Which of the conditions are necessary to determine the reason for the area?

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I have this statement:

My development was:

The $triangleBPC$ have an area of $fracah2$, where $h =$ Height

The $triangleDPA$ have an area of $fracat2$, where $t = $ Height

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

So, area of $triangleBPC + triangleDPA = fraca2(h+t)$

And the reason is $fracfraca(h+t)2ak$

This is only taking into account condition 1, but I could not achieve more.

The correct answer must be $D)$, but i cant understand how is it possible.

Then, ¿How can I get to the correct answer, and why?

geometry

share|cite|improve this question

edited Jul 30 at 3:23

asked Jul 30 at 2:14

Mattiu

759316

add a comment | 

up vote
1
down vote

favorite

I have this statement:

My development was:

The $triangleBPC$ have an area of $fracah2$, where $h =$ Height

The $triangleDPA$ have an area of $fracat2$, where $t = $ Height

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

So, area of $triangleBPC + triangleDPA = fraca2(h+t)$

And the reason is $fracfraca(h+t)2ak$

This is only taking into account condition 1, but I could not achieve more.

The correct answer must be $D)$, but i cant understand how is it possible.

Then, ¿How can I get to the correct answer, and why?

geometry

share|cite|improve this question

edited Jul 30 at 3:23

asked Jul 30 at 2:14

Mattiu

759316

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have this statement:

My development was:

The $triangleBPC$ have an area of $fracah2$, where $h =$ Height

The $triangleDPA$ have an area of $fracat2$, where $t = $ Height

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

So, area of $triangleBPC + triangleDPA = fraca2(h+t)$

And the reason is $fracfraca(h+t)2ak$

This is only taking into account condition 1, but I could not achieve more.

The correct answer must be $D)$, but i cant understand how is it possible.

Then, ¿How can I get to the correct answer, and why?

geometry

share|cite|improve this question

edited Jul 30 at 3:23

asked Jul 30 at 2:14

Mattiu

759316

I have this statement:

My development was:

The $triangleBPC$ have an area of $fracah2$, where $h =$ Height

The $triangleDPA$ have an area of $fracat2$, where $t = $ Height

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

So, area of $triangleBPC + triangleDPA = fraca2(h+t)$

And the reason is $fracfraca(h+t)2ak$

This is only taking into account condition 1, but I could not achieve more.

The correct answer must be $D)$, but i cant understand how is it possible.

Then, ¿How can I get to the correct answer, and why?

geometry

share|cite|improve this question

edited Jul 30 at 3:23

asked Jul 30 at 2:14

Mattiu

759316

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 3:23

edited Jul 30 at 3:23

edited Jul 30 at 3:23

asked Jul 30 at 2:14

Mattiu

759316

asked Jul 30 at 2:14

Mattiu

759316

asked Jul 30 at 2:14

Mattiu

759316

759316

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $,BC,$ and $,DA,$ so $,k=h+t,$.

And the reason is $;displaystylefrac;;fraca(h+t)2;;ak$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $,dfrac12,$.

share|cite|improve this answer

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $AC$ isn't diagonal
  – Mattiu
  Jul 30 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mattiu You are right, and the drawing is misleading. Answer corrected.
  – dxiv
  Jul 30 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
  – dxiv
  Jul 30 at 4:02
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
  – Mattiu
  Jul 30 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
  – Mattiu
  Jul 31 at 16:53
  
  

  
  
  
  

 | 
show 3 more comments

up vote
1
down vote

You found this expression:
$$fracleft(fraca(h+t)2right)ak.$$

You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.

To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$

To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try drawing
the height of triangle $triangle ADP$ on the base $AD,$
the height of triangle $triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$

share|cite|improve this answer

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited, the correct answer must be D
  – Mattiu
  Jul 30 at 3:23
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $,BC,$ and $,DA,$ so $,k=h+t,$.

And the reason is $;displaystylefrac;;fraca(h+t)2;;ak$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $,dfrac12,$.

share|cite|improve this answer

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $AC$ isn't diagonal
  – Mattiu
  Jul 30 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mattiu You are right, and the drawing is misleading. Answer corrected.
  – dxiv
  Jul 30 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
  – dxiv
  Jul 30 at 4:02
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
  – Mattiu
  Jul 30 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
  – Mattiu
  Jul 31 at 16:53
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote



accepted

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $,BC,$ and $,DA,$ so $,k=h+t,$.

And the reason is $;displaystylefrac;;fraca(h+t)2;;ak$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $,dfrac12,$.

share|cite|improve this answer

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $AC$ isn't diagonal
  – Mattiu
  Jul 30 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mattiu You are right, and the drawing is misleading. Answer corrected.
  – dxiv
  Jul 30 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
  – dxiv
  Jul 30 at 4:02
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
  – Mattiu
  Jul 30 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
  – Mattiu
  Jul 31 at 16:53
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $,BC,$ and $,DA,$ so $,k=h+t,$.

And the reason is $;displaystylefrac;;fraca(h+t)2;;ak$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $,dfrac12,$.

share|cite|improve this answer

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $,BC,$ and $,DA,$ so $,k=h+t,$.

And the reason is $;displaystylefrac;;fraca(h+t)2;;ak$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $,dfrac12,$.

share|cite|improve this answer

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 2:37

edited Jul 30 at 2:37

edited Jul 30 at 2:37

answered Jul 30 at 2:23

dxiv

53.8k64796

answered Jul 30 at 2:23

dxiv

53.8k64796

answered Jul 30 at 2:23

dxiv

53.8k64796

53.8k64796

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $AC$ isn't diagonal
  – Mattiu
  Jul 30 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mattiu You are right, and the drawing is misleading. Answer corrected.
  – dxiv
  Jul 30 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
  – dxiv
  Jul 30 at 4:02
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
  – Mattiu
  Jul 30 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
  – Mattiu
  Jul 31 at 16:53
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But $AC$ isn't diagonal
  – Mattiu
  Jul 30 at 2:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mattiu You are right, and the drawing is misleading. Answer corrected.
  – dxiv
  Jul 30 at 2:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
  – dxiv
  Jul 30 at 4:02
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
  – Mattiu
  Jul 30 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
  – Mattiu
  Jul 31 at 16:53
  
  

  
  
  
  

But $AC$ isn't diagonal
– Mattiu
Jul 30 at 2:29

But $AC$ isn't diagonal
– Mattiu
Jul 30 at 2:29

@Mattiu You are right, and the drawing is misleading. Answer corrected.
– dxiv
Jul 30 at 2:38

@Mattiu You are right, and the drawing is misleading. Answer corrected.
– dxiv
Jul 30 at 2:38

1

1

@Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
– dxiv
Jul 30 at 4:02

@Mattiu Because $h$ and $t$ are altitudes which are drawn perpendicular to $BC$ and $DA$ respectively. Since $BC || DA$ in a parallelogram, it follows that the altitudes are also parallel and, given that they have the common point $P$, they both lie on the same line. That line is perpendicular to $BC$ and $DA$ so the segment between the intercepts on $BC$ and $DA$ is precisely the height of the parallelogram i.e. $k$.
– dxiv
Jul 30 at 4:02

1

1

Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
– Mattiu
Jul 30 at 16:51

Well, and with the information of i) can be solved, so ii) implies the i) , so each one by itself. Thanks!
– Mattiu
Jul 30 at 16:51

1

1

An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
– Mattiu
Jul 31 at 16:53

An another answer can be: If you draw a parallel line that contains the point $P$, now you have two parallelogram with area $alpha + beta$, so the triangles, $triangleAPD, triangleBPC$ have a $(alpha)/2$ and $(beta)/2$ area, respectively, so the total area of triangles is $(alpha + beta)/2$
– Mattiu
Jul 31 at 16:53

 | 
show 3 more comments

up vote
1
down vote

You found this expression:
$$fracleft(fraca(h+t)2right)ak.$$

You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.

To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$

To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try drawing
the height of triangle $triangle ADP$ on the base $AD,$
the height of triangle $triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$

share|cite|improve this answer

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited, the correct answer must be D
  – Mattiu
  Jul 30 at 3:23
  

  
  
  
  

add a comment | 

up vote
1
down vote

You found this expression:
$$fracleft(fraca(h+t)2right)ak.$$

You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.

To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$

To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try drawing
the height of triangle $triangle ADP$ on the base $AD,$
the height of triangle $triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$

share|cite|improve this answer

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited, the correct answer must be D
  – Mattiu
  Jul 30 at 3:23
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

You found this expression:
$$fracleft(fraca(h+t)2right)ak.$$

You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.

To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$

To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try drawing
the height of triangle $triangle ADP$ on the base $AD,$
the height of triangle $triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$

share|cite|improve this answer

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

You found this expression:
$$fracleft(fraca(h+t)2right)ak.$$

You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.

To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$

To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try drawing
the height of triangle $triangle ADP$ on the base $AD,$
the height of triangle $triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$

share|cite|improve this answer

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 3:51

edited Jul 30 at 3:51

edited Jul 30 at 3:51

answered Jul 30 at 2:43

David K

48.1k340107

answered Jul 30 at 2:43

David K

48.1k340107

answered Jul 30 at 2:43

David K

48.1k340107

48.1k340107

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited, the correct answer must be D
  – Mattiu
  Jul 30 at 3:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited, the correct answer must be D
  – Mattiu
  Jul 30 at 3:23
  

  
  
  
  

Edited, the correct answer must be D
– Mattiu
Jul 30 at 3:23

Edited, the correct answer must be D
– Mattiu
Jul 30 at 3:23

add a comment | 

 

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