SOLUTION: Given that A(0, 0), B(4, 0), C(0, 2), D(0, 0), E(12, 0), and F(0, 6), which of the following proves that ΔABC ~ ΔDEF? A. http://my.thinkwell.com/questionbank/97001-980

Algebra ->  Triangles -> SOLUTION: Given that A(0, 0), B(4, 0), C(0, 2), D(0, 0), E(12, 0), and F(0, 6), which of the following proves that ΔABC ~ ΔDEF? A. http://my.thinkwell.com/questionbank/97001-980      Log On


   

Question 954729: Given that A(0, 0), B(4, 0), C(0, 2), D(0, 0), E(12, 0), and F(0, 6), which of the following proves that ΔABC ~ ΔDEF?
A. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244855.gif
B. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244856.gif
C. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244844.gif
D. ΔABC and ΔDEF are not similar.
Found 2 solutions by Fombitz, MathLover1:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

AB/DE=4/12=1/3
CA/FD=2/6=1/3
The triangles are similar.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

first find out what is the distance between given points A(0, 0), B(4, 0), C(0, 2)(vertices of a triangle ABC)
and the distance between given points D(0, 0), E(12, 0), and F(0, 6) (vertices of a triangle DEF)
these distances are the lengths of the sides
Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (0, 0), we can say (x1, y1) = (0, 0)
So x%5B1%5D+=+0, y%5B1%5D+=+0


Since the second point is (4, 0), we can also say (x2, y2) = (4, 0)
So x%5B2%5D+=+4, y%5B2%5D+=+0


Put this all together to get: x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+4, and y%5B2%5D+=+0

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (0, 0) and (4, 0)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%280+-+4%29%5E2+%2B+%280+-+0%29%5E2%29 Plug in x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+4, and y%5B2%5D+=+0


d+=+sqrt%28%28-4%29%5E2+%2B+%280%29%5E2%29


d+=+sqrt%2816+%2B+0%29


d+=+sqrt%2816%29


d+=+4

==========================================================

Answer:


The distance between the two points (0, 0) and (4, 0) is exactly 4 units




Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (0, 0), we can say (x1, y1) = (0, 0)
So x%5B1%5D+=+0, y%5B1%5D+=+0


Since the second point is (0, 2), we can also say (x2, y2) = (0, 2)
So x%5B2%5D+=+0, y%5B2%5D+=+2


Put this all together to get: x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+2

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (0, 0) and (0, 2)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%280+-+0%29%5E2+%2B+%280+-+2%29%5E2%29 Plug in x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+2


d+=+sqrt%28%280%29%5E2+%2B+%28-2%29%5E2%29


d+=+sqrt%280+%2B+4%29


d+=+sqrt%284%29


d+=+2

==========================================================

Answer:


The distance between the two points (0, 0) and (0, 2) is exactly 2 units




Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (4, 0), we can say (x1, y1) = (4, 0)
So x%5B1%5D+=+4, y%5B1%5D+=+0


Since the second point is (0, 2), we can also say (x2, y2) = (0, 2)
So x%5B2%5D+=+0, y%5B2%5D+=+2


Put this all together to get: x%5B1%5D+=+4, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+2

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (4, 0) and (0, 2)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%284+-+0%29%5E2+%2B+%280+-+2%29%5E2%29 Plug in x%5B1%5D+=+4, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+2


d+=+sqrt%28%284%29%5E2+%2B+%28-2%29%5E2%29


d+=+sqrt%2816+%2B+4%29


d+=+sqrt%2820%29


d+=+sqrt%284%2A5%29


d+=+sqrt%284%29%2Asqrt%285%29


d+=+2%2Asqrt%285%29


d+=+4.47213595499958

==========================================================

Answer:


The distance between the two points (4, 0) and (0, 2) is exactly 2%2Asqrt%285%29 units


The approximate distance between the two points is about 4.47213595499958 units



So again,


Exact Distance: 2%2Asqrt%285%29 units


Approximate Distance: 4.47213595499958 units




and
Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (0, 0), we can say (x1, y1) = (0, 0)
So x%5B1%5D+=+0, y%5B1%5D+=+0


Since the second point is (12, 0), we can also say (x2, y2) = (12, 0)
So x%5B2%5D+=+12, y%5B2%5D+=+0


Put this all together to get: x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+12, and y%5B2%5D+=+0

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (0, 0) and (12, 0)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%280+-+12%29%5E2+%2B+%280+-+0%29%5E2%29 Plug in x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+12, and y%5B2%5D+=+0


d+=+sqrt%28%28-12%29%5E2+%2B+%280%29%5E2%29


d+=+sqrt%28144+%2B+0%29


d+=+sqrt%28144%29


d+=+12

==========================================================

Answer:


The distance between the two points (0, 0) and (12, 0) is exactly 12 units




Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (0, 0), we can say (x1, y1) = (0, 0)
So x%5B1%5D+=+0, y%5B1%5D+=+0


Since the second point is (0, 6), we can also say (x2, y2) = (0, 6)
So x%5B2%5D+=+0, y%5B2%5D+=+6


Put this all together to get: x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+6

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (0, 0) and (0, 6)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%280+-+0%29%5E2+%2B+%280+-+6%29%5E2%29 Plug in x%5B1%5D+=+0, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+6


d+=+sqrt%28%280%29%5E2+%2B+%28-6%29%5E2%29


d+=+sqrt%280+%2B+36%29


d+=+sqrt%2836%29


d+=+6

==========================================================

Answer:


The distance between the two points (0, 0) and (0, 6) is exactly 6 units




Solved by pluggable solver: Distance Formula


The first point is (x1,y1). The second point is (x2,y2)


Since the first point is (12, 0), we can say (x1, y1) = (12, 0)
So x%5B1%5D+=+12, y%5B1%5D+=+0


Since the second point is (0, 6), we can also say (x2, y2) = (0, 6)
So x%5B2%5D+=+0, y%5B2%5D+=+6


Put this all together to get: x%5B1%5D+=+12, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+6

--------------------------------------------------------------------------------------------


Now use the distance formula to find the distance between the two points (12, 0) and (0, 6)



d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D+-+y%5B2%5D%29%5E2%29


d+=+sqrt%28%2812+-+0%29%5E2+%2B+%280+-+6%29%5E2%29 Plug in x%5B1%5D+=+12, y%5B1%5D+=+0, x%5B2%5D+=+0, and y%5B2%5D+=+6


d+=+sqrt%28%2812%29%5E2+%2B+%28-6%29%5E2%29


d+=+sqrt%28144+%2B+36%29


d+=+sqrt%28180%29


d+=+sqrt%2836%2A5%29


d+=+sqrt%2836%29%2Asqrt%285%29


d+=+6%2Asqrt%285%29


d+=+13.4164078649987

==========================================================

Answer:


The distance between the two points (12, 0) and (0, 6) is exactly 6%2Asqrt%285%29 units


The approximate distance between the two points is about 13.4164078649987 units



So again,


Exact Distance: 6%2Asqrt%285%29 units


Approximate Distance: 13.4164078649987 units





A. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244855.gif

AB=4
BC=2sqrt%285%29
CA=2
DE=12
EF=6sqrt%285%29
FD=6
therefore AB%2FDE=BC%2FEF=CA%2FFD=1%2F3
check:
AB%2FDE=4%2F12=1%2F3
BC%2FEF=2sqrt%285%29%2F6sqrt%285%29=2%2F6=1%2F3
CA%2FFD=2%2F6=1%2F3
so, the statement AB%2FDE=BC%2FEF=CA%2FFD=1%2F3 is true and therefore triangle ABC ~ to triangle DEF by SSS
B. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244856.gif
this image shows that
AB=16
BC=20
CA=4
DE=144
EF=180
FD=36
which is not true for given points
so, this is not an answer
C. http://my.thinkwell.com/questionbank/97001-98000/97647/img/244844.gif
this image shows that by the distance formula
AB=16
BC=20
and also
DE=144
EF=180
which is not true, this is not an answer either

so, your answer is: A