Practice Problems
By: Jack Eike
Problem #1
Find the length of all the lines combined.
Step 1:
First, we have to find all the lengths of the sides. We know that ABDC and FGHE are rectangles, so we have to set up our equations. The first equation would be making FE and GH equal to each other. So, 2x+4=8. We know this because of the properties of a rectangle. The next equation would be making EH and FG congruent to each other. So, 5y+5=10. The third equation would be making CD and AB equal. So, 10y+10=20. The fourth equation is making AC and BD equal. So, 6x+4=16. The last equation is for finding HD. That equation is 2^2+2^2=c^2. Now, let's line them all up and solve them.
2x+4=8 5y+5=10 10y+10=20 6x+4=16 2^2+2^2=c^2
-4 -4 -5 -5 -10 -10 -4 -4 4+4=c^2
2x=4 5y=5 10y=10 6x=12 8=c^2
x=2 y=1 y=1 x=2 c=2* square root of 2
Find the length of all the lines combined.
Step 1:
First, we have to find all the lengths of the sides. We know that ABDC and FGHE are rectangles, so we have to set up our equations. The first equation would be making FE and GH equal to each other. So, 2x+4=8. We know this because of the properties of a rectangle. The next equation would be making EH and FG congruent to each other. So, 5y+5=10. The third equation would be making CD and AB equal. So, 10y+10=20. The fourth equation is making AC and BD equal. So, 6x+4=16. The last equation is for finding HD. That equation is 2^2+2^2=c^2. Now, let's line them all up and solve them.
2x+4=8 5y+5=10 10y+10=20 6x+4=16 2^2+2^2=c^2
-4 -4 -5 -5 -10 -10 -4 -4 4+4=c^2
2x=4 5y=5 10y=10 6x=12 8=c^2
x=2 y=1 y=1 x=2 c=2* square root of 2
Step 2:
Now that we have all of the variables, we can solve this problem. First we have to input the variables back into the equations. So 2(2)+4, 6(2)+4, 5(1)+5, 10(1)+10. That gets us 8, 16, 10, 20. Now, we add up all of the side lengths. This gets us 108. But we are not done yet. We still have to add the little diagonals which are 2.8. So 2.8*4 gets us 11.2 and we add that to the previous total and we get 119.2.
Now that we have all of the variables, we can solve this problem. First we have to input the variables back into the equations. So 2(2)+4, 6(2)+4, 5(1)+5, 10(1)+10. That gets us 8, 16, 10, 20. Now, we add up all of the side lengths. This gets us 108. But we are not done yet. We still have to add the little diagonals which are 2.8. So 2.8*4 gets us 11.2 and we add that to the previous total and we get 119.2.
Problem #2
Find the lengths of the midsegments of all of the trapezoids.
Step 1:
First, we have to know the formula for this. It is m=1/2(b+b). Plus, the variables are x=2 and y=1. Now that we know that we can say that m=1/2(2x+4+6x+4). The next equation is m=1/2(5y+5+10y+10). The third equation is m=1/2(16+8). The final equation is m=1/2(10+20). Now let's set them all up.
1/2(2x+4+6x+4) 1/2(5y+5+10y+10) 1/2(16+8) 1/2(10+20)
1/2(8x+8) 1/2(15y+15) 1/2(24) 1/2(30)
1/2(8(2)+8) 1/2(15(1)+15) 12 15
1/2(24) 1/2(30)
12 15
So, now we know that the lengths of the midsegments of the trapezoid AFEC and BGHD are 12 and the midsegments of the trapezoid EHDC and FGBA are 15.
Find the lengths of the midsegments of all of the trapezoids.
Step 1:
First, we have to know the formula for this. It is m=1/2(b+b). Plus, the variables are x=2 and y=1. Now that we know that we can say that m=1/2(2x+4+6x+4). The next equation is m=1/2(5y+5+10y+10). The third equation is m=1/2(16+8). The final equation is m=1/2(10+20). Now let's set them all up.
1/2(2x+4+6x+4) 1/2(5y+5+10y+10) 1/2(16+8) 1/2(10+20)
1/2(8x+8) 1/2(15y+15) 1/2(24) 1/2(30)
1/2(8(2)+8) 1/2(15(1)+15) 12 15
1/2(24) 1/2(30)
12 15
So, now we know that the lengths of the midsegments of the trapezoid AFEC and BGHD are 12 and the midsegments of the trapezoid EHDC and FGBA are 15.