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Difficulty:
75%
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Question Stats:
50% (02:35) correct
50%
(02:31)
wrong
based on 34
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The area of the largest square which can be inscribed in a triangle with side lengths 3,4 and 5 is \(\frac{a}{b}\). Find a+b, where a and b are co-prime, positive integers.
A. 125
B. 153
C. 162
D. 171
E. 193
A. 125
B. 153
C. 162
D. 171
E. 193
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 24 Jul 2026
Posts: 616
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
Since we are given that the sides of the triangle are 3-4-5, we can consider it a right angled triangle.
Let the side length of the square be \(x\) so that Area of the square = \(x^2=\frac{a}{b}\)
Now from the figure, triangles ABC and FEC are similar.
Therefore, \(\frac{FC}{FE} = \frac{AC}{AB}\)
That is, \(\frac{(3-x)}{x} = \frac{3}{4}\)
\(x = \frac{12}{7}\) and so \(x^2=\frac{a}{b}=\frac{144}{49}\)
144 and 49 are co-primes since their only common factor is 1 and so we can say \(a=144, b=49\)
\(a+b = 193\)
Answer is (E)
Hit Kudos if this helped!
_________________
Let the side length of the square be \(x\) so that Area of the square = \(x^2=\frac{a}{b}\)
Now from the figure, triangles ABC and FEC are similar.
Therefore, \(\frac{FC}{FE} = \frac{AC}{AB}\)
That is, \(\frac{(3-x)}{x} = \frac{3}{4}\)
\(x = \frac{12}{7}\) and so \(x^2=\frac{a}{b}=\frac{144}{49}\)
144 and 49 are co-primes since their only common factor is 1 and so we can say \(a=144, b=49\)
\(a+b = 193\)
Answer is (E)
Hit Kudos if this helped!
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