Question

Q1 ABC, is right angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the sides opposite ∠A,∠B, ∠C respectively then prove that 1/p^2 =1/a^2 +1/b^2​

Answer 1

❥︎Question :-

• ABC, is right angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the sides opposite ∠A,∠B, ∠C respectively then prove that 1/p^2 =1/a^2 +1/b^2.

❥︎Answer :-

❥︎Given :-

• ABC, is right angled at C.
• p is the length of the perpendicular from C to AB
• a, b, c are the lengths of the sides opposite ∠A,∠B, ∠C respectively.
• It implies, AB = c, BC = a, CA = b.

❥︎To prove :-

$\bf \:\dfrac{1}{ {p}^{2} } = \dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2} }$

❥︎Concept used :-

Pythagoras Theorem

$\begin{gathered}{\boxed{\bf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

❥︎Solution :-

⟶ In right triangle ABC

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

$\bf\implies \:Area \: of \: triangle = \dfrac{1}{2} \times BC \times CA$

$\bf\implies \:Area \: of \: triangle = \dfrac{1}{2} \times a \times b \: ⟶ \: (1)$

❥︎Also, In triangle ABC,

$\bf \:Area \: of \: triangle = \dfrac{1}{2} \times AB \times p$

$\bf\implies \:Area \: of \: triangle = \dfrac{1}{2} \times c \times p \: ⟶ \: (2)$

❥︎On equating equation (1) and (2), we get

$\bf\implies \:\dfrac{1}{2} \times a \times b = \dfrac{1}{2} \times c \times p$

$\bf \:⟶ \: ab = pc$

$\bf\implies \:c = \dfrac{ab}{p} \: ⟶ \: (3)$

❥︎ In right triangle ABC,

⟶ Using pythagoras theorem,

$\bf \: {CA}^{2} + {BC}^{2} = {AB}^{2}$

$\bf\implies \: {a}^{2} + {b}^{2} = {c}^{2}$

⟶ On substituting the value of c from eq(3), we get

$\bf\implies \: {a}^{2} + {b}^{2} = ({\dfrac{ab}{p} })^{2}$

$\bf\implies \: {a}^{2} + {b}^{2} = \dfrac{ {a}^{2} {b}^{2} }{ {p}^{2} }$

⟶ Divide both sides by a²b², we get

$\bf\implies \:\dfrac{ {a}^{2} }{ {a}^{2} {b}^{2} } + \dfrac{ {b}^{2} }{ {a}^{2} {b}^{2} } = \dfrac{ {a}^{2} {b}^{2} }{ {p}^{2} {a}^{2} {b}^{2} }$

$\bf\implies \:\dfrac{1}{ {b}^{2} } + \dfrac{1}{ {a}^{2} } = \dfrac{1}{ {p}^{2} }$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

_________________________________________