Question

ABC is a right triangle right angled at C.P is the length of the perpendicular from C to AB .if AB = c BC = a and CA = b prove that 1/p²=1/a²=1/b² ​

Answer 1

Question:

ABC is a right triangle, right angled at C. If p is the length of the perpendicular from C to AB and AB =c, BC= a, and CA = b, prove that  1/p² = 1/a² + 1/b²

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Right triangle ABC
• CD perpendicular to AB
• BC= a
• CA = b
• AB = c

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

1/p² = 1/a² + 1/b²

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Area of Δ ABC = 1/2 × BC × AC

  Area of Δ ABC = 1/2 × a × b ------equation 1

→ But

  Area of Δ ABC = 1/2 × AB × DC

  Area of Δ ABC = 1/2 × c × p ------equation 2

→ From equation 1 and 2, LHS are equal, hence RHS must be equal

  1/2 × a × b = 1/2 × c × p

→ Cancelling 1/2 on both sides

  ab = cp

→ Now,

   p = ab/c

  1/p = c/ab

→ Squaring on both sides

  1/p² = c²/a²b²

→ But we know that by pythagoras theorem,

   c² = b² + a² ( AB ² = AC² + BC²)

→ Substitute this in the above equation,

  1/p² = b²+ a²/a²b²

  1/p²= b²/a²b² +  a²/a²b²

  1/p² = 1/a² + 1/b²

→ Hence proved.