What is the area of the irregular pentagon a. 43 b. 105 c) 108 in2 d) 114 in2

The area = 105 in² ⇒ answer (b)

Step-by-step explanation:

Lets divide the figure into:

* rectangle of dimensions 8 in and 9 in

* trapezium with parallel bases 8 in and 14 in with height 3 in (12 - 9)

∵ Area rectangle = 8 × 9 = 72 in²

∵ Area trapezium = (8 + 14)/2 × 3 = 11 × 3 = 33 in²

∴ The area of the irregular pentagon = 72 + 33 = 105 in²

b.

Step-by-step explanation:

We have been given an image of an irregular pentagon and we are asked to find the area of our given pentagon.

To find the area of our given pentagon, we will divide it into parts by dropping an altitude from the 9 inch side to the 12 inch side. This altitude will divide our given pentagon into one rectangle and one trapezoid.

The area of pentagon will be equal to area of rectangle plus area of trapezoid.

The sides of rectangle will be 9 inch and 8 inches.

, where, a and b represents the length of parallel sides of trapezoid.

h = height of the trapezoid.

When we drop an altitude to the 12 inch side, it will give us the height of trapezoid as: 12-9=3 inch.

Therefore, the area of our given pentagon will be 105 square inches and option 'b' is the correct choice.

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.

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!