A triangle has sides 6, 7, and 8. The line through its incenter
parallel to the shortest side is drawn to meet the other two
sides at P and Q. Then find the length of the segment PQ?
Answers
Answer:
Answer
AnswerΔ=r×s ∴221×r
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar.
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ ⇒1−hr=6PQ
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ ⇒1−hr=6PQ ⇒1−72=6PQ
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ ⇒1−hr=6PQ ⇒1−72=6PQ⇒75=6PQ
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ ⇒1−hr=6PQ ⇒1−72=6PQ⇒75=6PQ⇒PQ=730
AnswerΔ=r×s ∴221×r=21b×h=26×h=3h or hr=72Now APQ and ABC are similar. Thus, hh−r=6PQ ⇒1−hr=6PQ ⇒1−72=6PQ⇒75=6PQ⇒PQ=730
Answer:
Δ=r×s ∴
2
21×r
=
2
1
b×h=
2
6×h
=3h or
h
r
=
7
2
Now APQ and ABC are similar.
Thus,
h
h−r
=
6
PQ
⇒1−
h
r
=
6
PQ
⇒1−
7
2
=
6
PQ
⇒
7
5
=
6
PQ
⇒PQ=
7
30
Step-by-step explanation:
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