# What fraction of the square is the area of this quadrilateral? (diagram included)

• Aug 4th 2010, 02:12 AM
jgv115
What fraction of the square is the area of this quadrilateral? (diagram included)
Diagram:

Attachment 18426

U and T are midpoints of PQ and PS. V is the intersection of QT and US.

What fraction of the square is the area of the quadrilateral QVSR?

I've thought long and hard about this but I can never seem to get it. Could someone start me off?
• Aug 4th 2010, 02:31 AM
Quote:

Originally Posted by jgv115
Diagram:

Attachment 18426

U and T are midpoints of PQ and PS. V is the intersection of QT and US.

What fraction of the square is the area of the quadrilateral QVSR?

I've thought long and hard about this but I can never seem to get it. Could someone start me off?

If you begin to label angles,
you will see that triangle SVR and PVU are equiangular.
|SR|=2|PU|, hence SVR's perpendicular height is double that of PVU's
and so it's area is 4 times the area of PVU.

Similarly for triangles RVQ and TVP.

Finally examine triangles UVQ and TVS.
• Aug 4th 2010, 03:12 AM
jgv115
Ok, I understand. I still don't understand how to apply the knowledge though. May I ask for a bit more assistance? :)
• Aug 4th 2010, 03:52 AM
Quote:

Originally Posted by jgv115
Ok, I understand. I still don't understand how to apply the knowledge though. May I ask for a bit more assistance? :)

Sure,

label \$\displaystyle \displaystyle\angle PUS=\angle PUV\$ as angle B

then \$\displaystyle \angle USR=\angle VSR\$ also is B

\$\displaystyle \angle UPV=\angle UPR\$ is A

\$\displaystyle \angle SRP=\angle SRV\$ also is A

\$\displaystyle \angle PVU=\angle SVR\$ is C

Hence, triangles SVR and PVU are equiangular.
Now, since U is midway between P and Q, then |SR|=2|UP|

The base of triangle SVR is twice the length of the base of triangle PUV.
Since the triangles are equiangular, then triangle SVR is a magnified version of PUV.

Area of a triangle is 0.5(base)(perpendicular height)
Since all lengths (linear measurements) of SVR are twice the corresponding lengths of PVU, then it's perpendicular height is also twice that of PVU.

Hence, area of PVU=0.5|PU|h and area of SVR is 0.5[2|PU|2h]

where h is the perpendicular height of PVU.

Therefore the area of SVR is 4 times the area of PVU.

By the same analysis, triangle RVQ is 4 times the area of triangle PVT.

It remains to compare UVQ with TVS.
Using the line going from P to R, the sum of the areas of PVU+UVQ+QVR
equals the sum of the areas of PVT+TVS+SVR.

SVR and QVR are equiangular, since the shape is a square,
so the two angles at vertex R are 45 degrees.

\$\displaystyle |\angle VQR|=|\angle VSR|\$

since the shape is a square and U and T are the midpoints of sides.
If 2 pairs of angles in 2 triangles are equal, the third angles are equal.
Hence SVR and QVR are equiangular and since their base lengths are equal,
they are congruent (identical).

Therefore the sums of SVR+PVU and the sums of QVR and TVS are equal.
Hence the areas of UVQ and TVS are equal.
Since those triangles have the same bases and perpendicular heights as PVU and TVS, they have the same areas as UVQ and TVS.

Therefore the area of SVR+QVR equals twice the remaining area.
So, quadrilateral QVSR is two-thirds of the entire area.
• Aug 4th 2010, 04:25 AM
jgv115
Wow i'm officially confused o__O Thanks for your input though.

Quote:

Therefore the area of SVR+QVR equals twice the remaining area.
So, quadrilateral QVSR is two-thirds of the entire area.
How can you know that that quadrilateral is two thirds and SVR+QVE equals twice the remaining area of what?
• Aug 4th 2010, 04:50 AM
Quote:

Originally Posted by jgv115
Wow i'm officially confused o__O Thanks for your input though.

How can you know that that quadrilateral is two thirds and SVR+QVE equals twice the remaining area of what?

The remaining area within the square.

If SVR is 4 times PVU and if PVU is the same area as PVT and TVS,
then SVR is twice (PVT+TVS).
Those 3 triangles make up half the square.
Triangle SVR is twice triangle PVS.
Hence PVS+2(PVS)=3(PVS)
PVS is one-third of triangle PRS and SVR is two-thirds of triangle PRS.

The same proportions apply for the other half of the square inside PQR.

• Aug 4th 2010, 05:42 AM
sa-ri-ga-ma
Try this one.

PSU = PQT = 1/4*PQRS

PSVQ = PSU + PQT - PTVU = 1/2*PQRS - PTVU

Archie Meade already shown that PVU = 1/4*SVR and PTV = 1/4*QRV

SO PTVU = 1/4*SVQR

Now required area SVQR = PQRS - PSVQ= PQRS - (1/2*PQRS - PTVU) = PQRS - (1/2*PQRS - 1/4*SVQR)

SVRQ = PQRS - 1/2*PQRS + 1/4*SVQR

SVQR - 1/4*SVQR = 1/2*PQRS

3/4*SVQR = 1/2*PQRS

So SVQR = 2/3*PQRS.
• Aug 4th 2010, 06:04 AM
It can be solved without dealing with angles at all.

As sa-ri-ga-ma showed,

triangle SPU is a quarter of the area of the square (which can be seen if we drop a perpendicular to [SR] from U).
triangle QPT is also a quarter of the square.
triangles PVU and UVQ have the same base lengths (since [UP] is the same length as [UQ]) and the same perpendicular heights.
The same goes for triangles PVT and TVS.

Hence, since triangles PTQ and SPU contain 3 identical triangles,
and are 1/4 of the square area, then each small triangle is (1/3)(1/4)=1/12 of the entire square.

Since PRS is half the square, then 1/12 +1/12 +SVR is 1/2 of the square.

1/6 +SVR = 3/6 of the square, so SVR is 2/6 of the square.

Also RVQ is 2/6 of the square.

Hence the quadrilateral QVSR is 2/6+2/6=4/6=2/3 of the square.
• Aug 4th 2010, 12:07 PM
bjhopper
fraction of square
Hello all,
different similar triangles
1 draw TU meeting PR @M and SQ meeting PR @K.Triangle PTU similar to PSQ.TU parallel to SQ and = 1/2 SQ.Triangle TVU similar to SVQ. MV =1/2 VK
2 let PQ =1 Area square =1
5 VK = 2/3 MK = 1/3rad2
6Area QSR = 1/2
7 Area SVQ = MK x SQ x1/2
8 Area SVQR = QSR + SVQ
= 1/2 + 1/6= 2/3

bjh
• Aug 4th 2010, 08:22 PM
sa-ri-ga-ma
Further simplification.

V is the centroid of the triangle SPQ.

Hence VK = 1/3*PK and area of SVQ = 1/3*area of SPQ = 1/3*PQRS/2 = PQRS/6.

SQR = 1/2*PQRS

SO SVQR = SVQ + SQR = (1/6 + 1/2)*PQRS.
• Aug 5th 2010, 01:45 AM
jgv115
Thanks guys! After reading your working out 100 times, I think i finally get it!

I love this forum :)
• Aug 5th 2010, 05:13 AM
bjhopper
Excellent!

bjh