A wire was initially used to fence a right-triangular plot of land alo

Question

A wire was initially used to fence a right-triangular plot of land along its perimeter. The lengths, in metres, of the perpendicular sides of the plot were both integers and in the ratio 3:4. The wire was later removed from the triangular plot and used to fence a square plot along its perimeter. When the square plot had been completely fenced, 4 metres of wire still remained. Which of the following could represent the greatest distance between two points in the square plot?

I. \(\sqrt{2}\) metres
II. \(8\sqrt{2}\) metres
III. \(10\sqrt{2}\) metres

A. I only
B. II only
C. III only
D. I, II and III
E. None of the above

I.  \sqrt{2}  metres
II.  8\sqrt{2}  metres 
III.  10\sqrt{2}  metres

A. I only
B. II only
C. III only
D. I, II and III
E. None of the above

Archit3110 · Answer

sides of ∆ are 3:4:5 and length of wire is used for 3:4
so we have 3:4 ; 6:8 ; 9:12
total length 7; 14 ; 21
after using around square plot wire we are left with ; 3;10;17
we know that digonal is the greatest length so we can have 3√2 ; 10√2 ; 17√2
OPTION II matches ; B only

NitishJain · Answer

IMO  B

Let the perpendicular sides of the plot be 3x, 4x. (x is an integer constant)
The third side will be 5x. (Using Pythagoras theorem)

Total length of fence will be 5x + 4x + 3x = 12x

Let the side of square = y
Given : 12x = 4y +4
or: y = 3x -1

greatest distance between two points in the square = Diagonal
Diagonal = √2 side = √2 y = √2 (3x -1)

Substituting x =1, 2, 3 ,4
y = 2√2 , 5√2 , 8√2 , 11√2

From given options only 8√2 .

Shrey08 · Answer

This is how I did it

If the 2 legs are in the ratio 3:4 it is a Pythagorean triples whose side are in ratio 3:4:5
The length of the fence is  3x+4x+5x = 12x

The question is essentially asking the diagonal of this square.
Say the side of the square is  a 
Diagonal(let's call it AB) would then be =  a\sqrt{2}............(a^2+a^2 = Diagonal^2)

lets see which answer option fits in

I.  \sqrt{2}  metres
if AB= \sqrt{2}  =  a\sqrt{2} 
a = 1
which gives us the parameter of square = 4
that means the length of fence is 4+4(the part that remained after fencing the sq.)= 8
but the length of fence is 12x
 this option is not possible.

II.  8\sqrt{2}  metres 
if AB= 8\sqrt{2}  =  a\sqrt{2} 
a = 8
which gives us the parameter of square = 4*8 = 32
that means the length of fence is 32+4(the part that remained after fencing the sq.)= 36
the length of fence is 12x and 12*3 = 36
 so this is possible

III.  10\sqrt{2}  metres
if AB= 10\sqrt{2}  =  a\sqrt{2} 
a = 10
which gives us the parameter of square = 4*10 = 40
that means the length of fence is 40+4(the part that remained after fencing the sq.)= 44
the length of fence is 12x and 44 is not a multiple of 12
 so this is also not possible

Only II is possible, Option B is correct

A wire was initially used to fence a right-triangular plot of land alo

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A wire was initially used to fence a right-triangular plot of land alo

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards