# Quadrilateral Problem

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Let ABCD be a quadrilateral and P,Q,R,S be the midpoints of AB, BC, CD and DA respectively. Show that if the area of PQRS is 1, then the area of ABCD is 2.

Totally stumped, don't know where to start, except to draw a diagram.

Totally stumped, don't know where to start, except to draw a diagram.

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#2

(Original post by

Let ABCD be a quadrilateral and P,Q,R,S be the midpoints of AB, BC, CD and DA respectively. Show that if the area of PQRS is 1, then the area of ABCD is 2.

Totally stumped, don't know where to start, except to draw a diagram.

**theone**)Let ABCD be a quadrilateral and P,Q,R,S be the midpoints of AB, BC, CD and DA respectively. Show that if the area of PQRS is 1, then the area of ABCD is 2.

Totally stumped, don't know where to start, except to draw a diagram.

Doh I flopped!

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#3

You mean that there is a diamond inside a square?

Okay, if the smaller one is 1cm^2, then you have 4 triangles each corner from the bigger square, right? So divide the 1 by 4 = 0.25cm^2. this equals the area of those triangles, so 4 x 0.25 = 1 then add the area of the inside triangle and you have 2. Easy.

I think this is correct, hope it helps

Dave

Okay, if the smaller one is 1cm^2, then you have 4 triangles each corner from the bigger square, right? So divide the 1 by 4 = 0.25cm^2. this equals the area of those triangles, so 4 x 0.25 = 1 then add the area of the inside triangle and you have 2. Easy.

I think this is correct, hope it helps

Dave

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#4

(Original post by

You mean that there is a diamond inside a square?

Okay, if the smaller one is 1cm^2, then you have 4 triangles each corner from the bigger square, right? So divide the 1 by 4 = 0.25cm^2. this equals the area of those triangles, so 4 x 0.25 = 1 then add the area of the inside triangle and you have 2. Easy.

I think this is correct, hope it helps

Dave

**-=flyboy=-**)You mean that there is a diamond inside a square?

Okay, if the smaller one is 1cm^2, then you have 4 triangles each corner from the bigger square, right? So divide the 1 by 4 = 0.25cm^2. this equals the area of those triangles, so 4 x 0.25 = 1 then add the area of the inside triangle and you have 2. Easy.

I think this is correct, hope it helps

Dave

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#5

**theone**)

Let ABCD be a quadrilateral and P,Q,R,S be the midpoints of AB, BC, CD and DA respectively. Show that if the area of PQRS is 1, then the area of ABCD is 2.

Totally stumped, don't know where to start, except to draw a diagram.

Worked it out.

If PQRS are equal length then PQ*PQ =1 in a square a side * side = area : lets call this 'a'

Next, from drawing the problem we see that 'a' intersects the midpoints of DA and AP. Lets call the midpoint length 'b'. Therefore b^2=1, therefore b also equal 1.

Hence, if b = midpoint, then AB = 2b. To

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#6

But the question was about quadrilaterals in general. You people are answering for a parallelogram-type scenario.

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#7

Here's one I found on the net. Bit of a headf*ck but it works I suppose. Enjoy!

Let us consider a quadrilateral as you described it:

D

/ \

/ R

/ \

S C

/ |

/ Q

/ |

A-------P-------B

First note that, for instance, PQ is parallel to and half the measure

of AC, because in triangle ABC segment PQ is a midsegment.

The same can be said about RS.

So PQ and RS are parallel and congruent. From this we see that PQRS is

a parallelogram.

Now let X and Y be the intersections of AC with PS and QR,

respectively. And let E be the foot of the altitude from B on AC.

PQYX is a parallelogram too. When PQ is taken as base, then the

measure of the height of this parallelogram is half BE. We find:

Area PQYX = PQ * 0.5*BE

= 0.5*AC * 0.5*BE

= 0.5 * (0.5*AC*BE)

= 0.5 * Area ABC

In the same way we find that area XYRS = 0.5* area ACD.

Combining these two we find the desired result that:

Area PQRS = 0.5 * Area ABCD.

Let us consider a quadrilateral as you described it:

D

/ \

/ R

/ \

S C

/ |

/ Q

/ |

A-------P-------B

First note that, for instance, PQ is parallel to and half the measure

of AC, because in triangle ABC segment PQ is a midsegment.

The same can be said about RS.

So PQ and RS are parallel and congruent. From this we see that PQRS is

a parallelogram.

Now let X and Y be the intersections of AC with PS and QR,

respectively. And let E be the foot of the altitude from B on AC.

PQYX is a parallelogram too. When PQ is taken as base, then the

measure of the height of this parallelogram is half BE. We find:

Area PQYX = PQ * 0.5*BE

= 0.5*AC * 0.5*BE

= 0.5 * (0.5*AC*BE)

= 0.5 * Area ABC

In the same way we find that area XYRS = 0.5* area ACD.

Combining these two we find the desired result that:

Area PQRS = 0.5 * Area ABCD.

0

(Original post by

Here's one I found on the net. Bit of a headf*ck but it works I suppose. Enjoy!

Let us consider a quadrilateral as you described it:

D

/ \

/ R

/ \

S C

/ |

/ Q

/ |

A-------P-------B

First note that, for instance, PQ is parallel to and half the measure

of AC, because in triangle ABC segment PQ is a midsegment.

The same can be said about RS.

So PQ and RS are parallel and congruent. From this we see that PQRS is

a parallelogram.

Now let X and Y be the intersections of AC with PS and QR,

respectively. And let E be the foot of the altitude from B on AC.

PQYX is a parallelogram too. When PQ is taken as base, then the

measure of the height of this parallelogram is half BE. We find:

Area PQYX = PQ * 0.5*BE

= 0.5*AC * 0.5*BE

= 0.5 * (0.5*AC*BE)

= 0.5 * Area ABC

In the same way we find that area XYRS = 0.5* area ACD.

Combining these two we find the desired result that:

Area PQRS = 0.5 * Area ABCD.

**ZJuwelH**)Here's one I found on the net. Bit of a headf*ck but it works I suppose. Enjoy!

Let us consider a quadrilateral as you described it:

D

/ \

/ R

/ \

S C

/ |

/ Q

/ |

A-------P-------B

First note that, for instance, PQ is parallel to and half the measure

of AC, because in triangle ABC segment PQ is a midsegment.

The same can be said about RS.

So PQ and RS are parallel and congruent. From this we see that PQRS is

a parallelogram.

Now let X and Y be the intersections of AC with PS and QR,

respectively. And let E be the foot of the altitude from B on AC.

PQYX is a parallelogram too. When PQ is taken as base, then the

measure of the height of this parallelogram is half BE. We find:

Area PQYX = PQ * 0.5*BE

= 0.5*AC * 0.5*BE

= 0.5 * (0.5*AC*BE)

= 0.5 * Area ABC

In the same way we find that area XYRS = 0.5* area ACD.

Combining these two we find the desired result that:

Area PQRS = 0.5 * Area ABCD.

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#9

(Original post by

Cheers, it's the midsegment bit I never saw. Added to your rep

**theone**)Cheers, it's the midsegment bit I never saw. Added to your rep

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(Original post by

Oh ta. So you get it then? I didn't at first, though admittedly I haven't had a good look, but I have a vague understanding of what it's saying now.

**ZJuwelH**)Oh ta. So you get it then? I didn't at first, though admittedly I haven't had a good look, but I have a vague understanding of what it's saying now.

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(Original post by

You joking...

**Camford**)You joking...

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