Question

Given that rectangle ABCD where AB=10 and AD=8 is folded at line AP passing through vertex A, vertex B coincides with point Q on CD. Solve the following problems.

1) Find the lengths of AQ and DQ.

2) Let PQ=x. Find the length of PQ by using the Pythagorean Theorem in triangle PQC

Answer 1

1) Since the rectangle is folded at line AP, we can consider triangle PAQ and triangle PAD.

In triangle PAD, using the Pythagorean Theorem, we have:
PA^2 + AD^2 = PD^2
PA^2 + 8^2 = 10^2
PA^2 + 64 = 100
PA^2 = 36
PA = 6

In triangle PAQ, AQ = AP - PQ = 6 - PQ
Since B coincides with Q:
AD = DQ + AQ
8 = DQ + 6 - PQ
DQ = 2 + PQ

Therefore, AQ = 6 - PQ and DQ = 2 + PQ.

2) Now, let's calculate the length of PQ using the Pythagorean Theorem in triangle PQC:

PC^2 = PQ^2 + QC^2
PC^2 = PQ^2 + QC^2
PC^2 = PQ^2 + AD^2
PC^2 = PQ^2 + 8^2

Also, in triangle DAC:
AC^2 = AD^2 + DC^2
AC^2 = 8^2 + 10^2
AC = sqrt(64+100) = sqrt(164) = 2√41

Since PC = AC - AQ:
PC = 2√41 - (6 - PQ)
PC = 2√41 - 6 + PQ

Now, we can write the equation for PC^2:
(PC)^2 = (2√41 - 6 + PQ)^2
PC^2 = 4*41 - 24√41 + 36 - 12√41 + 36 - 12PQ + PQ^2
PC^2 = 4*41 - 36√41 + 12 - 12√41 + 36 - 12PQ + PQ^2
PC^2 = 164 - 48√41 + 48 - 12PQ + PQ^2

Now, let's replace PC^2 and PQ^2 in the equation:
164 - 48√41 + 48 - 12PQ + PQ^2 = PQ^2 + 8^2
164 - 48√41 + 48 - 12PQ = 64
212 - 48√41 - 12PQ = 64
148 - 12PQ = 48√41
12PQ = 148 - 48√41
PQ = (148 - 48√41)/12
PQ = 12.85

Therefore, the length of PQ is approximately 12.85 units.