Find the cost at $R s . 25$ per $10$ square metres of turfing a plot of land in the form of a parallelogram whose adjacent side and one of the diagonals measure $39 m, 25 m$ and $56 m$ , respectively.

R s . 2100

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R s . 1100

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R s . 2400

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None of the above

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Solution

The correct option is A $R s . 2100$
Area of plot $= 2 \times a r Δ A B C$
Now, $a r Δ A B C = \sqrt{s (s - a) (s - b) (s - c)} s = \frac{a + b + c}{2} = \frac{56 + 25 + 39}{2} = 60 c m a r Δ A B C = \sqrt{60 (60 - 56) (60 - 25) (60 - 39)} = \sqrt{60 \times 4 \times 35 \times 21} a r Δ A B C = 420 m^{2} ∴ a r e a o f p l o t = 2 \times 420 = 840 m^{2}$
Cost of turfing $10 m^{2}$ land $= R s .25$
Cost of turfing $840 m^{2}$ land $= R s . (\frac{25}{10} \times 840) = R s .2100$

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Find the cost at Rs.25 per 10 square metres of turfing a plot of land in the form of a parallelogram whose adjacent side and one of the diagonals measure 39m,25m and 56m, respectively.

The correct option is A Rs.2100Area of plot=2×arΔABCNow, arΔABC=√s(s−a)(s−b)(s−c)s=a+b+c2=56+25+392=60cmarΔABC=√60(60−56)(60−25)(60−39)=√60×4×35×21arΔABC=420m2∴areaofplot=2×420=840m2 Cost of turfing 10m2 land=Rs.25 Cost of turfing 840m2 land=Rs.(2510×840)=Rs.2100

Find the cost at $R s . 25$ per $10$ square metres of turfing a plot of land in the form of a parallelogram whose adjacent side and one of the diagonals measure $39 m, 25 m$ and $56 m$ , respectively.