Question

∆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 from C on AB.Prove that :
$\sf{1.CP=AB}$
$\sf{2.\frac{1}{ {p}^{2} } = \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } }$​

Answer 1

Answer:

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Step-by-step explanation:

Altitude: Take a Δ abc, Suppose, ab=bc=ca=a Let's drop a altitude AD from a on bc By Pythagoras theorem,we know: ab^2=1/2bd^2+ad^2 =>a^2=1/4a^2 + ad^2 =>3/4 a^2=ad^2 =>√3/4 a^2=ad=altitude Area: We know that, area of a triangle=1/2 x sinθ x product of sides containing angle suppose,side of equilateral triangle=a Then, Ar(triangle)=1/2 * sin60 * a^2 = a^2 √3/4

Answer 2

Step-by-step explanation:

Take a Δ abc, Suppose, ab=bc=ca=a Let's drop a altitude AD from a on bc By Pythagoras theorem,we know: ab^2=1/2bd^2+ad^2 =>a^2=1/4a^2 + ad^2 =>3/4 a^2=ad^2 =>√3/4 a^2=ad=altitude Area: We know that, area of a triangle=1/2 x sinθ x product of sides containing angle suppose,side of equilateral triangle=a Then, Ar(triangle)=1/2 * sin60 * a^2 = a^2 √3/4

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