# Desperately need help!

• Aug 5th 2009, 12:55 AM
jgv115
Desperately need help!
PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

A) 30/17
B) 12/7
C) 5/3
D) 60/37
E) 60/39

• Aug 5th 2009, 01:49 AM
alexmahone
Quote:

Originally Posted by jgv115
PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

A) 30/17
B) 12/7
C) 5/3
D) 60/37
E) 60/39

Construction: Draw a perpendicular RO from R to the hypotenuse PQ (such that O lies on PQ).

Area of the triangle PQR = $\frac{1}{2} b h$

$\frac{1}{2}*5*RO=\frac{1}{2}*4*3$

$RO=\frac{12}{5}$

-----------------------------------------------------------------------------
In $\triangle RQP$,

$VS \parallel QP$

By basic proportionality theorem,

$\frac{RV}{RQ}=\frac{RS}{RP}=\frac{SV}{PQ}$

$\frac{RV}{4}=\frac{RS}{3}=\frac{SV}{5}=k$

So, $RV=4k, RS=3k, SV=5k$

So the side of the square is $5k$.

------------------------------------------------------------------------------
In $\triangle RQO$,

$VU \parallel RO$

By basic proportionality theorem,

$\frac{VU}{RO}=\frac{VQ}{RQ}$

$\frac{5k}{\frac{12}{5}}=\frac{VQ}{4}$

$\frac{25k}{12}=\frac{VQ}{4}$

$VQ=\frac{25k}{3}$

-----------------------------------------------------------------------------

$RV+VQ=4\ cm$

$4k+\frac{25k}{3}=4$

$12k+25k=12$

$k=\frac{12}{37}$

So, the side of the square is $5k=\frac{60}{37}\ cm$
• Aug 5th 2009, 02:07 AM
Ephemerasylum
This is my first post in the forum, and
I'm here to provide an extremely complicated method.

I'll get you an easier one later.

Be noticed that the units are all omitted.

As illurstrated we know that

$PR = 3$
$QR = 4$

By applying the Pythagoras' theorem in the right-angled triangle,
we obtain that

$PQ = \sqrt{PR^2 + QR^2}= \sqrt{25} = 5$

Then we assume the length of the side to be $a$.
In $\triangle UVQ$, we can see that

$\tan{Q} = \frac{PR}{RQ} = \frac{UV}{UQ}$

Simplification of the equations above gives that

$UQ = \frac{UV*RQ}{PR} = \frac{4a}{3}$

Similarly, in $\triangle PTS$, calculation gives that

$PT = \frac{3a}{4}$

Considering that the segment $SV$ and the segment $PQ$ are parallel, we know that,

$\angle{SVR} = \angle{Q}$

Then we may see that

$\sin{\angle{SVR}} = \frac{SR}{SV} = \frac{PR}{PQ} = \sin{\angle{Q}}$

which gives that,

$SR = \frac{3a}{5}$

and similarly

$RV = \frac{4a}{5}$

By doing all the calculations above we see that

$S_{\triangle{PRQ}} = \frac{1}{2}PR*RQ$
$S_{\triangle{PRQ}} = \frac{1}{2}*[PT*ST+UQ*VU+SR*RV]+a^2$

By solving the equations we finally obtain that

$a = \frac{60}{37}$
• Aug 5th 2009, 02:22 AM
Ephemerasylum
Okay here it comes the easier one.

At first we need to find the length of $PQ$.
My work above gives that

$PQ = 5$

Obviously segments $SV$ and $PQ$ are parallel, thus

$\cos{\angle{SVR}} = \cos{\angle{Q}} = 0.8$
$\cos{\angle{UVQ}} = \sin{\angle{Q}} = 0.6$

Assume the length of the segment $VQ$ to be $x$, then it naturally satisifies that $RV$ equals to $4-x$.

Hence, we obtain that

$SV = \frac{RV}{\cos{\angle{SVR}}} = VU = VQ*\cos{\angle{UVQ}}$

which gives that

$x = \frac{48}{37}$

Therefore, the length of the side equals to

$\frac{5x}{4} = \frac{60}{37}$
• Aug 5th 2009, 03:17 AM
jgv115

alexmahone you wrote this:

$\frac{1}{2}*5*RO=\frac{1}{2}*4*3$

How do you know RO is 3?
• Aug 5th 2009, 03:21 AM
alexmahone
Quote:

Originally Posted by jgv115

alexmahone you wrote this:

$\frac{1}{2}*5*RO=\frac{1}{2}*4*3$

How do you know RO is 3?

Area of $\triangle PQR = \frac{1}{2}bh$

Area = $\frac{1}{2}*5*RO=\frac{1}{2}*4*3$

We equate these two values of the area to find RO.

Thus $RO=\frac{12}{5}$
• Aug 5th 2009, 03:26 AM
jgv115
Do you mean you make it up? You said that half times 5 times RO (the extra line you added) is the area.

What I don't understand it where the value of RO came from? You substituted 3 for RO but how did that get there?
• Aug 5th 2009, 03:40 AM
alexmahone
Quote:

Originally Posted by jgv115
Do you mean you make it up? You said that half times 5 times RO (the extra line you added) is the area.

What I don't understand it where the value of RO came from? You substituted 3 for RO but how did that get there?

Area of $\triangle PQR=\frac{1}{2}PQ*RO$

Also, Area of $\triangle PQR=\frac{1}{2}RQ*RP$

So, $\frac{1}{2}PQ*RO=\frac{1}{2}RQ*RP$

Now do you understand?
• Aug 5th 2009, 04:00 AM
Similar Triangles
Hello jgv115
Quote:

Originally Posted by jgv115
PQR is a right angled triangle with PR = 3cm and QR = 4cm. The square STUV is inscribed in triangle PQR. What is the length, in centimeters of the side of the square.

A) 30/17
B) 12/7
C) 5/3
D) 60/37
E) 60/39

The solutions you've been given are fine, but look a bit complicated, and most have left quite a bit of the working out. May I suggest a simpler way of setting out the answer?

Basically, all four of the triangles are $3,4,5$ right-angled triangles. This means, for example, that the longest side of any of them is $\tfrac53$ of the shortest, the middle-sized side is $\tfrac45$ of the longest, and so on. So, if we call the length of the sides of the square $x$ cm, we can work around the various triangles to get an equation for $x$, like this:

In $\triangle QUV, UQ = \tfrac43x$

In $\triangle RSV, SR = \tfrac35x$

In $\triangle PST, PS = 3 - SR = 3 - \tfrac35x$

$\Rightarrow PT = \tfrac35PS = \tfrac35(3 - \tfrac35x)$

So if we now add up the three lengths along $PQ$:

$PT + TU + UQ = \tfrac35(3 - \tfrac35x)+x+ \tfrac43x = PQ = 5$

Multiply both sides of this equation by $75\, (= 3\times 5\times 5)$:

$45(3 - \tfrac35x) + 75x + 75\times\tfrac43x = 5\times 75$

$\Rightarrow 135 - 27x + 75x + 100x = 375$

$\Rightarrow 148x = 240$

$\Rightarrow x = \frac{60}{37}$