Lesson 4: perimeters and areas of similar figures geometry foundations b unit 6: area connections academy write a comment or dm me if you

Part 1) 
area of triangle=b*h/2
b=2.9 cm
h=9 cm
so
area=2.9*9/2> 13.05 cm²

the answer part 1) is 13.05 cm²

Part 2) 
area of the figure=area of rectangle+area of triangle

find the area of rectangle
area rectangle=b*h
b=12 ft
h=?
tan 45=1
tan 45=h/(22-10)> h/10=1> h=10 ft
area of rectangle=12*10> 120 ft²

area of triangle=b*h/2
b=10 ft
h=10 ft
area of triangle=10*10/2> 50 ft²

area of the figure=120+50> 170 ft²

the answer Part 2) is 170 ft²

Part 3)
the area of kite is half the product of the diagonals. 
Area=(d1*d2)/2
d1=5+5---> 10 ft
d2=16+8> 24 ft
area=(10*24)/2> 120 ft²

the answer Part 3) is 120 ft²

Part 4)
the area of rhombus is half the product of the diagonals. 
Area=(d1*d2)/2
d1=6+6---> 12 m
d2=6+6> 12 m
area=(12*12)/2> 72 m²

the answer part 4) is 72 m²

Part 5)

area of the figure=6*area of one triangle

area of triangle=b*h/2
b=4 cm
is an equilateral triangle
applying the Pythagoras theorem
h²=4²-2²> h²=12> h=2√3 cm

area of triangle=4*2√3/2> 4√3 cm²

area of hexagon=6*(4√3)> 24√3 cm²

the answer Part 5) is 24√3 cm²