The sector (shaded segment + triangle) makes up 1/3 of the circle (which is evident from the fact that the labeled arc measures 120° and a full circle measures 360°). The circle has radius 96 cm, so its total area is π (96 cm)² = 9216π cm². The area of the sector is then 1/3 • 9216π cm² = 3072π cm².
The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines
b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°) ⇒ b = 96√3 cm
Call b the base of this triangle.
The vertex angle is 120°, so the other two angles have measure θ such that
120° + 2θ = 180°
since the interior angles of any triangle sum to 180°. Solve for θ :
2θ = 60°
θ = 30°
Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find
tan(30°) = h / (b/2) ⇒ h = 48 cm
Then the area of the triangle is
1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²
and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:
3072π cm² - 2304√3 cm² ≈ 5660.3 cm²
Answer:
w =< 70
(width is less or equal to 70 inches)
Step-by-step explanation:
Let l = length, w = width, h = height
Restrictions given in this question:
'sum of perimeter of the base and the height cannot exceed 130 inches'
perimeter of the base is 2 width and 2 length of the box
perimeter = 2w + 2l
Therefore, inequality involves here is
2w + 2l + h =< 130
(Note that =< here means less or equal)
Then a new condition given with
height, h = 60 in
and length is 2.5 times the width
l = 2.5w
Substitute this new condition into the equation will give us the following:
2w + 2(2.5w) + 60 =< 130
2w + 5w + 60 =< 130
7w + 60 =< 130
7w =< 130-60
7w =< 70
w =< 10
The metric system is based on power os 10’s- brainest pls xx
Well, from 2008 to 2010 is just 2 years, so let's say the initial amount is the 19300 from 2008, how many will it be in 2010?
