Check it (Question:251253.0)

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TRINA DAS

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· 1 Months ago

Check it (Question:251253.0)

Trapezium ABCD may not be isosceles.

Question:

In a trapezium ABCD a line EF parallel to AB is drawn in such a way that CF = 5 cm and FB = 4 cm. If AB = 3/2 CD = 6 cm, then find the length of EF.

Options:

$4 \frac{8}{9}$ cm

$5 \frac{1}{9}$ cm

$2 \frac{2}{9}$ cm

$4 \frac{2}{3}$ cm

Solution:

$\frac{F^{1} F}{C^{1} B} = (\frac{5}{9})$

$E F = E E^{1} + E^{1} F^{1} + F^{1} F$

$= \frac{5}{9} A D^{1} + C^{1} D^{1} + \frac{5}{9} C^{1} B$

$= \frac{5}{9} (A D^{1} + C^{1} B) + C^{1} D^{1}$

$= \frac{5}{9} (A B - C D) + C D = \frac{5}{9} (\frac{3}{2} - 1) C D + C D$

$= (\frac{5}{9} \times \frac{1}{2} + 1) C D$

$= \frac{23}{18} \times 6 \times \frac{2}{3}$

$= \frac{46}{9} = (5 \frac{1}{9})$

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