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Prove that the area of the trangles are equal.

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Prove that the area of the trangles are equal.

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4

$begingroup$

Prove that the area of all the traingles in the figure below are equal.

I tried using geogebra to determine an arbitrary values of $a$ , $b$, and $c$. I found out that the triangles have equal measure of area.

triangles area pythagorean-triples

share|cite|improve this question

asked 8 hours ago

MRAMRA

6677 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: what is the angle between sides "with two stripes" and "with three stripes" in your triangle $A_1$? How is it related to angles in the triangle $A_3$? How do we find are given two sides and an angle between them?
  $endgroup$
  – TZakrevskiy
  8 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Prove that the area of all the traingles in the figure below are equal.

I tried using geogebra to determine an arbitrary values of $a$ , $b$, and $c$. I found out that the triangles have equal measure of area.

triangles area pythagorean-triples

share|cite|improve this question

asked 8 hours ago

MRAMRA

6677 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: what is the angle between sides "with two stripes" and "with three stripes" in your triangle $A_1$? How is it related to angles in the triangle $A_3$? How do we find are given two sides and an angle between them?
  $endgroup$
  – TZakrevskiy
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Prove that the area of all the traingles in the figure below are equal.

I tried using geogebra to determine an arbitrary values of $a$ , $b$, and $c$. I found out that the triangles have equal measure of area.

triangles area pythagorean-triples

share|cite|improve this question

asked 8 hours ago

MRAMRA

6677 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

MRAMRA

6677 bronze badges

share|cite|improve this question

asked 8 hours ago

MRAMRA

6677 bronze badges

asked 8 hours ago

MRAMRA

6677 bronze badges

asked 8 hours ago

MRAMRA

6677 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

Let

• 
  $*sbh$ stands for triangles have same base and height.


• 
  $*c$ stands for triangles are congruent.

We have

1. $A_1 stackrel{*sbh}{=} B_1 stackrel{*c}{=} B_3 stackrel{*c}{=} A_3$,




2. $A_3 stackrel{*c}{=} B_3 stackrel{*c}{=} B_4 stackrel{*sbh}= A_4$,




3. $A_2 stackrel{*c}= A_3$

share|cite|improve this answer

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

It is useful to know that the area of a triangle can be calculated by
$$frac12absintheta$$
where $a$ and $b$ are two side lengths of the triangle and $theta$ is the angle between those two side lengths.

Let $alpha$ be the angle in $A_3$ formed by the side lengths $b$ and $c$. Then, the angle formed by the side lengths $b$ and $c$ in $A_4$ is $pi-alpha$. Since $sinalpha=sin(pi-alpha)$, $A_3$ and $A_4$ have the same area.

This can be applied to the other triangles in your diagram.

share|cite|improve this answer

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

Area of $A_2$ and $A_3$ is $ab/2$.

For $A_1$: draw a parallel to $b$ through the upper vertex of the square on $c$. The distance from this parallel to the side at bottom of $A_1$ is $b$ (Pythagoren theorem).

Can you do the same for $A_4$?

share|cite|improve this answer

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

Let

• 
  $*sbh$ stands for triangles have same base and height.


• 
  $*c$ stands for triangles are congruent.

We have

1. $A_1 stackrel{*sbh}{=} B_1 stackrel{*c}{=} B_3 stackrel{*c}{=} A_3$,




2. $A_3 stackrel{*c}{=} B_3 stackrel{*c}{=} B_4 stackrel{*sbh}= A_4$,




3. $A_2 stackrel{*c}= A_3$

share|cite|improve this answer

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Let

• 
  $*sbh$ stands for triangles have same base and height.


• 
  $*c$ stands for triangles are congruent.

We have

1. $A_1 stackrel{*sbh}{=} B_1 stackrel{*c}{=} B_3 stackrel{*c}{=} A_3$,




2. $A_3 stackrel{*c}{=} B_3 stackrel{*c}{=} B_4 stackrel{*sbh}= A_4$,




3. $A_2 stackrel{*c}= A_3$

share|cite|improve this answer

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let

• 
  $*sbh$ stands for triangles have same base and height.


• 
  $*c$ stands for triangles are congruent.

We have

1. $A_1 stackrel{*sbh}{=} B_1 stackrel{*c}{=} B_3 stackrel{*c}{=} A_3$,




2. $A_3 stackrel{*c}{=} B_3 stackrel{*c}{=} B_4 stackrel{*sbh}= A_4$,




3. $A_2 stackrel{*c}= A_3$

share|cite|improve this answer

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

answered 8 hours ago

achille huiachille hui

99.9k55 gold badges136136 silver badges271271 bronze badges

2

$begingroup$

It is useful to know that the area of a triangle can be calculated by
$$frac12absintheta$$
where $a$ and $b$ are two side lengths of the triangle and $theta$ is the angle between those two side lengths.

Let $alpha$ be the angle in $A_3$ formed by the side lengths $b$ and $c$. Then, the angle formed by the side lengths $b$ and $c$ in $A_4$ is $pi-alpha$. Since $sinalpha=sin(pi-alpha)$, $A_3$ and $A_4$ have the same area.

This can be applied to the other triangles in your diagram.

share|cite|improve this answer

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

It is useful to know that the area of a triangle can be calculated by
$$frac12absintheta$$
where $a$ and $b$ are two side lengths of the triangle and $theta$ is the angle between those two side lengths.

Let $alpha$ be the angle in $A_3$ formed by the side lengths $b$ and $c$. Then, the angle formed by the side lengths $b$ and $c$ in $A_4$ is $pi-alpha$. Since $sinalpha=sin(pi-alpha)$, $A_3$ and $A_4$ have the same area.

This can be applied to the other triangles in your diagram.

share|cite|improve this answer

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

It is useful to know that the area of a triangle can be calculated by
$$frac12absintheta$$
where $a$ and $b$ are two side lengths of the triangle and $theta$ is the angle between those two side lengths.

Let $alpha$ be the angle in $A_3$ formed by the side lengths $b$ and $c$. Then, the angle formed by the side lengths $b$ and $c$ in $A_4$ is $pi-alpha$. Since $sinalpha=sin(pi-alpha)$, $A_3$ and $A_4$ have the same area.

This can be applied to the other triangles in your diagram.

share|cite|improve this answer

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

answered 8 hours ago

Andrew ChinAndrew Chin

7511212 bronze badges

1

$begingroup$

Area of $A_2$ and $A_3$ is $ab/2$.

For $A_1$: draw a parallel to $b$ through the upper vertex of the square on $c$. The distance from this parallel to the side at bottom of $A_1$ is $b$ (Pythagoren theorem).

Can you do the same for $A_4$?

share|cite|improve this answer

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

Area of $A_2$ and $A_3$ is $ab/2$.

For $A_1$: draw a parallel to $b$ through the upper vertex of the square on $c$. The distance from this parallel to the side at bottom of $A_1$ is $b$ (Pythagoren theorem).

Can you do the same for $A_4$?

share|cite|improve this answer

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Area of $A_2$ and $A_3$ is $ab/2$.

For $A_1$: draw a parallel to $b$ through the upper vertex of the square on $c$. The distance from this parallel to the side at bottom of $A_1$ is $b$ (Pythagoren theorem).

Can you do the same for $A_4$?

share|cite|improve this answer

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges

answered 8 hours ago

ajotatxeajotatxe

57.3k22 gold badges4545 silver badges9393 bronze badges