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# In a figure above, triangle PQR is right-angled at angle R. If QR = 4

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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 [#permalink]

Since ∆ PQR is a right-angled triangle, $$a^{2}+ 16= b^{2}$$ ---> $$(b-a)(b+a )= 16$$

The perimeter of triangle PQT equals to the perimeter of triangle PTR:
---> $$a+x +3 = b+ (4-x) + 3$$
$$b -a = 2x -4$$
---> $$(b+a )= \frac{16}{(2x -4)}= \frac{8}{(x-2)}$$

One of features of ∆PRQ is $$b+a > 4$$
---> $$\frac{8}{(x-2)} >4$$
$$\frac{2}{(x-2)} > 1$$
$$\frac{(x-4)}{( x-2)} < 0$$

Since ∆ PTR is a right-angled triangle, $$x <3$$.

So, we got x ----> $$x> 2$$ and $$x <3$$

Well,
$$x(2)= \frac{4-x }{x} = \frac{4-2}{2}= 1$$
$$x(3) =\frac{4-x }{x} = \frac{4-3}{3}= \frac{1}{3}$$

$$\frac{1}{3} < \frac{4-x}{x} < 1$$

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In a figure above, triangle PQR is right-angled at angle R. If QR = 4 [#permalink]
lacktutor wrote:

One of features of ∆PRQ is $$b+a > 4$$
---> $$\frac{8}{(x-2)} >4$$
$$\frac{2}{(x-2)} > 1$$
$$\frac{(x-4)}{( x-2)} < 0$$

lacktutor How did you get from $$\frac{2}{(x-2)} > 1$$ to $$\frac{(x-4)}{( x-2)} < 0$$? Which substitution or identity allowed that? Of course it's true intuitively, since x > 0 must be true, but why does it even matter?
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Re: In a figure above, triangle PQR is right-angled at angle R. If QR = 4 [#permalink]
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Simple solution

There's a much easier and faster method of solving this. This involves picking values for TR that are suggested by our answer choices, instead of trying to work out the answer and pretending we don't already have information in the answer choices.

Let's call TR = x. Then QT = 4-x. We want to set bounds on x so that we can see a relationship in the fraction (4-x)/x.

- Can x be 2? If x = 2, then 4-x=2. For that to be true and for the relationship PR + TR = QT + PQ to also be true, then PR = PQ and the triangle is isosceles. HOWEVER, since we know that angle PRQ is 90 degrees, the triangle isn't isosceles. D is wrong, as TR and QT are not equal.

- Can x be less than 2? If TR=x<2, then for PR + TR = QT + PQ to be true, since TR<QT, PR>PQ. This also cannot be true, since PR is the height of a right triangle and PQ is the hypotenuse. E is wrong, as TR>2>QT.

Therefore x>2. A simpler way of saying this -- since by definition the hypotenuse PQ > height PR, for PR + TR = QT + PQ to be true, then TR > QT. Therefore TR=x>2.

- Next, can QT/TR = 1/3? In this case, QT=1 and TR=3. That's not possible, since TR is the base of a right triangle, and its hypotenuse PT is equal to 3. Therefore TR must be less than 3. B is wrong.

A or C? We just said that TR < 3, so QT>1 therefore QT/TR > 1/3. C is the correct answer.

TLDR;

TR = 4 - QT. By definition, the hypotenuse PQ > height PR, so for PR + TR = QT + PQ to be true, then TR > QT. Therefore TR>2 and QT<2. Then, TR cannot be 3 or larger since PT=3. So the bounds on TR are: 2 < TR < 3. So 1/3 < QT/TR < 1.

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Re: In a figure above, triangle PQR is right-angled at angle R. If QR = 4 [#permalink]
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Re: In a figure above, triangle PQR is right-angled at angle R. If QR = 4 [#permalink]
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