If 'a' is the area of …
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Sia ? 4 years, 10 months ago
Let QR = b
{tex}A = \operatorname { ar } ( \Delta P Q R ){/tex}
or, {tex}A = \frac { 1 } { 2 } \times b \times P Q{/tex}
or, {tex}P Q = \frac { 2 A } { b }{/tex} ...(i)
{tex}\Delta P N Q \sim P Q R{/tex} (AA similarity)
or, {tex}\frac { P Q } { P R } = \frac { N Q } { Q R }{/tex} ...(ii)
From {tex}\Delta P Q R{/tex}
{tex}P Q ^ { 2 } + Q R ^ { 2 } = P R ^ { 2 }{/tex}
or, {tex}\frac { 4 A ^ { 2 } } { b ^ { 2 } } + b ^ { 2 } = P R ^ { 2 }{/tex}
or, {tex}P R = \sqrt { \frac { 4 A ^ { 2 } + b ^ { 4 } } { b ^ { 2 } } } = \frac { \sqrt { 4 A ^ { 2 } + b ^ { 4 } } } { b }{/tex}
Equation (ii) becomes
{tex}\frac { 2 A } { b \times P R } = \frac { N Q } { b } \text { or, } N Q = \frac { 2 A } { P R }{/tex}
Altitude {tex}N Q = \frac { 2 A b } { \sqrt { 4 A ^ { 2 } + b ^ { 4 } } }{/tex}
Hence proved.
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