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13 Nov 2015, 04:57
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
46% (02:00) correct
54%
(02:09)
wrong
based on 140
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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.
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.
ENGRTOMBA2018
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Concentration: Finance, Strategy
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13 Nov 2015, 06:11
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excelingmat
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.
General Discussion
04 Jul 2017, 07:52
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Engr2012: Can you tell how did you arrive at Formula 1?
