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17 Jan 2016, 11:28
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Difficulty:
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52% (00:56) correct
48%
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wrong
based on 161
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What is the perimeter of triangle MNO?
(1) The area of triangle MNO is 24.
(2) Angle MON equals 90°.
Attachment:
2016-01-17_2227.png [ 3.24 KiB | Viewed 3409 times ]
ENGRTOMBA2018
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
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17 Jan 2016, 12:56
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Bunuel
What is the perimeter of triangle MNO?
(1) The area of triangle MNO is 24.
(2) Angle MON equals 90°.
Attachment:
2016-01-17_2227.png
In the given triangle, let the 3 sides be a,b,c with a=8 and b=10. For calculating perimeter, we need to know the value of c.
Per statement 2, \(\angle{MON} = 90\) ---> triangle MNO is a right angled triangle ----> c = 6. Sufficient.
Per statement 1, This statement is sufficient and will require some knowledge of trigonometry to see that with 2 sides, the included angle and the area of the triangle, this statement is sufficient.
Via trigonometry: Area of the triangle = \(24 = 0.5*a*b*sin (\angle{ONM})\), with a =9 and b=10 and from this equation, you can calculate the included angle as well. Once you calculate \(sin (\angle{ONM})\), you can calculate \(cos (\angle{ONM})\).
Relation between \(cos (\angle{ONM})\) and 3 sides of a triangle ---> \(cos (\angle{ONM})\) = \(\frac{a^2+b^2-c^2}{2ab}\). Thus with all values except 'c' provided to us, you can calculate c and hence calculate the perimeter of the triangle.
By Heron's formula: if the sides of the triangle are a,b, and c and Area = A, then
\(A = 0.25*\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}\), with values of A,a and b already provided to you. Thus you can clearly calculate the value of c and hence the perimeter.
D is thus the correct answer.
rachitbechem
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17 Jan 2016, 23:42
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Engr2012
Bunuel
What is the perimeter of triangle MNO?
(1) The area of triangle MNO is 24.
(2) Angle MON equals 90°.
Attachment:
2016-01-17_2227.png
In the given triangle, let the 3 sides be a,b,c with a=8 and b=10. For calculating perimeter, we need to know the value of c.
Per statement 2, \(\angle{MON} = 90\) ---> triangle MNO is a right angled triangle ----> c = 6. Sufficient.
Per statement 1, This statement is sufficient and will require some knowledge of trigonometry to see that with 2 sides, the included angle and the area of the triangle, this statement is sufficient.
Proof: Area of the triangle = \(24 = 0.5*a*b*sin (\angle{ONM})\), with a =9 and b=10 and from this equation, you can calculate the included angle as well. Once you calculate \(sin (\angle{ONM})\), you can calculate \(cos (\angle{ONM})\).
Relation between \(cos (\angle{ONM})\) and 3 sides of a triangle ---> \(cos (\angle{ONM})\) = \(\frac{a^2+b^2-c^2}{2ab}\). Thus with all values except 'c' provided to us, you can calculate c and hence calculate the perimeter of the triangle.
D is thus the correct answer.
Is there any other way apart from Trignometry to prove stat 1 correct. I tried to use 3rd side concept. The 3rd side should be 2 < 3rd side < 18. And so the Perimeter could be anything but obviously the area = 24 has a role to play. How do we know what are the two legs of triangle to find base and height? If it is 6 8 10 then the max area is 24, but if the angle is not 90 it could be the third side is <6 but then I suppose we violate the area = 24 rule?
Can we use Heron's formula instead. We will get the 3rd side I suppose since we know the area and (s = a+b+c/2) substitute c in terms of s and find s and then get c?
