Question

ABC is a right triangle rt angled at C. If p is the length of the perpendicular from C to AB and a,b,c have their usual meanings. prove that :1by a square +1 by b square = 1 by p square.

Answer 1

hope the solution helps you...

Answer 2

Use Pythagoras Theorem to Prove

Step-by-step explanation:

• Let us suppose ABC as triangle being right angled at C, such that BC be 'a', CA be 'b' and AB be 'c'.

(i) Area of Δ ABC = $\frac {1}{2} \times Base \times Height = \frac {1}{2} \times BC \times AC = \frac {1}{2} ab$

   Area of Δ ABC = $\frac {1}{2} \times Base \times Height = \frac {1}{2} \times AB \times CD = \frac {1}{2} cp$

   ⇒ $\frac {1}{2} ab = \frac {1}{2} cp$

   ⇒ ab = cp

   Hence proved.

(ii) In right angled triangle ABC,

$AB^2 = BC^2 + AC^2$

$c^2 = a^2 + b^2$

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

$\frac {a^2b^2}{p^2} = a^2 + b^2$ -------- From proof (1)

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

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

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

$\frac {1}{p^2} = (\frac {1}{a^2} + \frac {1}{b^2})$ Proved !