Question

ABC is a triangle right angled at A and p is the
length of the perpendicular from A on BC. Show that

(i) pa= bc. Hence deduce that.
(ii)
$\frac{1}{ {p}^{2} } = \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} }$
​

Answer 1

Answer:

Step-by-step explanation:

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

• Right angled triangle ABC
• AD ⊥ BC
• AD = p, AC = b, DC = a

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

• pa = bc
• 1/p² = 1/b² + 1/c²

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

⇝ Consider Δ ABC

⇝ We know that area of a triangle is given by,

    Area of a triangle = 1/2 × base × height

⇝ Hence,

    Area of Δ ABC = 1/2 × AB × AC

    Area of Δ ABC = c × b

    Area of Δ ABC = bc -----(1)

⇝ But Area of Δ ABC can also be given by,

    Area of Δ ABC = BC × AD

    Area of Δ ABC = a × p

    Area of Δ ABC = ap----(2)

⇝ From equations 1 and 2 , LHS are equal, hence RHS must also be equal.

    pa = bc

⇝ Hence proved.

⇝ We know that,

    pa = bc

    p = bc/a

    1/p = a/bc

    1/p² = a²/b²c²

⇝ But we know that by pythagoras theorem,

    a² = b² + c²

⇝ Hence,

    1/p² = (b² + c²)/b² c²

    1/p² = b²/b² c² + c²/b² c²

     1/p² = 1/c² + 1/b²

⇝ Hence proved.

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