(2.5,.87)); label("6", (0.75,1.3), NW); label("2", (2.5, 0), S); label("2", (2.75,.44), NE); label("A", (1.5,2.6), N); label("B", (3,0), S); label("C", (0,0), W); label("D", (2.5,.87), NE); label("E", (2,0), S); [/asy]

Answer: $16$
If $\angle \text{CBD}$ is a right angle, then this protractor indicates that the measure of $\angle \text{ABD}$ is

[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]

Answer: $140$
Four rectangular paper strips of length $10$ and width $2$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?

[asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(4,5)--(5,5)--(5,4)--(4,4)); [/asy]

Answer: $64$
An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below. The owners desire to move the table to the position shown in Figure $2$. The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); label("Figure 2", (E+F)/2, S); label("10'", (I+J)/2, S); label("8'", (12,12)); label("8'", (L+M)/2, S); label("10'", (42,11)); label("table", (5,12)); label("table", (36,11)); [/asy]

Answer: $9\sqrt{2}$
$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac{1}{6}$ the altitude of $\triangle ABC$. The area of $\triangle ABC$ divided by the area of $\triangle A'B'C'$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]

Answer: $4$
Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$.

[asy] defaultpen(fontsize(10)+0.8); size(175); pair A,B,C,D,M,P,Q; C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P=midpoint(A--B); Q=midpoint(C--D); draw(B--C--D--B--A--C^^A--D); draw(D--P--C^^P--Q, gray+dashed+0.5); pen p=fontsize(12)+linewidth(3); dot("$A$",A,down,p); dot("$B$",B,right,p); dot("$C$",C,left,p); dot("$D$",D,up,p); dot("$M$",P,dir(-45),p); dot("$N$",Q,0.2*(Q-P),p); label("$27$",B--D,2*dir(30),fontsize(10)); label("$7$",A--C,2*dir(210),fontsize(10)); label("$18$",A--D,1.5*dir(30),fontsize(10)); label("$36$",(3,0),up,fontsize(10)); [/asy]

Answer: $137$
The side length of the square is 4, what is the shaded area of the square?

[asy] draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,2)--(2,2)--(2,0)); draw((0,1)--(1,1)--(1,0)); draw((0,0)--(3,3)); fill((0,0)--(0,1)--(1,1)--cycle,grey); fill((1,0)--(1,1)--(2,2)--(2,0)--cycle,grey); fill((0,2)--(2,2)--(3,3)--(0,3)--cycle,grey); [/asy]

Answer: $8$
The area of this figure is $400\text{ cm}^2$. Its perimeter is

[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]

Answer: $100$
Each corner of a rectangular prism is cut off. Two (of the eight) cuts are shown. How many edges does the new figure have?  <i>Assume that the planes cutting the prism do not intersect anywhere in or on the prism.</i>

[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewidth(1)); [/asy]

Answer: $36$