Question

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a field is in the shape of a trapezium whose parallel sides are 22m and 10m the non parallel sides are given as 13m and 14m find the area of the field

Answer 1

Answer:

Let ABCD be a trapezium with,

Let ABCD be a trapezium with,AB∥CD

Let ABCD be a trapezium with,AB∥CDAB=25m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+c

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21m

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)=21(7)(8)(6)

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)=21(7)(8)(6)=7056

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)=21(7)(8)(6)=7056=84m2

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)=21(7)(8)(6)=7056=84m2Also, area of ΔBCE is,

Let ABCD be a trapezium with,AB∥CDAB=25mCD=10mBC=14mAD=13mDraw CE∥DA. So, ADCE is a parallelogram with,CD=AE=10mCE=AD=13mBE=AB−AE=25−10=15mIn ΔBCE, the semi perimeter will be,s=2a+b+cs=214+13+15s=21mArea of ΔBCE,A=s(s−a)(s−b)(s−c)=21(21−14)(21−13)(21−15)=21(7)(8)(6)=7056=84m2Also, area of ΔBCE is,A=21×