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\(ABCD\) is a rectangle and \(P\) is any point on \(AC\) (except \(A\) and \(C\)). Through \(P\) is plotted a line parallel to \(BC\) which cuts \(AB\)  and \(DC\) at \(R\) and \(S\), respectively. Also, through \(S\) is plotted a line parallel to \(AC\) which cuts \(AD\) at \(T\).

Find \(\dfrac{\text{Area }TSPA}{\text{Area } PRB}\).

 

 Jan 15, 2021
 #1
avatar+112523 
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I have not looked for a general solution but the question implies that the ratio is always the same.

 

If you make ABCD a square and let P be the midpoint of AC then the solution is super easy to find.

 Jan 15, 2021
 #2
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[TSPA] / [PRB] = 2

 Jan 16, 2021
 #3
avatar+112523 
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Refer to the diagram below:

The area of the original rectangle can be assigned any value l since we are only interested in the area ratio of the 2 internal sections.

So I have let the area be 2 units

The width is w

The length is 2/w

The original brown rectangle and the smaller red rectangle are similar figures.  So I have said the ratio of the sides is delta where delta is a positive number less than 1.

From this all the given dimensions and be found.

 

Are of blue parallelogram \(=\frac{2\delta}{w}*w(1-\delta)=2\delta(1-\delta)\)


Area of green triangle =\( \frac{1}{2}*\frac{2}{w}(1-\delta)*\delta w =\delta(1-\delta) \)

 

 

\(\dfrac{\text{Area }TSPA}{\text{Area } PRB} =2\)

 Jan 16, 2021

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