What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
9514 1404 393
Answer:
x ≈ 136.7 ft
Step-by-step explanation:
As you have marked, the relevant sides are the adjacent side and the hypotenuse. The mnemonic SOH CAH TOA reminds you ...
Cos = Adjacent/Hypotenuse
For this triangle, that means ...
cos(43°) = (100 ft)/x
x = (100 ft)/cos(43°) . . . . . . . . . . multiply both sides by x/cos(43°)
x ≈ 136.73 ft
The length x is about 136.7 feet.
Circumference + perimeter =38
x = circumference of circle
find r in terms of x
2
r=x
r=
Area=
r^2
so..
A=
(
)^2
=
(
)
=
(38-x) is perimeter
then
(
)^2
+ (
)^2
then you graph it and it equals
16.716 cm the circumference of the circle
The ans should be ASA because angle AVR is equal to angle EVN (opposite angles equal)
Answer:
10=7
Step-by-step explanation:
6+4=10
2x5=10
10-3=7
Hope this helps :)
