In the figure, $s e g P R ⊥ s e g B C, s e g A S ⊥ s e g B C a n d s e g Q T ⊥ s e g B C$ , then find the ratio $\frac{A (△ A B C)}{A (△ P B C)}$ .

\frac{A S}{P R}

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\frac{P R}{Q T}

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\frac{A S}{B C}

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\frac{P T}{Q R}

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Solution

The correct option is D $\frac{A S}{P R}$
Area of triangle = $\frac{1}{2}$ base $\times$ height

$\frac{A (△ A B C)}{A (△ P B C)} = \frac{\frac{1}{2} \times A S \times B C}{\frac{1}{2} \times P R \times B C}$

$\frac{A (△ A B C)}{A (△ P B C)} = \frac{A S}{P R}$

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s e g P R ⊥ s e g B C, s e g A S ⊥ s e g B C a n d s e g Q T ⊥ s e g B C

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s e g P R ⊥ s e g B C, s e g A S ⊥ s e g B C a n d s e g Q T ⊥ s e g B C

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