The similarities are;
- Compass and a straight edge required for both construction
- Both construction includes a line drawn from the intersection of arcs to bisect a segment or an angle
- The bases for the construction of both bisector are the ends of segment and the angle to be bisected
- The width of the compass when drawing intersecting arcs, is more than half the width of the segment or angle being bisected
The differences are;
- Two points of intersection of arcs are used in the segment bisector while only one is requited in an angle bisector
- The bisecting line crosses the segment in a segment bisector, while it stops at the vertex of the angle being bisected in an angle bisector
The sources of the above equations are as follows;
The steps to construct a segment bisector are;
- Place the needle of the compass at one of the ends of the line segment to be bisected
- Widen the compass so as to extend more than half of the length of the segment to be bisected
- Draw two arcs, one above, and the other below the line
- Place the compass needle at the other end and with the same compass width draw arcs that intersects with the arcs drawn in the above step
- Draw a line segment by placing the ruler on the points of intersection of the arcs above and below the line
The steps to construct an angle bisector are;
- With the compass needle at the vertex, open the pencil end such that arcs can be drawn on the rays (lines) forming the angle
- Draw an arc on both lines forming the angle
- Place the compass needle at one of the intersection points and draw an arc in between the lines forming the angle
- Repeat the above step with the same compass width from the other intersection point with the rays forming the angle
- Join the point of intersection of the two arcs to the vertex of the angle to bisect the angle
Therefore, we have;
The similarities are;
- A compass and a straight edge can be used for both construction
- A straight line is drawn from the point of intersection of arcs to bisect the segment or the angle
- The arcs are drawn from the ends of the segment or angle to be bisected
- The width of the compass is more than half the width of the line or angle when drawing the arcs
The differences are;
- In a segment bisector, the intersection point is above and below the line, while in an angle bisector only one pair of arcs are drawn to intersect above the line
- The bisecting line passes through the segment being bisected, while the line stops at the vertex in an angle bisector
Learn more about the construction of segment and angle bisectors here;
brainly.com/question/17335869
brainly.com/question/12028523
Answer:
here
Step-by-step explanation:
9-3=6
9+-3=6
Answer:
y=8
n=1.5
Step-by-step explanation:
Find the area of the squares.
4 ×4 = 16
Find the area of the wall.
20 × 10 =200
Divide 200 by 16.
200 ÷ 16 = 12.5
You need 12.5 squares.
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.
![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)
![\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o ) \right] \\\\\\ A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\ -------------------------------\\\\ \textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}](https://tex.z-dn.net/?f=%5Cbf%20A_y%3D%5Ccfrac%7B6%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%5Ccdot%20120%20%7D%7B180%7D~-~sin%28120%5Eo%20%29%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0AA_y%3D18%5Cleft%5B%5Ccfrac%7B2%5Cpi%20%7D%7B3%7D~-~%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright%5D%5Cimplies%20%5Cboxed%7BA_y%3D12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bshaded%20area%7D%5Cqquad%20%5Cstackrel%7BA_x%7D%7B6%5Cpi%20%7D~~%2B~~%5Cstackrel%7BA_y%7D%7B12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5Cimplies%2018%5Cpi%20-9%5Csqrt%7B3%7D)
