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In the figure given below, DE∥BC and the ratio of the areas of △ADE and trapezium DBCE is 4:5. Find the ratio of DE:BC

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An important question from ML Aggarwal of similarity chapter in which we have been asked to find the ratio of DE:BC by using the basic proportionality theorem if DEBC and the ratio of the areas of ADE and trapezium DBCE is 4:5.

ML Aggarwal Avichal Publication chapter 13.3, question no 13.a


1 Answer

  1. From the question it is given that,
    The ratio of the areas of ADE and trapezium DBCE is 4:5
    Now, consider the ABC and ADE
    A=A … [common angle for both triangles]
    D=B and E=C  … [because corresponding angles are equal]
    Therefore, ADEABC
    So, area of ADE/area of ABC=(DE)2/(BC)2 … [equation (i)]
    Then, area of ADE/area of trapezium DBCE=4/5
    area of trapezium DBCE/area of ADE=5/4
    Add 1 for both LHS and RHS we get,
    (area of trapezium DBCE/area of ADE)+1=(5/4)+1
    (area of trapezium DBCE+ area of ADE)/area of ADE=(5+4)/4
    area of ABC/area of ADE=9/4
    area of ADE/area of ABC=4/9
    From equation (i),
    area of ADE/area of ABC=(DE)2/(BC)2
    area of ADE/area of ABC=(DE)2/(BC)2=42/92
    area of ADE/area of ABC=(DE)2/(BC)2=2/3
    Therefore, DE:BC=2:3

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