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9th Class Mathematics Areas of Parallelograms and Triangles Question Bank Areas of Parallelograms and Triangles

  • question_answer
    ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD such that the area of \[\Delta BED=K\] area of \[\Delta ABC\]. Find K

    A)  2                                

    B)         \[\frac{1}{4}\]             

    C)         4                                

    D)         \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

    Median of a triangle divides it into two triangles of equal areas. \[\because \]AD is a median of\[\Delta \Alpha \Beta C.\] \[\therefore \]\[ar(\Delta \Alpha \Beta D)=ar(\Delta ADC)=\frac{1}{2}ar(\Delta \Alpha \Beta C)\]?(i) Again, BE is a median of\[\Delta \Alpha \Beta D,\] \[\therefore \]\[ar(\Delta \Beta \Epsilon \Alpha )=ar(\Delta BED)=\frac{1}{2}ar(\Delta \Alpha BD)\] \[=ar(\Delta \Beta \Epsilon D=\frac{1}{2}\times \frac{1}{2}\,ar\,(\Delta \Alpha \Beta C)\](From (i)) \[=ar(\Delta \Beta \Epsilon D)=\frac{1}{4}ar(\Delta \Alpha \Beta C)\] \[\therefore \]\[K=\frac{1}{4}\]


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