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 - Geometry Mathematics 2

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Sum

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`

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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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2013-2014 (March)

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