The triangle $\triangle ABC$ is an isosceles triangle where $AB = 4\sqrt{2}$ and $\angle B$ is a right angle. If $I$ is the incenter of $\triangle ABC,$ then what is $BI$? Express your answer in the form $a + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect square other than $1.$

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)

step-by-step explanation: