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In a figure, BD is parallel to AE, ED = 2 cm and DC = 6 cm. Given that the area of...

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syedajmain | Student | (Level 3) Honors

Posted January 10, 2010 at 1:33 AM via web

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In a figure, BD is parallel to AE, ED = 2 cm and DC = 6 cm. Given that the area of triangle CBD = 9 cm^2, Calculate the area of ABDE.

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neela | High School Teacher | (Level 3) Valedictorian

Posted January 10, 2010 at 2:26 AM (Answer #1)

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We pressume that ABC is a straight line.

If you draw the figure, AEC is tringle with EC =2+6=8 cm and D is a point on AC such that ED =2cm.  Bis a point on AC and BD || AE. (Sorry, no environment to draw the figure.)

So, consider, triangles CAE and CBD.

AE || BD and AC is transval. So the  angle CAE = angle CBD, being corresponding angles.

|||ly, AE || BD and EC is a transval. So angle CEA = angle CBD.

Therefore, all three angles of triangles CAE and CBD are equal, as the 2 angles of triangles being equal, the remaing angle of the triangle have to be equal. So, triangleCAE and CBD are similar. Therefore,their areas are proportional to the square of their sides.

Therefore,

Area of CAE/area of CBD = EC^2/DC^2 = (2+6)^2/6^2 = 16/9. or

Area CAE = (area CBD)(16/9) = 9(16/9) =16 cm^2

Area Area ABDE +area CBD =16cm^2, as CAE = ABDE+CBDE.

Area ABDE =16-area CBD =16-9 = 7cm^2

 

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