triangle BCD is similar to triangle ACE (side, angle C, side)
AC = 2 BC
therefore AE = 2 BD
therefore
4x+20 = 2 (3x+5) solve that
B is the midpoint of segment AC and D is the midpoint of segment CE. Solve for x, given BD=3x+5 and AE=4x+20.
Sorry I couldn't get the link for the picture but, the diagram is a picture of a triangle with a C at the top, and A and E on
the bottom angles, with segment BD going through the middle of the triangle.
Thank You ^0^
15 answers
Thank You So Much! ^0^
What's the answer?
whats answer
What's the answer?
BD= 1/2 AE.
So the real equation is 6x+10=4x+20
isolate the variable and you will get your answer.
the answer is x=5
So the real equation is 6x+10=4x+20
isolate the variable and you will get your answer.
the answer is x=5
1. b
2. a
3. a
4. a
5. c
6. b
7. b
8. c
9. d
10. c
11. midsegment
12. equidstant
13. perpendicluar
14. concurrent lines
2. a
3. a
4. a
5. c
6. b
7. b
8. c
9. d
10. c
11. midsegment
12. equidstant
13. perpendicluar
14. concurrent lines
is pickle right?
Is pickle right?
No, pickle is not right on all the answers.
1. b
2. b
3. b
4. a
5. b
6. d
7. b
8. a
9. d
10. a
2. b
3. b
4. a
5. b
6. d
7. b
8. a
9. d
10. a
x=5
B is the midpoint of AC and D is the midpoint of CE Solve for x, given BD = 5x + 4 and AE = 4x + 50.
AE is divided into two pairs of equal lengths. BD contains one of each pair.
So, BD is half the length of AE.
2(3x+5) = 4x+20
x = 5
So, BD is half the length of AE.
2(3x+5) = 4x+20
x = 5
AE = 2(BD)
4x + 20 = 2(3x + 5)
4x + 20 = 6x + 10
20 = 2x +10
10 = 2x
5 = x
4x + 20 = 2(3x + 5)
4x + 20 = 6x + 10
20 = 2x +10
10 = 2x
5 = x