Question

Find the area of pentagonal field shown alongside in figure. Dimensions are in metres.

• A. $7000{\text{m}}^2$
• B. $7550{\text{m}}^2$
• C. $5650{\text{m}}^2$
• D. $6500{\text{m}}^2$

Answer 1

Answer: (B)

$7550{\text{m}}^2$

The area of the pentagonal field,
$ar\left(ABCDE\right)=ar\left(△ABC\right)+ar\left(△AFE\right)+ar\left(EFDH\right)+ar\left(△DHC\right)$

Area of $△ABC=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times AC\times BG[AC=AF+FG+GH+HC=50+50+40+30=170$]
$=\frac{1}{2}\times 170\times 50$
$=85\times 50$
$=4250{\text{m}}^2$

Area of $△AFE=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times FA\times EF$
$=\frac{1}{2}\times 50\times 30$
$=50\times 15$
$=750{\text{m}}^2$

Area of trapezium $EFDH=\frac{1}{2}\times \left(\text{sum of parallel sides}\right)\times \text{height}$
$=\frac{1}{2}\times (EF+DH)\times HF[HF+HG+GF=50+40=90]$
$=\frac{1}{2}\times (30+20)\times 90$
$=\frac{1}{2}\times 50\times 90$
$=50\times 45$
$=2250{\text{m}}^2$

Area of $△DHC=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times DH\times CH$
$=\frac{1}{2}\times 20\times 30$
$=10\times 30$
$=300{\text{m}}^2$

Therefore,
$ar\left(ABCDE\right)=ar\left(△ABC\right)+ar\left(△AFE\right)+ar\left(EFDH\right)+ar\left(△DHC\right)$
$=4250+750+2250+300$
$=7550{\text{m}}^2$

Hence, the area of the pentagonal field is $7550{\text{m}}^2$.