Help I'm taking a k12 quiz 2.04: composite figures but, I can't figure these out can anyone help me?Question. 1 What is the area of this polygon? (Picture 1)Question. 2 What is the area of this polygon? 42.5 Units? 41.5 Units? 35.5 Units? 29.5 Units? (Picture 2)Question. 3 This figure is made up of a rectangle and parallelogram.What is the area of this figure? (Picture 3)Question. 4 What is the area of this composite shape? (Picture 4)Thank you angels who help me with these
Accepted Solution
A:
All of these can be solved by separating complex figur into simpler one. Needed formulas are: Area of rectangle A=a*b Area of triangle A=a*h/2 Area of paralelogram A= a*h
QUESTION 1. When we connect points F and S we get a rectangle and a triangle. Dimensions of sides of a rectangle are a=9 and b=2. A=9*2=18 [tex]units^{2} [/tex] For a triangle we need to take one side and it's height. Let's take side FS because it's easiest to find the height. a=9 h=6 A=9*6/2=27[tex]units^{2} [/tex]
Tot+al area of a figure: A=18+27=45[tex]units^{2} [/tex]
QUESTION 2. We connect point (1,1) with (1,-2) and (-6,1). This way we get a rectangle and three triangles. Rectangle: a=3 b=4 A=3*4=12[tex]units^{2} [/tex]
Top right triangle: a=4 h=3 A=4*3/2=6[tex]units^{2} [/tex]
Top left triangle: a=7 h=2 A=7*2/2=7[tex]units^{2} [/tex]
Bottom left triangle: a=7 h=3 A=7*3/2=10.5[tex]units^{2} [/tex]
Total area of a figure: A=12+6+7+10.5=35.5[tex]units^{2} [/tex]
QUESTION 3. We connect points (-4,2) and (-1,3). This way we get a paralelogram and a rectangle.
Total area of a figure: A=[tex]10 \sqrt{2} [/tex] + 6 [tex]units^{2} [/tex]
QUESTION 4. We coonect missing side of a triangle. That side has a length of: 6in-3in=3in. Vertical side of triangle is: 13in-7in=6in Area: a=3 h=6 A=3*6/2=9[tex] in^{2} [/tex]