A field is in the shape of a trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 13m and 14m. What is the area of the field ?

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CW Share
Aug 10, 2017

Answer:

#Area = 196 " m"^2#

Explanation:

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Draw a line #AE#, which is parallel to #BC#
#=> ABCE# is a parallelogram
#=># Area of trapezium #=# Area #DeltaADE# + Area #ABCE#
In #DeltaADE#, as the lengths of the three sides are #13, 14 and 15#, we can use Heron's Formula to find its area.
Heron's formula : #A=sqrt(s(s-a)(s-b)(s-c))#
where #s="perimeter"/2#, and #a,b,c# are the three known sides.
#=> s=(13+14+15)/2=21#
#=># Area of #DeltaADE = sqrt(21(21-13)(21-14)(21-15)#
#= sqrt((21*8*7*6))=sqrt7056=84#
#84=1/2*15*h#
#=> h=(84*2)/15=11.2#
Area #ABCE=10xxh=10xx11.2=112#
Hence Area of trapezium #ABCD=# Area #DeltaADE# + Area #ABCE#
#=84+112=196 " m"^2#

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dk_ch Share
Aug 10, 2017

drawn

Let #ABCD# be the trapezium whose parallel sides are #AB=10m and CD =25m# The non parallel sides are #BC=14m and DA=13m#

#BE# is drawn parallel to #AD#.It intersects #CD# at E.#BF# is the perpendicular drawn from #B# to #CE# . Hence #BE# is the height of the triangle #BCE# and trapezium #ABCD#

Now for triangle #BCE#

#BE=13m,BC=14m and CE=15m#

So semi perimeter of #DeltaBCE# is #s=1/2(13+14+15)=21m#

Hence area of #DeltaBCE#

#=sqrt(21xx(21-13)xx(21-14)xx(21-15))m^2#

#=sqrt(21xx8xx7xx6)m^2=84m^2#

Now
#1/2xxCExxBF=84#

#=>1/2xx15xxBF=84#

#=>BF=(84xx2)/15=11.2m#

So area of the trapezium #ABCD#

#=1/2xx(AB+CD)xxBF#

#=1/2xx(10+25)xx11.2m^2#

#=196m^2#

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