Question-217495

Question Number 217495 by Hanuda354 last updated on 14/Mar/25

Answered by mr W last updated on 17/Mar/25

Commented by Hanuda354 last updated on 18/Mar/25

Thanks a lot, Prof .

Commented by mr W last updated on 17/Mar/25

e q n . o f P R :

y = \frac{7 x}{2 (a - 3)}

e q n . o f S W :

y = a - \frac{x}{2}

y_{Q} = a - \frac{x_{Q}}{2} = \frac{7 x_{Q}}{2 (a - 3)}

\Rightarrow x_{Q} = \frac{2 a (a - 3)}{(a + 4)}

A = \frac{a}{2} \times \frac{2 a (a - 3)}{(a + 4)} = \frac{a^{2} (a - 3)}{a + 4}

A + B = \frac{(2 a - 6) (a + a - 7)}{2} = (a - 3) (2 a - 7)

\frac{A + B}{A} = 1 + \frac{B}{A} = 1 + \frac{41}{50} = \frac{91}{50}

\frac{(a - 3) (2 a - 7)}{\frac{a^{2} (a - 3)}{a + 4}} = \frac{91}{50}

9 a^{2} + 50 a - 1400 = 0

a = \frac{- 50 + \sqrt{50^{2} + 4 \times 9 \times 1400}}{18} = 10

S W = \sqrt{a^{2} + {(2 a)}^{2}} = 10 \sqrt{5}

\frac{S Q}{S W} = \frac{x_{Q}}{2 a} = \frac{a - 3}{a + 4} = \frac{10 - 3}{10 + 4} = \frac{1}{2}

Q W = S Q = \frac{10 \sqrt{5}}{2} = 5 \sqrt{5}

\frac{P Q}{P R} = \frac{x_{Q}}{2 a - 6} = \frac{a}{a + 4} = \frac{10}{14} = \frac{5}{7}

\Rightarrow \frac{Q R}{P R} = \frac{2}{7}

\frac{a}{a} \times \frac{x}{y} \times \frac{Q R}{R P} = 1

\frac{x}{y} \times \frac{2}{7} = 1

\Rightarrow \frac{x}{y} = \frac{7}{2}

\Rightarrow \frac{x}{x + y} = \frac{7}{9}

\Rightarrow x = \frac{7 \times Q W}{9} = \frac{7 \times 5 \sqrt{5}}{9} = \frac{35 \sqrt{5}}{9} ✓

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