Question

In triangle ABC, angle ACB = 90, AB = c unit, BC = a unit, AC = b unit, C perpendicular to AB and CD = p unit. Prove that : 1/p^2 = 1/a^2 + 1/b^2​

Answer 1

Step-by-step explanation:

Area of ΔABC =  1/2×base×height

Therefore, Area ΔABC = 1/2×a×b

Also, p is the height of the Δ

we can write area as

= 1/2×p×c

now, 1/2×p×c= 1/2×a×b

p= ab/c

Also, p^2=  (ab)^2/c^2

Also, c^2 = a^2 +b^2 (  ABC is right angle Δ)

therefore,

$p^2= \frac{(ab)^2}{a^2+b^2}$

$\frac{1}{p^2} = \frac{1}{a^2} +\frac{1}{b^2}$

Answer 2

Answer:

view below

Step-by-step explanation:

Area of triangle(1) =1/2×pc (1)

Area of triangle (2)= 1/2×ab (2)

pc=ab

p=ab/c (3)

Squaring both the sides

p^2=(ab)^2/c^2

c^2=a^2+b^2

p^2=(ab)^2+a^2+b^2

On dividing by a^2b^2

c^2/a^2b^2=a^2/a^2 b^2+b^2/a^2b^2

c^2/c^2p^2=1/a^2+1/b^2 (pc=ab)

1/p^2=1/a^2+1/b^2

Hence,proved