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 = 
= 
= 3075 square meter
After donation of the land the triangle formed is ΔDBE.
In ΔABC,

tan(∠ABC) = 
= 0.4065
∠ABC = 
= 22.12°
In ΔEFC,
tanC = 
0.4065 = 
CF = 
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) = 
sin(22.12) = 
DB = 102.48×0.3765
= 38.59 m
Similarly, cos(22.12) = 
0.9264 = 
BE = 102.48×0.9264
= 94.94m
Now area of ΔDBE = 
= 
= 1831.87 square meter
Area of remaining triangle with the villager = Area of ΔABC - Area of ΔDBE
= 3075 - 1831.87
= 1243.13 square meter
Answer:

Step-by-step explanation:
Given:
m<FGK = (7w + 3)°
m<FGH = 104°
angle bisector of <FGH = GK
Required:
Value of w
SOLUTION:
Since GK bisects angle FGH, it divides the angle into two equal parts. Therefore, the following equation can be generated to find the value of w:
m<FGH = 2*m<FGK
(substitution)
Divide both sides by 2


Subtract 3 from each side


Divide both sides by 7



Answer:
x=1, x=8
Step-by-step explanation:
4x2−36x+32=0
Step 1: Factor left side of equation.
4(x−1)(x−8)=0
Step 2: Set factors equal to 0.
x−1=0 or x−8=0
(+1) (+8)
x=1 or x=8
All others did not satisfy the inequality simultaneously except #4 and #6
For #4:
2 < 5(1) + 2 i.e. 2 < 7 (true)
2 >= 1/2(1) + 1 i.e. 2 >= 3/2 (true)
For #6
2 < 5(2) + 2 i.e. 2 < 12 (true)
2 >= 1/2(2) + 1 i.e. 2 >= 2