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)
<span>the area of kite is half the product of the diagonals.
</span>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
Answer:
451. No, the angles are wrong.
Step-by-step explanation:
450. AB = 15, BC = 10, and CD= 7. Find the length DA.
This cannot be done without additional information about the sort of figure that ABCD is. If these are points on a line segment, we need to know their order. If these are points on a quadrilateral, we need to know its description in more detail.
If these are points ordered ABCD on a line, then AD = 15+10+7 = 32.
__
451. See the attached figure. BPQD is not a parallelogram: BCQ is not a straight line. (The internal angles of a pentagon are 108°, but would need to be 120° for BCQ to be a straight line, making BP parallel to DQ.) Instead, BPQD is an isosceles trapezoid.
2a 3x4a +3x-5
2a+12a-15
14a-15
i think thats the answer
Answer:
10
Step-by-step explanation:
c²=b²+a²=100
so
c=10
jjthj
