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Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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Triangle PQR has angle PRQ equal to 90 degrees. What is the value of PR + RQ?

A. Diameter of the inscribed circle of the triangle PQR is equal to 10 cm.
B. Diameter of the circumscribed circle of the triangle PQR is equal to 18 cm.
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Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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excelingmat wrote:
Triangle PQR has angle PRQ equal to 90 degrees. What is the value of PR + RQ?

A. Diameter of the inscribed circle of the triangle PQR is equal to 10 cm.
B. Diameter of the circumscribed circle of the triangle PQR is equal to 18 cm.

When you select the tag, "source-others please specify", make sure to specify the source at the end of your question.

Given that $$\angle{PRQ}$$ = 90 degrees.

This question uses a very non standard relation between the incircle's radius (r) and the sides (a,b,c, c being the hypotenuse). Let A be the area of the triangle = 0.5*a*b,

Thus, $$r = \frac{2*Area of the triangle}{Perimter of the triangle}$$ --->$$r = \frac {2*0.5*a*b}{a+b+c}$$ ...(1)

Per statement 1, r is given but without a,b and c , we wont be able to find a+b. Not sufficient.

Per statement 2, the diameter of the circumcircle will become the hypotenuse (--> c=18) of the triangle but without a or b , this statement is not sufficient.

Combining, you get, r=5, c=18, and from the relation (1), you can clearly calculate a+b. C is the correct answer.

For the sake of completeness, once you get r=5, c=18, substituting in (1) you get,

--->$$r = \frac {2*0.5*a*b}{a+b+c}$$ ---> $$5 = \frac {2*0.5*a*b}{a+b+18}$$ ...(2) and from the pythogoras theorem for triangle PQR, $$a^2+b^2=c^2$$---> $$(a+b)^2-2ab = c^2$$ ...(3)

Solving, 2 and 3, will give you a quadratic equation, $$(a+b)^2-10(a+b)-504 = 0$$ ---> a+b = 28 or -18 (rejecting negative as sides can not sum to a ngeative number).

Thus the only value you get is a+b = 28.

Hope this helps.
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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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Engr2012: Can you tell how did you arrive at Formula 1?
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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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I think the answer should be B. The length pq should be equal to the diameter (18cm) of the circumscribed circle. Which then means that the right triangle is 30:60:90 triangle with sides equal to 2x9:3^0.5x9:9

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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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Sebastian Shoaib wrote:
I think the answer should be B. The length pq should be equal to the diameter (18cm) of the circumscribed circle. Which then means that the right triangle is 30:60:90 triangle with sides equal to 2x9:3^0.5x9:9

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Let me ask you a question: why should it be 30:60:90 triangle? Why not 10-80-90 triangle? Or any other type of a right triangle?
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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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gauravprashar17 wrote:
Engr2012: Can you tell how did you arrive at Formula 1?

Refer to http://www.mathalino.com/reviewer/deriv ... f-incircle
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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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Bunuel wrote:
Sebastian Shoaib wrote:
I think the answer should be B. The length pq should be equal to the diameter (18cm) of the circumscribed circle. Which then means that the right triangle is 30:60:90 triangle with sides equal to 2x9:3^0.5x9:9

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Let me ask you a question: why should it be 30:60:90 triangle? Why not 10-80-90 triangle? Or any other type of a right triangle?

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Yes I see...dia of inscribed circle (10) = sum of two sides minus the hypo...so the answer is C. Thanks

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Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P  [#permalink]

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_________________ Re: Triangle PQR has angle PRQ equal to 90 degrees. What is the value of P   [#permalink] 16 Oct 2018, 05:55
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