Question

ABC is a right angled at C. If p is the length of the perpendicular from C to AB and a.b,c are the length
of the sides opposite angle a ,angle b, angle c respectively, then prove that
Solve it quickly... ​

Answer 1

Solution :-

According to the given information draw a figure.

[ Refer the attachment ]

Given :-

→ Δ ABC is Right angled triangle.

→ ∠C = 90°

→ BC = a

→ AC = b

→ AB = c

→ Perpendicular CD = p is drawn from C to AB

To prove :-

1 / p² = ( 1 / a² ) + ( 1 / b² )

Proof :-

Finding ar( ΔABC ) when AB is considered as the the base

$ar( \triangle ABC) = \dfrac{1}{2} \times base \times height \\ \\ \\ \implies ar( \triangle ABC ) = \dfrac{1}{2} \times AB \times CD \\ \\ \\ \implies ar( \triangle ABC) = \dfrac{1}{2} cp \rightarrow eq(1)$

Finding the area of the triangle when BC is considered as the base.

$ar( \triangle ABC) = \dfrac{1}{2} \times base \times height \\ \\ \\ \implies ar( \triangle ABC ) = \dfrac{1}{2} \times BC \times AC \\ \\ \\ \implies ar( \triangle ABC) = \dfrac{1}{2} ab \rightarrow eq(1)$

$\text{From eq(1) and eq(2)} \\ \\ \\ \implies \dfrac{1}{2} cp = \dfrac{1}{2} ab \\ \\ \\ \implies cp = \dfrac{1}{2}ab \times 2 \\ \\ \\ \implies cp = ab \\ \\ \\ \implies c = \dfrac{ab}{p} \rightarrow eq(3)$

ΔABC is a right angled triangled triangle

By Pythagoras theorem

⇒ AB² = BC² + AB²

⇒ c² = a² + b²

$\implies \bigg(\dfrac{ab}{p} \bigg)^{2} = {a}^{2} + {b}^{2} \qquad \{ \because from \ eq(3) \} \\ \\ \\ \implies \dfrac{ {(ab)}^{2} }{ {p}^{2} } = {a}^{2} + {b}^{2} \\ \\ \\ \implies \dfrac{ {a}^{2} {b}^{2} }{ {p}^{2} } = {a}^{2} + {b}^{2} \\ \\ \\ \implies \dfrac{1}{ {p}^{2} } = \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} {b}^{2} } \\ \\ \\ \implies \dfrac{1}{ {p}^{2} } = \dfrac{ {a}^{2} }{ {a}^{2} {b}^{2} } + \dfrac{ {b}^{2} }{ {a}^{2} {b}^{2} }$

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

Hence proved.

Answer 2

AnswEr :

• Given :

[ Check the Given Attachment ]

• To Proof :

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

• Prove :

In ∆ABC, there is two Right Angle.

• First at ∠C.
• Second at ∠E $\perp$ on AB.

⋆ Area of ∆ ABC considering BC as Base ;

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times Base \times Height}$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times BC \times AC }$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times a \times b}$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{ab}{2}} \: \: \: \frac{ \: \: \: \: }{} \frak{eq(1)}$

⋆ Area of ∆ ABC considering AB as Base ;

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times Base \times Height}$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times AB \times CE}$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{1}{2} \times c \times p}$

$\longrightarrow \sf{Ar. \: of \: (\triangle ABC) = \dfrac{cp}{2}} \: \: \: \frac{ \: \: \: \: }{} \frak{eq(2)}$

⋆ From eq(1) and eq(2), we get :

$\leadsto \sf{ \dfrac{ab}{ \cancel2} = \dfrac{cp}{ \cancel2} }$

$\leadsto \sf{ab = cp}$

$\leadsto \sf{c = \dfrac{ab}{p} }$

◑ In ∆ ABC, By Pythagoras Theorem :

$\leadsto \sf{(AB)^{2} = {(BC)}^{2} + {(AC)}^{2} }$

$\leadsto \sf{(c)^{2} = {(a)}^{2} + {(b)}^{2} }$

$\leadsto \sf{ \bigg( \dfrac{ab}{p} \bigg)^{2} = {(a)}^{2} + {(b)}^{2} }$

$\leadsto \sf{ \dfrac{a^{2} {b}^{2} }{p^{2} } = {a}^{2} + {b}^{2} }$

$\leadsto \sf{ \dfrac{1 }{p^{2} } = \dfrac{{a}^{2} + {b}^{2}}{a^{2} {b}^{2}}}$

$\leadsto \sf{ \dfrac{1 }{p^{2} } = \dfrac{ \cancel{{a}^{2}}}{ \cancel{a^{2}} {b}^{2}}} +\dfrac{\cancel{{b}^{2}}}{a^{2} \cancel{{b}^{2}}}$

$\leadsto \sf{ \dfrac{1}{ {p}^{2} } = \dfrac{1}{ {b}^{2} } + \dfrac{1}{ {a}^{2} } }$

$\boxed{\leadsto \large\sf{ \dfrac{1}{ {p}^{2} } = \dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2} } }}$