Question.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
Draw a line BE parallel to AD and draw a perpendicular BF on CD.
It can be observed that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 − ED = 15 m
For $\triangle B E C$,
Semi-perimeter, $s=\frac{(13+14+15) \mathrm{m}}{2}=21 \mathrm{~m}$
By Heron’s formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of $\triangle B E C=[\sqrt{21(21-13)(21-14)(21-15)}] \mathrm{m}^{2}$
$=[\sqrt{21(8)(7)(6)}] \mathrm{m}^{2}=84 \mathrm{~m}^{2}$
Area of $\triangle B E C=\frac{1}{2} \times C E \times B F$
$\Rightarrow 84=\frac{1}{2} \times 15 \times \mathrm{BF}$
$\Rightarrow \mathrm{BF}=\frac{168}{15}=11.2 \mathrm{~m}$
Area of $A B E D=B F \times D E=11.2 \times 10=112 \mathrm{~m}^{2}$
Area of the field $=84+112=196 \mathrm{~m}^{2}$
Draw a line BE parallel to AD and draw a perpendicular BF on CD.
It can be observed that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 − ED = 15 m
For $\triangle B E C$,
Semi-perimeter, $s=\frac{(13+14+15) \mathrm{m}}{2}=21 \mathrm{~m}$
By Heron’s formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of $\triangle B E C=[\sqrt{21(21-13)(21-14)(21-15)}] \mathrm{m}^{2}$
$=[\sqrt{21(8)(7)(6)}] \mathrm{m}^{2}=84 \mathrm{~m}^{2}$
Area of $\triangle B E C=\frac{1}{2} \times C E \times B F$
$\Rightarrow 84=\frac{1}{2} \times 15 \times \mathrm{BF}$
$\Rightarrow \mathrm{BF}=\frac{168}{15}=11.2 \mathrm{~m}$
Area of $A B E D=B F \times D E=11.2 \times 10=112 \mathrm{~m}^{2}$
Area of the field $=84+112=196 \mathrm{~m}^{2}$
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