Mohan inherited a triangular field with sides 240m,200m, and 360m, where he grew wheat . in another triangular field with sides 240m,320m, and 400m adjacent to the previous field, he wanted to grow potatoes and onions.
he divided the field into 2 parts by joining the mid point f the longest side to the opposite vertex and grew potatoes in one part and onions in other part. how much area(in hectares) has been used for wheat, potatoes and onions?

See the attached figure with letters

1) apply the Pythagorean theorem to verify if there is any rectangular triangle
triangle ABC
360²=200²+240²-------------> 19600=40000+57600--> is not rectangular triangle
triangle BCD
400²=320²+240²----> 160000=102400+57600----> 160000=160000----- > ok
the triangle BCD is a rectangular triangle
the area of triangle BCD=(1/2)*320*240=38400 m²
Area for potatoes--------------- > 38400/2=19200  m²
and  Area for onions=Area for potatoes=19200  m²

2) by the formula of Heron finds the area of a triangle of which all its sides are known
triangle ABC
s-----------> semiperimeter=[360+200+240]/2=400 m
A=[s*(s-a)*(s-b)*(s-c)]^(1/2)-----> [400*(400-360)*(400-200)*(400-240)]^(1/2)
A= [400*(40)*(200)*(160)]^(1/2)=22627.42 m²
Area for wheat-------- > 22627.42 m²

the total area is
Area for potatoes=19200  m²
Area for onions=19200  m²
Area for wheat=22627.42 m²

total area=19200+19200+22627.42=61027.42 m²

in hectares
1 hectares---------------------- >10000 m²
   X------------------------------- > 61027.42 m²

X=61027.42 /10000=6.10 hectares

total area------> 6.10 Ha
Area for potatoes=1.92 Ha
Area for onions=1.92 Ha
Area for wheat=2.26 Ha

Ibelieve its 0 but i am not sure

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).