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Question
In triangle ABC, ∠C=90°. Let BC= a, CA= b, AB= c and let 'p' be the length of the perpendicular from 'C' on AB, prove that:
1. cp = ab
2. `1/p^2=1/a^2+1/b^2`
Solution
1. Area of a triangle = (1/2) x Base x Height
A(ΔABC) = (1/2) x AB x CD
A(ΔABC) = (1/2) x cp .......(i)
Area of right angle triangle ABC = A(ΔABC) = (1/2) x AC x BC
A(ΔABC) = (1/2) x ba ........(ii)
From (i) and (ii)
cp=ba⇒ cp⇒ab ..........(iii)
2. We have,
cp=ab ..........From(iii)
p = ab/c
Square both sides of the equation.
We get, `p^2=(a^2b^2)/c^2`
`1/p^2=c^2/(a^2b^2)" ..................(iv).....[By invertendo]"`
In right angled triangle ABC,
AB2 = AC2 + BC2 ................[By Pythagoras’ theorem]
c2 = b2 + a2 ............(v)
`c^2/(a^2b^2) = b^2/(a^2b^2) + a^2/(a^2b^2)`..........[Dividing throughout by `a^2b^2`]
`c^2/(ab)^2 = 1/a^2 + 1/b^2` .........(iii)
`c^2/(cp)^2 = 1/a^2 + 1/b^2` ...........[From (ii) and (iii)]
`c^2/(c^2p^2) = 1/a^2 + 1/b^2`
`1/p^2 = 1/a^2 + 1/b^2`
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