Question

The following diagram shows a pentagonal field $ABCDE$ in which the lengths of $AF,FG,GH$ and $HD$ are $50m,40m,15m$ and $25m$ respectively; and the lengths of perpendiculars $BF,CH$ and $EG$ are $50m,25m$ and $60m$ respectively. Determine the area of the field.

• A. $7900$ sq. m
• B. $7425$ sq. m
• C. $7750$ sq.m
• D. $7525$ sq. m

Answer 1

The correct option is D $7525$ sq. m

ABCDE is a pentagon field in which,
$AF=50cm$, $FG=40cm$, $GH=15cm$
$HC=25cm$, $BF=50cm$, $HD=25cm$
$GD=40cm$, $EG=60cm$
Area of ABCDE is a pentagon field$=$Area of $△ABF+$Area of trapezium $BCHF+$Area of $△CDH+$Area of $△ADE$

$=\frac{1}{2}\times \text{base}\times \text{height}+\frac{1}{2}\times \text{(sum of parallel sides)}\times \text{height}+\frac{1}{2}\times \text{base}\times \text{height}+\frac{1}{2}\times \text{base}\times \text{height}$

$=\frac{1}{2}\times AF\times BF+\frac{1}{2}\times (BF+CH)\times FH+\frac{1}{2}\times HD\times CH+\frac{1}{2}\times AD\times EG$

$=\frac{1}{2}\times 50\times 50+\frac{1}{2}\times (50+25)\times FG+GH\frac{1}{2}\times 25\times 25+\frac{1}{2}\times (AF+FG+GH+HD)\times 60$

$=25\times 50+\frac{1}{2}\times \left(75\right)\times 40+15+\frac{1}{2}\times 625+\frac{1}{2}\times (50+40+15+25)\times 60$

$=1250+\frac{1}{2}\times \left(75\right)\times 55+312.5+\frac{1}{2}\times \left(130\right)\times 60$

$=1250+\frac{1}{2}\times 4125+312.5+65\times 60$

$=1250+2062.5+312.5+3900$
$=7525m^2$

Hence, Area of pentagon $ABCDE$ is $7525sq.m$