If D -1-2-5-2 - E 7-3 and F 7-2-7-2 are the midpoints of sides of - ABC- find the area of the - ABC-

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Distance Formula

If D -1/2,5...

Question

If D $(\frac{- 1}{2}, \frac{5}{2})$ , E (7,3) and F $(\frac{7}{2}, \frac{7}{2})$ are the midpoints of sides of $Δ$ ABC, find the area of the $Δ$ ABC.

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Solution

$L e t t h e m i d p o i n t s o f t h e g i v e n Δ A B C a r e D, E & F a n d l e t t h e Δ D E F h a v e v e r t i c e s D (x_{1}, y_{1}) = (- \frac{1}{2}, \frac{5}{2}), E (x_{2}, y_{2}) = (7, 3) & F (x_{3}, y_{3}) = (\frac{7}{2}, \frac{7}{2}) . ∴ A p p l y i n g t h e a r e a f o r m u l a o f a t r i a n g l e w e g e t a r Δ D E F = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] . ∴ a r Δ D E F = \frac{1}{2} [- \frac{1}{2} (3 - \frac{7}{2}) + 7 (\frac{7}{2} - \frac{5}{2}) + \frac{7}{2} (\frac{5}{2} - 3)] s q . u n i t s . = \frac{11}{4} s q . u n i t s . N o w t h e a r e a o f t h e Δ D E F j o i n i n g t h e m i d - p o i n t s o f a Δ A B C = \frac{1}{4} a r Δ A B C . ⟹ a r Δ A B C = 4 a r Δ D E F = 4 \times \frac{11}{4} s q . u n i t s . = 11 s q . u n i t s . A n s - a r Δ A B C = 11 s q . u n i t s .$

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If D $(\frac{- 1}{2}, \frac{5}{2})$ , E (7,3) and F $(\frac{7}{2}, \frac{7}{2})$ are the midpoints of sides of $Δ$ ABC, find the area of the $Δ$ ABC.