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12 May 2015, 14:45
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
54% (01:22) correct
46%
(01:34)
wrong
based on 181
sessions
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What is the perimeter of a triangle with sides 6 and 3 that has the maximum possible area?
a. 9 +3√ 2
b. 9+3√ 3
c. 15
d. 9+3√ 5
e. 9+3√ 6
a. 9 +3√ 2
b. 9+3√ 3
c. 15
d. 9+3√ 5
e. 9+3√ 6
12 May 2015, 21:59
Kudos
Bookmarks
Hi Turkish,
This question is based on a specific Geometry rule: When given 2 sides of a triangle, the MAXIMUM area will occur when the two sides are the LEGS of a RIGHT triangle.
So, with legs of 3 and 6, the Pythagorean Theorem will help us figure out the third side (the hypotenuse):
3^2 + 6^2 = C^2
9 + 36 = 45
45 = C^2
(Root45) = C
(Root45) = 3(Root5)
Thus, the perimeter of the triangle is 3+6+3(Root5) = 9 + 3(Root5)
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This question is based on a specific Geometry rule: When given 2 sides of a triangle, the MAXIMUM area will occur when the two sides are the LEGS of a RIGHT triangle.
So, with legs of 3 and 6, the Pythagorean Theorem will help us figure out the third side (the hypotenuse):
3^2 + 6^2 = C^2
9 + 36 = 45
45 = C^2
(Root45) = C
(Root45) = 3(Root5)
Thus, the perimeter of the triangle is 3+6+3(Root5) = 9 + 3(Root5)
Final Answer:
GMAT assassins aren't born, they're made,
Rich
General Discussion
12 May 2015, 22:07
Kudos
Bookmarks
Let's consider the diagram below to find out the maximum possible area of a triangle with two of the sides given
We are given length of two sides BC = b and AC = a. We know that area of triangle = \(\frac{1}{2}\) * base * height = \(\frac{1}{2}\) * b * AD
Since we know the lengths of two sides as a and b, we need to express AD in terms of 'a' as 'b' is already present in the expression of the area.
We see that triangle ADC is a right angled triangle, so we can express AD in terms of 'a' using the trigonometric ratios.
We can write Sin ∠ACD = AD/a which gives us AD = a * Sin ∠ACD.
Hence, area of triangle ABC = \(\frac{1}{2}\) * b * a * Sin ∠ACD. This expression would be maximum when Sin ∠ACD is maximum which is when angle ∠ACD = 90 and Sin ∠ACD = 1.
Thus, it tells us that sides AC and BC are perpendicular to each other and hence form the two legs of a right angled triangle.
For two sides given the maximum area would be if these two sides form the legs of a right angled triangle
Coming back to the question, the sides given to us are 6 and 3 and we need to find the perimeter of the this right angled triangle with 6 & 3 being the legs of the right angled triangle. Using the phythagoras theorem we can calculate the hypotenuse which is the third side as 3√5.
Thus the perimeter will be 6+3+3√5 = 9 + 3√5
Answer is option D
Hope it helps
Regards
Harsh
We are given length of two sides BC = b and AC = a. We know that area of triangle = \(\frac{1}{2}\) * base * height = \(\frac{1}{2}\) * b * AD
Since we know the lengths of two sides as a and b, we need to express AD in terms of 'a' as 'b' is already present in the expression of the area.
We see that triangle ADC is a right angled triangle, so we can express AD in terms of 'a' using the trigonometric ratios.
We can write Sin ∠ACD = AD/a which gives us AD = a * Sin ∠ACD.
Hence, area of triangle ABC = \(\frac{1}{2}\) * b * a * Sin ∠ACD. This expression would be maximum when Sin ∠ACD is maximum which is when angle ∠ACD = 90 and Sin ∠ACD = 1.
Thus, it tells us that sides AC and BC are perpendicular to each other and hence form the two legs of a right angled triangle.
For two sides given the maximum area would be if these two sides form the legs of a right angled triangle
Coming back to the question, the sides given to us are 6 and 3 and we need to find the perimeter of the this right angled triangle with 6 & 3 being the legs of the right angled triangle. Using the phythagoras theorem we can calculate the hypotenuse which is the third side as 3√5.
Thus the perimeter will be 6+3+3√5 = 9 + 3√5
Answer is option D
Hope it helps
Regards
Harsh
