Answer:
The radius of the circle ,r= 1.43 m
The length of the side of square ,a= 2.77 m
Step-by-step explanation:
Given that
L= 20 m
Lets take radius of the circle =r m
The total parameter of the circle = 2π r
Area of circle ,A=π r²
The side of the square = a m
The total parameter of the square = 4 a
Area of square ,A'=a²
The total length ,L= 2π r+ 4 a
20 = 2π r+ 4 a
r=3.18 - 0.63 a
The total area = A+ A'
A" =π r² +a²
A"= 3.14(3.18 - 0.63 a)² + a²
For minimize the area
3.14 x 2(3.18 - 0.63 a) (-0.63) + 2 a = 0
3.14 x (3.18 - 0.63 a) (-0.63) + a = 0
-6.21 + 1.24 a + a=0
2.24 a = 6.21
a=2.77 m
r= 3.18 - 0.63 a
r= 3.18 - 0.63 x 2.77
r=1.43 m
Therefore the radius of the circle ,r= 1.43 m
The length of the side of square ,a= 2.77 m
Problem 2
Plot point L anywhere that isn't on segment JK. Draw a line through point L. I find it helps to make the lines parallel.
Next, use a compass to measure the width of segment JK. Keeping this same width, transfer the nonpencil end of the compass to point L. Draw an arc that crosses the line through L.
Mark this intersection point M. Lastly, use a pen or marker to form segment LM and erase everything else of that line.
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Problem 3
The ideas of the previous problem will be used here. We copied segment JK to form congruent segment LM. So JK = LM.
The same steps will be used to form segment GN where GN = EF. In other words, segment GN is a perfect copy of segment EF.
If you repeat these steps again, you'll get another segment of the same length. This segment goes from point N to point H. So NH = GN = EF
Then we can say,
GH = GN + NH
GH = EF + EF
GH = 2*EF
Answer:
1.28/7. 2.-17/7. 3.-11/-3. 4.not sure. 5. cant remeber rest
2. WXZ=YXZ
3. Reflexive property of equality
(if they are angles)-
4. WXZ=YXZ ~~Reason- Congruent angles have equal number of degrees
(if they are triangles-more likely judging by the problem)-
4. Triangle WXZ=YXZ ~ Reason- SAS Congruence Postulate
