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
The air resistance acting on the un-crumbled sheet is greater.
Answer:
51 74
Step-by-step explanation:
x = smaller
x +23 = larger
added together (x + x+23) equal to 28 less than 3x = 3x-28
x + x+23 = 3x-28 subtract 2x from both sides of the equation
23 = x -28 add 28 to both sides
51 = x then the larger = 51 + 23 = 74
Answer:
x=1; y=2; z=3
Step-by-step explanation:
Rearrange for convince:
2y-4z+6x = -2
-4y+2z-3x=-5
-6y+3z+4x=1
Now we take to pairs and eliminate the same var.:
2y-4z+6x = -2 *2
-4y+2z-3x=-5
and
-4y+2z-3x=-5 *-3
-6y+3z+4x=1. *2
We get:
4y-8z+12y=-4
-4y+2z-3x=-5
= -6z+9x=-9
and
12y-6x+9x=15
-12y+6x+8x=2
= 17x=17
We now know x = 1, so z is:
-6z+9=-9
-6z=-18
z=3
Let’s grab the first formula and find y:
6+2y-12=-2
6+2y=10
2y=4
y = 2
