Mixing
Triangle Area Ratio Theorem Problems?
Having a hell of a lot of issues with these problems, supposed to be on the topic of triangle area ratio theorem (ratio area of triangles = ratio of triangles' heights x ratio of triangles' bases.)
Here are that are causing me trouble:
Any help would be greatly appreciated.
triangle area ratio
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
add a comment |
Having a hell of a lot of issues with these problems, supposed to be on the topic of triangle area ratio theorem (ratio area of triangles = ratio of triangles' heights x ratio of triangles' bases.)
Here are that are causing me trouble:
Any help would be greatly appreciated.
triangle area ratio
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
add a comment |
Having a hell of a lot of issues with these problems, supposed to be on the topic of triangle area ratio theorem (ratio area of triangles = ratio of triangles' heights x ratio of triangles' bases.)
Here are that are causing me trouble:
Any help would be greatly appreciated.
triangle area ratio
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
Having a hell of a lot of issues with these problems, supposed to be on the topic of triangle area ratio theorem (ratio area of triangles = ratio of triangles' heights x ratio of triangles' bases.)
Here are that are causing me trouble:
Any help would be greatly appreciated.
triangle area ratio
triangle area ratio
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
asked Jan 20 '15 at 4:37
Lydia FinkelLydia Finkel
32
32
add a comment |
add a comment |
2 Answers
2
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oldest
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I'll get you started:
For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.
For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
add a comment |
Without using any trigonometry, here is a solution to (2):
To start off, I created a new point called F that lies on segment AC so that AF = 12.
Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].
△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.
7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].
Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅
So, [ABC] = 126.93̅.
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
I'll get you started:
For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.
For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
add a comment |
I'll get you started:
For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.
For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
add a comment |
I'll get you started:
For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.
For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
I'll get you started:
For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.
For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
answered Jan 20 '15 at 4:53
PlateheadPlatehead
70836
70836
add a comment |
add a comment |
Without using any trigonometry, here is a solution to (2):
To start off, I created a new point called F that lies on segment AC so that AF = 12.
Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].
△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.
7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].
Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅
So, [ABC] = 126.93̅.
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
add a comment |
Without using any trigonometry, here is a solution to (2):
To start off, I created a new point called F that lies on segment AC so that AF = 12.
Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].
△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.
7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].
Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅
So, [ABC] = 126.93̅.
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
add a comment |
Without using any trigonometry, here is a solution to (2):
To start off, I created a new point called F that lies on segment AC so that AF = 12.
Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].
△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.
7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].
Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅
So, [ABC] = 126.93̅.
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
Without using any trigonometry, here is a solution to (2):
To start off, I created a new point called F that lies on segment AC so that AF = 12.
Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].
△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.
7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].
Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅
So, [ABC] = 126.93̅.
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
answered Nov 21 '18 at 21:40
Trent ConleyTrent Conley
1
1
add a comment |
add a comment |
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