Step-by-step explanation:
in the given figure ,
the two sides of the triangle are equal , so their opposite angles are equal .
so , we get
=》Angle RPQ = Angle PRQ = 65°
and some of interior angels of the triangle measures 180°
so,
=》RPQ + PRQ + RQP = 180°
=》65° + 65° + RQP = 180°
(Because RPQ = PRQ = 65°)
=》130° + RQP = 180°
=》RQP = 50°
Answer: The answer is 0.41 or .41
Step-by-step explanation:
Move the
Answer:
- 12. x = 6, side = 83
- 13. x = 18, side = 29
- 14. x = 11, sides = 74, 74 and 37
- 15. x = 23, sides = 95, 95 and 108
Step-by-step explanation:
<em>11 is incomplete, can't solve</em>
12
<u>The triangle is equilateral, so all sides are equal, using one pair to find x:</u>
- 13x + 5 = 17x - 19
- 17x - 13x = 5 + 19
- 4x = 24
- x = 6
<u>Each side is:</u>
13.
<u>Sides are equal as triangle is equilateral</u>
- QR = 2x - 7
- RS = 5x - 61
- QS = x + 11
<u>Finding x by comparing two sides</u>
- 2x - 7 = 5x - 61
- 5x - 2x = 61 - 7
- 3x = 54
- x = 18
<u>Sides are equal</u>
14.
<u>Equal sides of isosceles triangle:</u>
- CD = DE
- 9x - 25 = 6x + 8
- 9x - 6x = 8 + 25
- 3x = 33
- x = 11
<u>Sides are</u>
- CD = DE = 9*11 - 25 = 99 - 25 = 74
- CE = 10*11 - 73 = 110 - 73 = 37
15.
<u>Equal sides of isosceles triangle WXY, WX = WY</u>
- WX = 4x + 3
- WY = 7x - 66
- XY = 5x - 7
- 4x + 3 = 7x - 66
- 7x - 4x = 3 + 66
- 3x = 69
- x = 23
<u>Sides are:</u>
- WX=WY = 4*23 + 3 = 95
- XY = 5*23 - 7 = 108
Check the picture below.
since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.
now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.
![\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20semi-circle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%5Cqquad%20r%3Dradius%7D~%5Chspace%7B10em%7D%5Cstackrel%7B%5Ctextit%7Barea%20of%20an%20equilateral%20triangle%7D%7D%7BA%3D%5Ccfrac%7Bs%5E2%5Csqrt%7B3%7D%7D%7B4%7D%5Cqquad%20s%3D%5Cstackrel%7Bside%27s%7D%7Blength%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Blarger%20figure%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%208%5E2~~%2B~~%5Ccfrac%7B16%5E2%5Csqrt%7B3%7D%7D%7B4%7D%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%28%20%5Ccfrac%7B1%7D%7B3%7D%20%5Cright%29%5E2%20%2B%5Ccfrac%7B%5Cleft%28%20%5Cfrac%7B2%7D%7B3%7D%20%5Cright%29%5E2%5Csqrt%7B3%7D%7D%7B4%7D%5Cright%5D%7D)
![\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%2032%5Cpi%20%2B64%5Csqrt%7B3%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%7D%7B18%7D%2B%5Ccfrac%7B%5Cfrac%7B4%7D%7B9%7D%5Csqrt%7B3%7D%7D%7B4%7D%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%20%5Cleft%5B%2032%5Cpi%20%2B64%5Csqrt%7B3%7D%20%5Cright%5D%5Cqquad%20-%5Cqquad%20%5Cleft%5B%20%5Ccfrac%7B%5Cpi%20%7D%7B18%7D%2B%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B9%7D%20%5Cright%5D~~%5Capprox~~%20211.38%20-%200.37~~%5Capprox~~%20211.01)
