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.
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.
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?Therefore the area of SVR+QVR equals twice the remaining area.
So, quadrilateral QVSR is two-thirds of the entire area.
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.
What are you confused about?
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.
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.
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
3 PK=1/rad2,PR=2/rad2=SQ
4 PM = MK=1/2rad2
5 VK = 2/3 MK = 1/3rad2
6Area QSR = 1/2
7 Area SVQ = MK x SQ x1/2
=1/3rad2 x2/rad2 x1/2 = 1/6
8 Area SVQR = QSR + SVQ
= 1/2 + 1/6= 2/3
bjh