Find the Area of the figure below, composed of an isosceles trapezoid and one semicircle. Rounded to the nearest tenths place
10

Area of given figure = 48.28 unit² (Approx.)

Step-by-step explanation:

Given:

Length of parallel lines are 10 , 4 respectively

Diameter of semi-circle = 4

Height of parallelogram = 6

Find:

Area of given figure

Computation:

Radius of semi-circle = Diameter of semi-circle / 2

Radius of semi-circle = 4 / 2

Radius of semi-circle = 2

Area of given figure = [1/2][Sum of parallel line][Height] + [πr² / 2]

Area of given figure = [1/2][10 + 4][6] + [(3.14)(2)² / 2]

Area of given figure = [1/2][14][6] + [(3.14)(4)/ 2]

Area of given figure = 42 + 6.28

Area of given figure = 48.28 unit² (Approx.)

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

answer: 1475

step-by-step explanation:

becuase you have to know what they are asking you