Problem:
Square pyramid ABCDE has base ABCD, which measures 3cm on a side, and altitude AE perpendicular to the base, which measures 6cm. Point P lies on BE, one third of the way from B to E; point Q lies on DE, one third of the way from D to E; and point R lies on CE, two thirds of the way from C to E. What is the area, in square centimeters, of △PQR ?
Answer Choices:
A. 232
B. 233
C. 22
D. 23
E. 32 Solution:
Using the given data, we can label the points A(0,0,0),B(3,0,0),C(3,3,0),D(0,3,0), and E(0,0,6). We can also find the points P=B+31BE=(3,0,0)+31(−3,0,6)=(3,0,0)+(−1,0,2)=(2,0,2). Similarly, Q=(0,2,2) and R=(1,1,4).
Using the distance formula, PQ=(−2)2+22+02=22,PR=(−1)2+12+22=6, and QR=12+(−1)2+22=6. Using Heron's formula, or by dropping an altitude from P to find the height, we can then find that the area of △PQR is (C)22.
OR
As in Solution 1, let A=(0,0,0),B=(3,0,0),C=(3,3,0),D=(0,3,0), and E=(0,0,6), and calculate the coordinates of P,Q, and R as P=(2,0,2),Q=(0,2,2),R=(1,1,4). Now notice that the plane determined by △PQR is perpendicular to the plane determined by ABCD. To see this, consider the bird's-eye view, looking down upon P,Q, and R projected onto ABCD :
Additionally, we know that PQ is parallel to the plane determined by ABCD, since P and Q have the same z-coordinate. Hence the height of △PQR is equal to the z-coordinate of R minus the z-coordinate of P, giving 4−2=2. By the distance formula, PQ=22, so the area of △PQR is 21⋅22⋅2=(C)22.
OR
By the Pythagorean Theorem, we can calculate EB=ED=35,EC=36,ER=6, and EP=EQ=25. Now by the Law of Cosines in △BEC, we have cos(∠BEC)=2⋅EB⋅ECEB2+EC2−BC2=305.
Similarly, by the Law of Cosines in △EPR, we have PR2=ER2+EP2−2⋅ER⋅EP⋅cos(∠BEC)=6, so PR=6. Observe that △ERP≅△ERQ (by side-angle-side), so QR=PR=6.
Next, notice that PQ is parallel to DB, and therefore △EQP is similiar to △EDB. Thus we have DBQP=EBEP=32. Since DB=32, this gives PQ=22.
Now we have the three side lengths of isosceles △PQR:PR=QR=6,PQ=22. Letting the midpoint of PQ be S,RS is the perpendicular bisector of PQ, and so can be used as a height of △PQR (taking PQ as\
the base). Using the Pythagorean Theorem again, we have RS=PR2−PS2=2, so the area of △PQR is 21⋅PQ⋅RS=21⋅22⋅2= (C) 22.
