what is the area of the irregular pentagon
a. 43
b. 105
C)
108 in2
D)
114 in2
![what is the area of the irregular pentagon a 43 b 105C 108 in2 D 114 in2 class=](https://us-static.z-dn.net/files/d44/05840cf7024314b45ba616082cee9b7a.png)
Answer:
Answer is 105 in^2
Step-by-step explanation: Area of rectangle= length X breadth
= 8X12
=96in^ 2
Area of triangle= 1/2 X base X height
= 1/2 X 3 X 6
= 9 in^2
Area of irregular pentagon = area of triangle + area of rectangle
= 96+9
=105 in ^2
Answer:
Answer is 105 in^2
Step-by-step explanation:
Area of rectangle= length X breadth = 8X12=96in^ 2Area of triangle= 1/2 X base X height= 1/2 X 3 X 6= 9 in^2 Area of irregular pentagon = area of triangle + area of rectangle = 96+9=105 in ^2