2x+x-7+3x-20=87cm
6x-27=87
6x=114
x=19
Therefore the sides are:12cm, 38cm, and 37cm
a. Length of the fence around the field = perimeter of quarter circle = 892.7 ft.
b. The area of the outfield is about 39,584 sq. ft..
<h3>What is the Perimeter of a Quarter Circle?</h3>
Perimeter of circle = 2πr
Perimeter of a quarter circle = 2r + 1/4(2πr).
a. The length of the fence around the field = perimeter of the quarter circle fence
= 2r + 1/4(2πr).
r = 250 ft
Plug in the value
The length of the fence around the field = 2(250) + 1/4(2 × π × 250)
= 892.7 ft.
b. Size of the outfield = area of the full field (quarter circle) - area of the infield (cicle)
= 1/4(πR²) - πr²
R = radius of the full field = 250 ft
r = radius of the infield = 110/2 = 55 ft
Plug in the values
Size of the outfield = 1/4(π × 250²) - π × 55²
= 49,087 - 9,503
= 39,584 sq. ft.
Learn more about perimeter of quarter circle on:
brainly.com/question/15976233
The area is given by
A = (5 + 2x) * (10 + 2x) - 50
Rewriting we have:
A = 50 + 10x + 20x + 4x ^ 2 -50
Rewriting we have:
A = 4x ^ 2 + 30x
Substituting the value of the area we have:
54 = 4x ^ 2 + 30x
Rewriting:
4x ^ 2 + 30x - 54 = 0
We look for the roots:
x1 = -9
x2 = 3/2
We take the positive root:
x2 = 3/2
Answer:
the value of x is:
x = 3/2
21/30 as a percent would be 70%.
Since triangle BDE is right, we can check it for special rules: BD is hypotenuse = 20, and BE = 12, both divisible by 4: get 5 and 3... so it's a 3-4-5 special right triangle, the 4×4 = 16 m for side DE.
Since AD bisects angle A, I think then side DG should be congruent with DG. So then
DG = 16 m