Answer:
perimeter is 4 sqrt(29) + 4pi cm
area is 40 + 8pi cm^2
Step-by-step explanation:
We have a semicircle and a triangle
First the semicircle with diameter 8
A = 1/2 pi r^2 for a semicircle
r = d/2 = 8/2 =4
A = 1/2 pi ( 4)^2
=1/2 pi *16
= 8pi
Now the triangle with base 8 and height 10
A = 1/2 bh
=1/2 8*10
= 40
Add the areas together
A = 40 + 8pi cm^2
Now the perimeter
We have 1/2 of the circumference
1/2 C =1/2 pi *d
= 1/2 pi 8
= 4pi
Now we need to find the length of the hypotenuse of the right triangles
using the pythagorean theorem
a^2+b^2 = c^2
The base is 4 ( 1/2 of the diameter) and the height is 10
4^2 + 10 ^2 = c^2
16 + 100 = c^2
116 = c^2
sqrt(116) = c
2 sqrt(29) = c
Each hypotenuse is the same so we have
hypotenuse + hypotenuse + 1/2 circumference
2 sqrt(29) + 2 sqrt(29) + 4 pi
4 sqrt(29) + 4pi cm
Answer:
B
Step-by-step explanation:
The profit made is equal to 2.25-2.25 which is 0.
Answer: The area is 572.5566
Step-by-step explanation:
27÷2=r=13.5
A=πr²
A=3.1416(π) · 13.5²(r)
A=572.5566
The triangle pay $32 more for that day than it paid per day during the first period of time.
Step-by-step explanation:
The given is,
Triangle Construction pays Square Insurance $5,980
To insure a construction site for 92 days
To extend the insurance beyond the 92 days costs $97 per day
Triangle extends the insurance by 1 day
Step:1
Insurance per day from the 92 days period,

Where, Total insurance for 92 days = $ 5,980
Period = 92 days
From the values, equation becomes,

= $ 65 per day
Step:2
Insurance per day after the 92 days,
= $ 97
Amount Pay for that day than it paid per day during the first period of time,

= $32
Result:
The triangle pay $32 more for that day than it paid per day during the first period of time, if the Triangle Construction pays Square Insurance $5,980
to insure a construction site for 92 days and to extend the insurance beyond the 92 days costs $97 per day.
<u><em>Answer:</em></u>
AC = 10sin(40°)
<u><em>Explanation:</em></u>
The diagram representing the question is shown in the attached image
Since the given triangle is a right-angled triangle, we can apply the special trig functions
<u>These functions are as follows:</u>
sin(θ) = opposite / hypotenuse
cos(θ) = adjacent / hypotenuse
tan(θ) = opposite / adjacent
<u>Now, in the given diagram:</u>
θ = 40°
AC is the side opposite to θ
AB = 10 in is the hypotenuse
<u>Based on these givens</u>, we will use the sin(θ) function
<u>Therefore:</u>
