ABC is a right angle triangle at C, let BC=a , CA=b and AB=c, and let p be the length of perpendicular from C on AB. prove thay 1/p^2= 1/a^2 + 1/b^2.
Answers
Answered by
1
Answer:
Let CD⊥AB. Then, CD=p
∴ Area of ΔABC=
2
1
(Base×Height)
=
2
1
(AB×CD)=
2
1
cp
Also,
Area of ΔABC=
2
1
(BC×AC)=
2
1
ab
∴
2
1
cp=
2
1
ab
⇒cp=ab
(ii) Since ΔABC is a right triangle, right angled at C.
∴AB
2
=BC
2
+AC
2
⇒c
2
=a
2
+b
2
⇒(
p
ab
)
2
=a
2
+b
2
[∵cp=ab⇒c=
p
ab
]
⇒
p
2
a
2
b
2
=a
2
+b
2
⇒
p
2
1
=
b
2
1
+
a
2
1
⇒
p
2
1
=
a
2
1
+
b
2
1
Step-by-step explanation:
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Answered by
0
Answer:
1.Area is 1/2 base * height . So for the same triangle area can be 1/2*b*a and can be 1/2*p*c. Hence cp = ab
2 . From above p2 = a2b2/c2
1/p2 = c2/ a2b2
but for right triangle c2 = a2 +b2
Hence 1/p2 = (a2+b2)/ a2b2
Hence 1/p2 = 1/a2+1/b2
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