In ΔABC seg AP is a median. If BC=18,AB²+AC²=260 Find AP.
Answers
Answered by
138
Given :
In ∆ ABC , AP is a median , BC = 18 & AB² + AC² = 260
In ∆ ABC , seg AP is a median. P is the midpoint of seg BC
BP = PC = 1/2BC
BP = PC = ½(18) = 9
AB² + AC² = 2AP² + 2PC²
[ By APOLLONIUS THEOREM, this theorem tells us the relation among the sides and medians of a triangle.]
AB² + AC² = 2(AP² + PC²)
260 = 2(AP² + 9²)
260/2 = AP² + 81
130 = AP² +81
130 - 81 = AP²
49 = AP²
AP = √ 49 = 7
AP = 7
Hence, the value of AP is 7 .
HOPE THIS WILL HELP YOU...
In ∆ ABC , AP is a median , BC = 18 & AB² + AC² = 260
In ∆ ABC , seg AP is a median. P is the midpoint of seg BC
BP = PC = 1/2BC
BP = PC = ½(18) = 9
AB² + AC² = 2AP² + 2PC²
[ By APOLLONIUS THEOREM, this theorem tells us the relation among the sides and medians of a triangle.]
AB² + AC² = 2(AP² + PC²)
260 = 2(AP² + 9²)
260/2 = AP² + 81
130 = AP² +81
130 - 81 = AP²
49 = AP²
AP = √ 49 = 7
AP = 7
Hence, the value of AP is 7 .
HOPE THIS WILL HELP YOU...
Attachments:
Answered by
6
Answer:
AP = 7
Step-by-step explanation:
Hence value of AP = 7
- HOPE THIS ANSWER HELPED YOU
Attachments:
Similar questions