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  • PA, QB and RC each is perpendicular to AC such that PA = x, RC = y QB = z. <img alt="" src="https://cdn.entrance360.com/media/uploads/2020/07/20/q9.png" style="height:159px; width:21

PA, QB and RC each is perpendicular to AC such that PA = x, RC = y QB = z.

Then,

Option: 1

\frac{1}{x}+\frac{1}{z}=\frac{1}{y}


Option: 2

\frac{1}{x}+\frac{1}{y}=\frac{1}{z}


Option: 3

\frac{1}{z}+\frac{1}{y}=\frac{1}{x}


Option: 4

None of the above


Answers (1)

best_answer

\\P A \perp A C \text { and } Q B \perp A C \Rightarrow Q B \| P A \\ \\\text {Thus, in } \triangle P A C, Q B \| P A . \\\\\text{So, }\triangle Q B C \sim \triangle P A C \text { . }

\therefore \frac{Q B}{P A}=\frac{B C}{A C} \Rightarrow \frac{z}{x}=\frac{b}{a+b} \cdot \ldots (i)\text{ [by the property of similar triangle ]}

\\\text { In } \triangle R A C, Q B \| R C . \\\\\text {So, } \triangle Q B A \sim \triangle R C A

\therefore \frac{Q B}{R C}=\frac{A B}{A C} \Rightarrow \frac{z}{y}=\frac{a}{a+b} \quad \ldots(ii)

\\\text{From (i) and (ii), we get} \\\\\frac{z}{x}+\frac{z}{y}=\left(\frac{b}{a+b}+\frac{a}{a+b}\right)=1

\\\Rightarrow \quad \frac{z}{x}+\frac{z}{y}=1 \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z} \\ \\\text { Hence, } \frac{1}{x}+\frac{1}{y}=\frac{1}{z}

Posted by

Ajit Kumar Dubey

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