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Applications of Cross Product

In the figure...

Question

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
$\frac{A (△ A B S)}{A (△ A S C)}$

\frac{P R}{S C}

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\frac{B S}{S C}

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\frac{B S}{P R}

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\frac{A S}{S C}

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Solution

The correct option is B $\frac{B S}{S C}$
Area of triangle = $\frac{1}{2}$ base $\times$ height

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

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

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Similar questions

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

\frac{A (△ A B C)}{A (△ P B C)}

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

\frac{A (△ B P R)}{A (△ C Q T)}

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

\frac{A (△ P R C)}{A (△ B Q T)}

2 A D = A B

P

A B

Q

D R

P R ∥ B S

A Q ∥ B S

S c

=

S c^{3 +}, S c^{2 +}

S c^{+}

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