Arancher has 180 meters of fence with which to enclose three sides of a rectangular field (the fourth side is a cliff wall and will not require fencing). find the dimensions of the field with the largest possible area. (for the purpose of this problem, the width will be the smaller dimension (needing two sides); the length with be the longer dimension (needing one

Step-by-step explanation:

We are given that

Total length of fence=180 m

Let length of rectangular field=x

Width of rectangular field=y

According to question

Length of fencing=x+2y

Area of rectangular field=

Area of rectangular field=

Differentiate w.r.t y

A'(y)=180-4y

A'(y)=0

180-4y=0

Again differentiate w.r.t y

A''(y)=-4<0

Areas of rectangular field is minimum at y=45

Substitute the value then, we get

x=180-2(45)=90

Dimensions of rectangular field are 90 m by 45 m

a

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.