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We can see that there are 5 CDs, each of radius 9 cm
<u>Area occupied by 1 disc:</u>
Area of a circle = πr²
Area of disc = π(9)²
Area of disc = 3.14 * 81 = 254 cm²
<u>Area occupied by 5 discs:</u>
Area occupied by 5 discs = Area occupied by 1 disc * 5
Area occupied by 5 discs = 254 * 5
Area occupied by 5 discs = 1270 cm²
Answer:
Step-by-step explanation:
The width of rectangle is the diameter of the semi-circle part
Area of one semicircle is given by 
Total area of semi circle will be 
Substituting 74 m for d and 
as 3.14 we obtain
Total area semi-circle=
Area of rectangle is given by the product of length and width
Rectangular area=
Total area of rectangular and semi-circles will be
Therefore, area of training field is
Answer:
20) MAD = 60°
21) <LPN = 57°
22) Second and third images, XY is the tangent of circle Z
23) x = 11.2
Step-by-step explanation:
20)
<B = <C
11x - 3 = 8x + 15
3x = 18
x = 3
so
<B = <C = 11(3) - 3 = 33 - 3 = 30
mAD = 2(30) = 60
21)
a) <LPN = 1/2(360 - 102 - 144)
= 1/2(114)
= 57°
22)
The tangent to a circle is perpendicular to the radius at the point of tangency.
a)
(7+5)^2 = 7^2 + 10^2
(12)^2 = 49 + 100
144 ≠ 149
First image, XY is not tangent of circle Z
b)
20^2 = 12^2 + (8+8)^2
400 = 144 + 256
400 = 400
Second image, XY is the tangent of circle Z
c)
(11.7 +7.8)^2 = 11.7^2 + 15.6^2
380.25 = 136.89 + 243.36
380.25 = 380.25
Third image, XY is the tangent of circle Z
23)
x^2 = (8.4+5.6)^2 - (8.4)^2
x^2 = 196 - 70.56
x^2 = 125.44
x = 11.2
3(x-7)=6(x-10)
3x-21=6x-60
3x-21+21=6x-60+21
3x=6x-39
3x-6x=6x-39-6x
-3x=-39
-3x/-3=-39/-3
x=13
