Question

The given figure is a pentagonal field which is ploughed at rate of $₹ 5.50 per m^{2}$ . Find the cost of ploughing it if $AH = 120 m, AF = 80 m, AG = 50 m, HC = 30 m, BF = 50 m, EG = 30 m$ and $HD = 20 m$ .

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Accepted Answer

In the above figure,

$AH = 120 m, AF = 80 m, AG = 50 m, HC = 30 m, BF = 50 m, EG = 30 m$ and $HD = 20 m$ .

$HD = 20 m$ is a triangle with height $BF$ and base $AC$ .

Now, $AC = AH + CH = 120 m + 30 cm = 150 cm$

Area of triangle $= \frac{1}{2} \times base \times height$

Area of $△ ABC = \frac{1}{2} \times BF \times AC$

$= \frac{1}{2} \times 150 m \times 50 m$

$= 3750 m^{2}$

Area of $△ AGE = \frac{1}{2} \times AG \times GE$

$= \frac{1}{2} \times 30 m \times 50 m$

$= 750 m^{2}$

Area of $△ CHD = \frac{1}{2} \times CH \times HD$

$= \frac{1}{2} \times 30 m \times 20 m$

$= 300 m^{2}$

$GHDE$ is a trapezium with parallel sides $EG, HD$ and height $GH$ .

$GH = AH - AG = 120 m - 50 m = 70 m$

Area of the trapezium $GHDE = \frac{1}{2} \times (EG + HD) \times GH$

$= \frac{1}{2} \times 70 m \times (30 + 20) m$

$= 1750 m^{2}$

Area of $ABCDE =$ Area of $△ ABC +$ Area of $△ AGE +$ Area of $△ CHD +$ Area of $GHDE$

$= (3750 + 750 + 300 + 1760) m^{2}$

$= 6550 m^{2}$

Cost of ploughing the field $per m^{2} = ₹ 5.50$

Therefore, cost of ploughing the field of $6550 m^{2} = ₹ (5.50 \times 6550) = ₹ 36025$

Monica Capoor Solutions for Chapter: Mensuration I Area of Polygons, Exercise 4: EXERCISE

Monica Capoor Mathematics Solutions for Exercise - Monica Capoor Solutions for Chapter: Mensuration I Area of Polygons, Exercise 4: EXERCISE

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