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RandomStoryteller
Joined: 13 Jun 2017
Last visit: 08 Apr 2022
Posts: 22
Given Kudos: 94
Location: India
Schools: LBS '21 INSEAD Jan '19 IIMB EPGP"20 Yale '22 IMD '22 Kellogg '24 NUS Wharton '24 Said '23 (II)
GMAT 1: 740 Q49 V41
Schools: LBS '21 INSEAD Jan '19 IIMB EPGP"20 Yale '22 IMD '22 Kellogg '24 NUS Wharton '24 Said '23 (II)
GMAT 1: 740 Q49 V41
Posts: 22
26 Sep 2021, 23:16
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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67% (02:14) correct
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based on 55
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An isosceles triangle is such that the length of its sides is an integral value. If the two equal
sides measure 12 units, what fraction of all such possible triangles are obtuse angled triangles?
A ) 7/23
B) 16/23
C) 1/3
D) 2/3
E ) 18/23
sides measure 12 units, what fraction of all such possible triangles are obtuse angled triangles?
A ) 7/23
B) 16/23
C) 1/3
D) 2/3
E ) 18/23
27 Sep 2021, 00:01
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Step 1: find all the integer values that the unknown side (call it X) can take
(12 - 12) < X < (12 + 12)
0 < X < 24
X can take integer values from 1 to 23, inclusive. That’s is 23 possible isosceles triangles with the 3rd side having an integer length.
Step 2: determining how many of these triangles are obtuse
It must be true that for any obtuse triangle the Square of the Largest Side must exceed the sum of the Squares of each shorter side.
There are 2 scenarios to look at:
Scenario 1: X takes a value of 13 or greater, making X the longest side
(X)^2 > (12)^2 + (12)^2
X > sqrt(288)
Since 16^2 = 256 —— and ——- 17^2 = 289
The first integer value that X can take when it is the longest side is 17.
17, 18, 19, 20, 21 , 22 , and 23 are all integer side lengths of X that would make the isosceles triangle and obtuse triangle.
Scenario 2: X takes a value less than 12, such that X is not the longest side
(12)^2 > (X)^2 + (12)^2
0 > (X)^2
Since the square of a number can never take a negative value, there are no valid lengths of X that fit this scenario.
For the triangle to be obtuse and isosceles, the unequal side must be the largest side (makes logical sense as well when you consider that any triangle can only have 1 angle that exceeds 90 degrees)
Therefore, the probability that the isosceles triangle with integer side lengths is obtuse is
7 / 23
Posted from my mobile device
(12 - 12) < X < (12 + 12)
0 < X < 24
X can take integer values from 1 to 23, inclusive. That’s is 23 possible isosceles triangles with the 3rd side having an integer length.
Step 2: determining how many of these triangles are obtuse
It must be true that for any obtuse triangle the Square of the Largest Side must exceed the sum of the Squares of each shorter side.
There are 2 scenarios to look at:
Scenario 1: X takes a value of 13 or greater, making X the longest side
(X)^2 > (12)^2 + (12)^2
X > sqrt(288)
Since 16^2 = 256 —— and ——- 17^2 = 289
The first integer value that X can take when it is the longest side is 17.
17, 18, 19, 20, 21 , 22 , and 23 are all integer side lengths of X that would make the isosceles triangle and obtuse triangle.
Scenario 2: X takes a value less than 12, such that X is not the longest side
(12)^2 > (X)^2 + (12)^2
0 > (X)^2
Since the square of a number can never take a negative value, there are no valid lengths of X that fit this scenario.
For the triangle to be obtuse and isosceles, the unequal side must be the largest side (makes logical sense as well when you consider that any triangle can only have 1 angle that exceeds 90 degrees)
Therefore, the probability that the isosceles triangle with integer side lengths is obtuse is
7 / 23
Posted from my mobile device
19 Jul 2023, 15:46
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