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Easy

Question

Find the area of the figure.GE equals 14 cm comma HF equals 10 cm comma AD equals 24 cm straight & BC equals 12
.

hintHint:

The figure consists of 1 rhombus and 1 trapezium
So, Area of figure = area of rhombus EFGH + area of trapezium ABCD .

The correct answer is: 340 cm2


    Ans :- 340 cm2.
    Explanation :-
    By observing the diagram EFGH is a quadrilateral where all sides are equal so
    EFGH is a rhombus .
    ABCD is quadrilateral with AD // BC .
    so, ABCD is trapezium with AD and BC as parallel sides.
    STEP 1:- Find the area of rhombus EFGH
    GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)
    Area of rhombus = 1 half cross times d subscript 1 cross times d subscript 2 text  where  end text d subscript 1 text  and  end text d subscript 2 text  are lengths of diagonals.  end text
    Area of rhombus =1 half cross times 14 cross times 10 equals 7 cross times 10 equals 70 cm squared
    We get area of rhombus EFGH is 70cm2.
    STEP 2:- Find the area of trapezium ABCD
    AD =  24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 15 right parenthesis left parenthesis 12 plus 24 right parenthesis equals 1 half left parenthesis 15 right parenthesis left parenthesis 36 right parenthesis equals 15 cross times 18
    = 270 cm2
    Area of trapezium ABCD is 270 cm2
    Step 3:- Find the area of figure
    Area of figure = area of rhombus EFGH + area of trapezium ABCD .
    Area of figure = 70+270 = 340 cm2.
    Therefore , the area of the entire figure is 340cm2.

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