8th Class Mathematics Sample Paper Mathematics Sample Paper - 3

  • question_answer The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are equal, each being 26 cm, find the area of the trapezium.

    Answer:

    Let ABCD be the trapezium such that AB = 40 cm and CD = 20 cm and AD = BC = 26 cm.                                                                                                                                                              

    Now, draw CL || AD
    Then, ALCD is a parallelogram
    So, AL=CD=20cmandCL=AD=26cm.                                                                                                                                        
    In\[\Delta CLB\], we have
    CL = CB = 26 cm
    Therefore, \[\Delta CLB\]is an isosceles triangle.                                                                                                                                                                             
    Draw altitude CM of \[\Delta CLB.~\]                                                                                                                                                                                               
    Since \[\Delta CLB\] is an isosceles triangle.
    So, CM is also the median.
    Then, LM = \[MB=\frac{1}{2}BL=\frac{1}{2}\text{ }\times 20cm=10cm\]
    [as BL = AB -AL = (40 - 20) cm = 20 cm].
    Applying Pythagoras theorem in \[\Delta CLM,\]
    we have,                     \[C{{L}^{2}}=C{{M}^{2}}+L{{W}^{2}}\]
    \[{{26}^{2}}=C{{M}^{2}}+\text{1}{{\text{0}}^{2}}\]
    \[C{{M}^{2}}={{26}^{2}}-{{10}^{2}}\]
    \[=\left( 26-10 \right)\left( 26+10 \right)\]
    \[=16\times 36=576\]
    CM = \[\sqrt{576}\] = 24 cm
    Hence,   the area of the trapezium\[=\frac{1}{2}\times \] (sum of parallel sides) \[\times \] Height
    \[=\frac{1}{2}(20+40)\times 24\]
    \[=30\times 24=720\text{ }c{{m}^{2}}~\]


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