# Equilateral Triangle

• Jul 31st 2009, 11:06 PM
jgv115
Equilateral Triangle
PQR is an equilateral triangle. The point U is the mid-point of PR. Points T and S divide QP and QR in the ratio 1:2. The point of intersection of PS, RT and QU is X. If the area of QSX is 1 sqaure unit, what is the area in square units of PQR.

http://i22.photobucket.com/albums/b3...ntitled-13.jpg
the middle point is X
• Aug 1st 2009, 12:18 AM
red_dog
We have $\displaystyle A[QSX]=1\Rightarrow A[SRX]=2\Rightarrow A[QPX]=3$

Let $\displaystyle ST\cap QU=M$.

$\displaystyle QM=\frac{1}{3}QU, \ ST=\frac{1}{3}PR, \ MX=\frac{1}{3}XU$

$\displaystyle \frac{MX}{XU}=\frac{1}{3}\Rightarrow\frac{MX}{MU}= \frac{1}{4}$

Then $\displaystyle MX=\frac{1}{4}MU=\frac{1}{4}(QU-QM)=\frac{1}{6}QU$

$\displaystyle QX=QM+MX=\frac{1}{2}QU\Rightarrow A[XUR]=A[QXR]=3$

Then $\displaystyle A[QUR]=6\Rightarrow A[PQR]=12$
• Aug 1st 2009, 12:21 AM
pacman
Hint: Use Area-Ratio Method or maybe 2-pole problem

reddog is very fast
• Aug 1st 2009, 12:22 AM
CaptainBlack
Quote:

Originally Posted by jgv115
PQR is an equilateral triangle. The point U is the mid-point of PR. Points T and S divide QP and QR in the ratio 1:2. The point of intersection of PS, RT and QU is X. If the area of QSX is 1 sqaure unit, what is the area in square units of PQR.

http://i22.photobucket.com/albums/b3...ntitled-13.jpg
the middle point is X

QX = (1/2)QU

QR =(1/3) perp height from S on to QX

therefore area QXS is (1/6) area of QUR.

Hence area of PQR=12.

CB
• Aug 1st 2009, 01:13 AM
jgv115
Quote:

Originally Posted by red_dog
We have $\displaystyle A[QSX]=1\Rightarrow A[SRX]=2\Rightarrow A[QPX]=3$

Let $\displaystyle ST\cap QU=M$.

$\displaystyle QM=\frac{1}{3}QU, \ ST=\frac{1}{3}PR, \ MX=\frac{1}{3}XU$

$\displaystyle \frac{MX}{XU}=\frac{1}{3}\Rightarrow\frac{MX}{MU}= \frac{1}{4}$

Then $\displaystyle MX=\frac{1}{4}MU=\frac{1}{4}(QU-QM)=\frac{1}{6}QU$

$\displaystyle QX=QM+MX=\frac{1}{2}QU\Rightarrow A[XUR]=A[QXR]=3$

Then $\displaystyle A[QUR]=6\Rightarrow A[PQR]=12$

mm i only understand parts of that.

$\displaystyle \frac{MX}{XU}=\frac{1}{3}\Rightarrow\frac{MX}{MU}= \frac{1}{4}$

this part.

I'm assuming your made up "M" is if you draw a straight line from T to S right in the middle? Correct me if I'm wrong.

How do you get that?