The playing area for a certain outdoor game requires markings accordin

The amount of paint required in meters would be the combined perimeter of the three figures- Triangle ABC, Sq. OQBR and the circle O:

Chord QR= Diagonal of the square OQBR = 3√2
Therefore, the side of the sq. = 3 and perimeter= 4*3=12 meters

Given, AB = 3*(QB) = 3*3 = 9
The sides AB:BC:AC are in the ratio 3:4:5. Since AB=9, so BC=12 and AC= 15.
Therefore perimeter of triangle ABC = 9+12+15= 36 meters

Radius of the circle = OR = 3
Circumference = 6π=19 meters (approx.)

Total Perimeter= 12 + 36 + 19 = 67 meters

Ans. D

Yeah! I thought of it. But according to this you have paint on BQ and BR twice which is not logical.

Bunuel What went wrong?
Suppose the sides of the triangle are AB = 3x, BC = 4x, & AC =5x
Also OBQR is a square --> OQ = OR = BR = QB = 1/3 AB = 1/3 * 3x = x
In Triangle OQR, right angle at O,
QR = 3√2 --> (3√2)² = x² + x² --> x = 3
AB = 3x = 3*3 = 9
BC = 4x = 4*3 =12
AC = 5x = 5*3 = 15

Paint required for the marking = Perimeter of ▲ ABC + OQ + OR + Circumference of Circle with radius O
= (9+12+15) + (3+3) + (2*π*3) = 42+6π ≈ 61

That's correct. But 61 is not an option. So I decided to stick with my original ans. Hoping someone can give the correct ans. or explanation

Radius of the circle: 3

Perimeter of the circle: 2πr = 6π
Perimeter of the square: 3*4 = 12
Perimeter of the triangle: 9+12+15 = 36

12+36+6π = 48+6π and this is approximately 67. Answer D

Just a minor correction above: 12+36+6π = 48+6π, and not 54π. And 48+6π=67

Of course, thanks for noticing. edited

Bunuel: Last line of the question seems out of sync. It should ask only the area of the field or Perimeter of all markings or should give paint needed for every unit area of playing field to paint the field. Please let me know if I am wrong.

QB = QR (Tangents drawn from the same point on same circle)
Since Tangents make 90 degreed when point f tangency is joined with Centre of Circle therefore QBRO is a square

Given: Diagonal of Square, QR = 3√2 = Side√2
i.e. Side of Square = QB = QR = radius of Circle = 3

QB = (1/3) AB
i.e. AB = 3*3 = 9
i.e. BC = 3*4 = 12
i.e. AC = 3*5 = 15

The length of the all Markings = 3+3+9+12+15+2*(pi)*3 = 6+36+6*3.14 = 6+18.84+36 = 61 meters

Area of Triangular Field = (1/2)*9*12 = 54 Square meters

GMATinsight, this question is badly written. For the last statement, we need to find the perimeter of all the lines to be painted. We dont need to consider the thickness of the lines.

Even when you do calculate the perimeter, it should be equal to 42+6\(\pi\) = approx. 61. (=triangle ABC+circle + OQ+OR)

Answer is D:

let the sides be 3x, 4x and 5x respectively.
From problem we know QR = 3√2
In triangle OQR, using pythogorous theorem, we know the radius of the circle is 3[m].
from this we also have x =3. (as QB =1/3= x/3 = x = 3)

from this we know the sides of the triangle are 9,12,15 and the radius of the circle is 3.
as all values of known, its just finding perimeter.

Answer D

The playing area is marked by the lines, so we are to find perimeter of the figures.
Let a be the side of square,
Diagonal of a square is given by \(\sqrt{2}\) * side
So,
\(3\sqrt{2}\) = \(\sqrt{2}\) * a
or a=3
perimeter of square= 4*3 = 12
AB=3QB
or AB = 3*3= 9
then according to the given ratio,
BC = 12 and AC = 15
the perimeter of triangle will be 9+12+15 = 36
The circumference of the circle will be \(2\pi*3\) = \(6\pi\) = 18.84 or 19 approx
Total perimeter will be

12+36+19 = 67

Answer:- D

I get 42 + 6pi. Approx 61. Not in the answer choice?

The note says figures are not drawn to scale.
I guess that is the only thing which accounts for calculation and addition of all the perimeters.
If we exclude QB and BR from the total perimeter, we would get 61, which is close to 67. This approximation would be extreme though.
The question seems to be flawed.

Why are they painting the part QB, RB twice? Is not this part of the 9 and 12 side?

The playing area for a certain outdoor game requires markings according to the figure shown. The ratio of AB:BC:AC is 3:4:5 ; O is the midpoint of the circle, R and Q are both points on the circle with center O and corners of square OBQR. If QB is 1/3 of AB and a person running in a straight line from Q to R travels 3√2 meters, approximately how many meters of paint are required to paint the playing area on a surface (assume the lines have negligible width)?

(A) 54π
(B) 97
(C) 72
(D) 67
(E) 45
how many meters of paint are required to paint the playing area on a surface . As per me .. it is asking us to calculate the area of the triangle .

From the question ... we get the side of the square as 3 ... because a person running in a straight line from Q to R travels 3√2 meters ...nothing but diameter of the square .

and side AB is 9 because QB = (1/3)AB , and QB is side of the square .

hence AB : BC :AC ratio given =3:4:5 = 9:12:15

hence the area of the triangle = 54

option A. ANS

Dont know why pie is there .( 54 pie)

GROCKIT OFFICIAL SOLUTION:

We know that the ratio of the lengths of the sides x of a square to its diagonal are the same as that of a 45:45:90 triangle: x:x:x√2. You could also use the Pythagorean Theorem, setting 3√2 as the hypotenuse, so (3√2)² = 2*(side length)², so 2s²=18, and s²=9, so the sides are 3 meters each. This is also the radius of the circle, so from here we can figure out:

Perimeter of circle: the circle’s radius is 3, and the perimeter (circumference) is 2πr, or 6π.

Perimeter of square: the square’s perimeter is 4s, or 12.

Perimeter of triangle: since QB is 1/3 of AB and QB = 3, AB = 9. Triangle ABC is in the 3:4:5 ratio, so the other sides are 12 and 15, respectively. 9 + 12 + 15 = 36.

Circle + square + triangle = 6π + 12 + 36 = 48 + 6π, or approximately 48+19 = 67 meters.

but Bunuel, aren't we calculating the line QB and BR twice. once in perimeter of triangle and then for perimeter of square. but it will be painted only once. please clear my doubt.