ABC is a right triangle, right angled at C. let BC=a,CA=b,AB=c and let p be the length of the perpendicular frm C on AB. Prove that 1)cp=ab 2) 1/p²=1/a² + 1/b²
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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
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