Question

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 . Ii) What are the values involved here ?

Answer 1

Answer:

Area of the remaining triangle with the villager is 1243.13 m²

Step-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 = $\frac{1}{2}\times {\text{Height}}\times \taxt{Base}$

= $\frac{1}{2}\times (123)\times (50)$

= 3075 square meter

After donation of the land the triangle formed is ΔDBE.

In ΔABC,

$tan(ABC)=(\frac{AB}{BC})$

tan(∠ABC) = $\frac{50}{123}$

                 = 0.4065

∠ABC = $tan^{-1}(0.4065)$

          = 22.12°

In ΔEFC,

tanC = $\frac{EF}{CF}$

0.4065 = $\frac{6}{CF}$

CF = $\frac{6}{0.4065}$

CF = 14.76 m

Since DE = AC - (CF + AG)

               = 132.8 - (2×14.76)

               = 132.8 - 29.52

               = 102.48 m

Now in ΔDBE,

sin(∠DEB) = $\frac{BE}{DE}$

sin(22.12) = $\frac{BE}{102.48}$

DB = 102.48×0.3765

     = 38.59 m

Similarly, cos(22.12) = $\frac{BE}{DE}$

0.9264 = $\frac{BE}{102.48}$

BE = 102.48×0.9264

     = 94.94m

Now area of ΔDBE = $\frac{1}{2}(DB)(BE)$

                                = $\frac{1}{2}(38.59)(94.94)$

                                = 1831.87 square meter

Area of remaining triangle with the villager = Area of ΔABC - Area of ΔDBE

= 3075 - 1831.87

= 1243.13 square meter