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
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
Answers
GPT 3.5
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.
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.