Rampur Sarpanch requested one of his villager to donate a 6m wide land adjusted to his 132.8m long side of his right triangular plot outside the village . The other sides of the plot is 123m and 50m .On his donated land , the Sarpanch wants to construct a link road which provides the connectivity with the other villages and towns . The villager agreed at once . i) Find the area of the triangular plot remaining with the villager .

Answer:Area of the remaining triangle with the villager is 1243.13 m²Step-by-step explanation:Triangle ABC is the triangular plot of a villager shown in the figure attached.Sarpanch requested the villager to donate land which is 6 m wide and along the side AC which measures 132.8m.Other sides of the plot has been given as AB = 50m and BC = 123 m.Now area of this land before donation = [tex]\frac{1}{2}\times {\text{Height}}\times \taxt{Base}[/tex]= [tex]\frac{1}{2}\times (123)\times (50)[/tex]= 3075 square meterAfter donation of the land the triangle formed is ΔDBE.In ΔABC,[tex]tan(ABC)=(\frac{AB}{BC})[/tex]tan(∠ABC) = [tex]\frac{50}{123}[/tex]                  = 0.4065∠ABC = [tex]tan^{-1}(0.4065)[/tex]           = 22.12°In ΔEFC,tanC = [tex]\frac{EF}{CF}[/tex]0.4065 = [tex]\frac{6}{CF}[/tex]CF = [tex]\frac{6}{0.4065}[/tex]CF = 14.76 mSince DE = AC - (CF + AG)                = 132.8 - (2×14.76)                = 132.8 - 29.52                = 102.48 mNow in ΔDBE,sin(∠DEB) = [tex]\frac{BE}{DE}[/tex]sin(22.12) = [tex]\frac{BE}{102.48}[/tex]DB = 102.48×0.3765      = 38.59 mSimilarly, cos(22.12) = [tex]\frac{BE}{DE}[/tex]0.9264 = [tex]\frac{BE}{102.48}[/tex]BE = 102.48×0.9264      = 94.94mNow area of ΔDBE = [tex]\frac{1}{2}(DB)(BE)[/tex]                                 = [tex]\frac{1}{2}(38.59)(94.94)[/tex]                                 = 1831.87 square meterArea of remaining triangle with the villager = Area of ΔABC - Area of ΔDBE = 3075 - 1831.87 = 1243.13 square meter