Part A:
The total area = x² + 12x + 20 = (x+2)(x+10)
<span>The width = x+2
∴ Length = area / width = </span>[ <span>(x+2)(x+10) ] / (x+2) = x+10
Part B: </span>
<span>A 2-foot walkway is built around the fountain
</span>
<span>∴ The width = (x+2) + 2 + 2 = x + 6
The length = (x+10) + 2 + 2 = x +14
Part C:
</span><span>the total area covered by the fountain and walkway.
</span><span>
= (x+6)(x+14)
= x² + 20 x + 84
See the attache figure for more explanation
</span>
Answer:
4.62 cm to nearest hundredth.
Step-by-step explanation:
If the parallel sides are x and y then:
x + y = 2*8 = 16
x + y = 16
If we drop a perpendicular line from one of the upper points on the trapezoid we have the height. Let the upper point be C and the point on the base be A. Let the point on right of the base be B.
AC is the height of the trapezoid. AB is the baseline of the triangle CAB.
In triangle CAB the angle B is 30 degrees.
As this is a 30-60-90 degree triangle
AC/AB = 1/√3 so AC = AB/ √3.
As the trapezoid is isosceles:
AB = x + 0.5(y - x)
AB = 0.5x + 0.5y
So AC = 1 /√3 (0.5x + 0.5y)
= 1 /√3 (0.5x + 0.5(16 - x)) (Substituting for x)
= 1 /√3 (0.5x + 8 - 0.5x)
=8 / √3
. = 4.6188 cm
Answer:
plz give brainlyst anwser but it is depentendent
Step-by-step explanation:
Answer:
10r²
Step-by-step explanation:
The following data were obtained from the question:
Radius (r) = 5r
Length of arc (L) = 4r
Area of sector (A) =?
Next, we shall determine the angle θ sustained at the centre.
Recall:
Length of arc (L) = θ/360 × 2πr
With the above formula, we shall determine the angle θ sustained at the centre as follow:
Radius (r) = 5r
Length of arc (L) = 4r
Angle at the centre θ =?
L= θ/360 × 2πr
4r = θ/360 × 2π × 5r
4r = (θ × 10πr)/360
Cross multiply
θ × 10πr = 4r × 360
Divide both side by 10πr
θ = (4r × 360) /10πr
θ = 144/π
Finally, we shall determine the area of the sector as follow:
Angle at the centre θ = 144/π
Radius (r) = 5r
Area of sector (A) =?
Area of sector (A) = θ/360 × πr²
A = (144/π)/360 × π(5r)²
A = 144/360π × π × 25r²
A = 144/360 × 25r²
A = 0.4 × 25r²
A = 10r²
Therefore, the area of the sector is 10r².