The triangle QPR having inscribed triangle STU will allow the artisan to divided his glass piece into four equal triangular pieces.
In order to divide an equilateral triangle into four equal triangular glass pieces, the artisan must;
- Take S as the mid-point on PA, T as the mid-point on PR, and U as the mid-point on QR. Thus, S, T, and U are the three mid-points on each side of the equilateral triangle QPR.
- Now, by joining these mid-points S, T, and U, four equal triangles are made(as shown in the figure).
Since the triangle is equilateral,
PQ = QR = RP
Mid-point divides the lines into equal parts. So,
PS = SQ = QU = UR = RT = TP
Thus, it is proved that
ΔPST = ΔSTU = ΔTUR = ΔQSU
Learn more about 'Equilateral Triangle' here:
brainly.com/question/2855144
Answer:
Plates are 2.80 Dollars and cups are 1.40 dollars
Step-by-step explanation:
Cups = c
Plates = p
6p + 5c = 23.80 ==> multiply by 6 ==> 36p + 30c = 142.80
7p + 6c = 28.00 ==> multiply by 5 ==> 35p + 30c = 140.00
Now use algebra to have the 30c be on one side and the rest on the other:
36p + 30c = 142.80 | -36p
30c = 142.80 - 36p
35p + 30c = 140.00 | -35p
30c = 140.00 - 35p
Now set them equal to each other about the 30c:
142.80 - 36p = 140.00 - 35p
Use algebra to solve for p:
142.80 - 36p = 140.00 - 35p | +36p
142.80 = 140.00 + p | - 140
2.80 = p
Go back to one of the "30c" equations and plug in the value for p:
30c = 140 - 35p
30c = 140 - 35(2.80)
30c = 140 - 98
30c = 42 | /30
c = 42/30
c = 1.4
Answer:
a)400V2 plus 3,600 m2
Step-by-step explanation:
if the diameter is 20, the its radius must be half that or 10.
![\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%5E2%7D%7B360%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20A%3D5%5Cpi%20%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%205%5Cpi%20%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20%2810%29%5E2%7D%7B360%7D%5Cimplies%205%5Cpi%20%3D%5Ccfrac%7B5%5Cpi%20%5Ctheta%20%7D%7B18%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B5%5Cpi%20%7D%7B5%5Cpi%20%7D%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%201%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%2018%3D%5Ctheta%20%5Cend%7Barray%7D)
