From E draw a segment to P. Since AE=EQ and AQ=BP, $\angle PAE = \angle PQE$, so $\triangle PAE = \triangle PQE$ (SAS). And since $\triangle PAE = \triangle PQE$,
$PE=EQ=CP=10$. Also, since $\triangle PQA = \triangle PQB$ (SAS) we must have $QA=QB=12$, so the perimeter of the pentagon is $10+10+12+10+12=54$ units, which is not what the problem stated.
But notice that $AP+PB=22$. To have perimeter 52, we need $EA+AQ+PB+BC+CP=52$, or, rearranging, $EF+FA+AB+BC+CD=30$. Since $FA=EF$, we then have $FE+2AB+BC+CD=30$. Between every two of these terms, we put in the side lengths stated above:
$FE+2AB+BC+CD=10+10+12+10+12=44$
$(10+10)+2AB+BC+CD=10+10+12+10+12=54$
Subtracting these equations gives $AB=10$. Then $FE+BC+CD=30-10=20$.
From the angles $\angle BCQ=\angle CBQ$ we know that $\triangle BCQ$ is isosceles, so $BC=DQ=10$. And from $\angle EAD = \angle ADE$ we know $\triangle ADE$ is isosceles, so $FE=CD=5$.
Now here is where we make our transition from an "illegal" problem to a legal one. We don't want five sides; we want six so we can divide it into triangles. So we draw $FD$. $\triangle FDE$ is an isosceles triangle, so by symmetry $FD=DE=5$.
$\triangle ABE$ is isosceles, with $AB=10$. Notice now that $\triangle ABF \sim \triangle EDF$; therefore $10/5=BF/5$, or $BF=10$. Now the perimeter is $10 + 10 + 10 + 5 + 5 = 40$. The two missing sides have a sum of 12, so they are 3 and 9, and the perimeter is $40+3+9=52$ as required.
The three triangles we have are a 5-5-5 equilateral triangle, a 10-10-10 equilateral triangle, and an isosceles triangle with base 10 made by '$ABF$', with $BF=10$ and $AF=5$. This third triangle I will refer to as our "special triangle," and will highlight it on the diagram below:
[asy]
size(170);
pathpen=black+linewidth(0.65);
pointpen=black;
pen d = linetype("4 4")+linewidth(0.65);
real r = 10/3;
pair A=(0,0),E=(10,0),Q=(12,0),B=(22,0), C=(16,-6), P=(22,-10), D=(10,-10), F=(5,-6);
pair[] dots = {A,E,Q,B,C,P,D,F};
D(CR(D(MP("A",A))));D(CR(D(MP("E",E))));D(CR(D(MP("Q",Q))));D(CR(D(MP("B",B))));D(CR(D(MP("C",C))));D(CR(D(MP("P",P))));MP("D",D);MP("F",F,S);
D(A--MP("B",B,W)--MP("C",C)--P--E,Nd,d);
D(A--MP("Q",Q)--MP("E",E,N),Nd,d);
D(F--MP("D",D,NW)--MP("C",C),Nd,d);
MP("10",(B+C)/2,NE);
MP("10",(C+D)/2,SW);
MP("10",(A+B)/2,NW);
MP("10",(A+Q)/2,NE);
MP("5",(D+E)/2,NE);
D(MP("5",(E+F)/2) -- F -- D--cycle);
[/asy]
The large equilateral triangle has height 5 and area $(5\sqrt{3})(5)/2=25\sqrt{3}/2$.
The smaller has height 10 and area $(10\sqrt{3})(10)/2=50\sqrt{3}$.
And the area of the special triangle is $(10)(5)/2 = 25$.
The area of the pentagon is the sum of these areas, which I leave to you.
In this figure, AE = EQ = BC = CP = 10 units, and AQ = BP = 12 units. The points A, P,
Q and B are collinear. If the perimeter of the concave pentagon ABCDE is 52 units, what is its area? Express your answer as a common fraction.
1 answer
1 answer