This discussion is closed.
theone
Badges: 0
Rep:
?
#1
Report Thread starter 17 years ago
#1
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.
0
Juwel
Badges: 18
Rep:
?
#2
Report 17 years ago
#2
(Original post by 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.
(PQ)^2=(AP)^2+(AQ)^2

Doh I flopped!
0
-=flyboy=-
Badges: 0
#3
Report 17 years ago
#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
0
Bhaal85
Badges: 9
Rep:
?
#4
Report 17 years ago
#4
(Original post by -=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
I keep getting 4.
0
Bhaal85
Badges: 9
Rep:
?
#5
Report 17 years ago
#5
(Original post by 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
0
Juwel
Badges: 18
Rep:
?
#6
Report 17 years ago
#6
But the question was about quadrilaterals in general. You people are answering for a parallelogram-type scenario.
0
Juwel
Badges: 18
Rep:
?
#7
Report 17 years ago
#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.
0
theone
Badges: 0
Rep:
?
#8
Report Thread starter 17 years ago
#8
(Original post by 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.
Cheers, it's the midsegment bit I never saw. Added to your rep
0
Juwel
Badges: 18
Rep:
?
#9
Report 17 years ago
#9
(Original post by theone)
Cheers, it's the midsegment bit I never saw. Added to your rep
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.
0
theone
Badges: 0
Rep:
?
#10
Report Thread starter 17 years ago
#10
(Original post by 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.
Yeah I've just gone through it thoroughly now and it makes sense...
0
Camford
Badges: 12
Rep:
?
#11
Report 17 years ago
#11
Is that a SMC Question?
0
theone
Badges: 0
Rep:
?
#12
Report Thread starter 17 years ago
#12
(Original post by Camford)
Is that a SMC Question?
It's a BMO question.
0
Camford
Badges: 12
Rep:
?
#13
Report 17 years ago
#13
It's a BMO question
You joking...
0
theone
Badges: 0
Rep:
?
#14
Report Thread starter 17 years ago
#14
(Original post by Camford)
You joking...
No, why would I be?
0
theone
Badges: 0
Rep:
?
#15
Report Thread starter 17 years ago
#15
(Original post by Camford)
You joking...
FYI, it's from the 1997 BMO 1 - and I found it the hardest question on the paper.
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Which of these would you use to help with making uni decisions?

Webinars (78)
11.91%
Virtual campus tours/open days (159)
24.27%
Live streaming events (52)
7.94%
Online AMAs/guest lectures (59)
9.01%
A uni comparison tool (153)
23.36%
An in-person event when available (154)
23.51%

Watched Threads

View All