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555-605 (Medium)|   Geometry|      
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Bunuel
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Three carpet piecesin the shapes of a square, a triangle, and a semic 29 Sep 2021, 01:30
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25% (medium)

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80% (01:33) correct 20% (01:59) wrong based on 49 sessions

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Three carpet pieces—in the shapes of a square, a triangle, and a semicircle—are attached to one another, as shown in the figure above, to cover the floor of a room. If the area of the square is 144 feet and the perimeter of the triangle is 28 feet, what is the perimeter of the room’s floor, in feet?

(A) \(32+12π\)

(B) \(40+6π\)

(C) \(34+12π\)

(D) \(52+6π\)

(E) \(52+12π\)



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Three carpet piecesin the shapes of a square, a triangle, and a semic Updated on: 29 Sep 2021, 03:46
Originally posted by GMATWhizTeam on 29 Sep 2021, 02:05.
Last edited by GMATWhizTeam on 29 Sep 2021, 03:46, edited 1 time in total.
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Bunuel

Three carpet pieces—in the shapes of a square, a triangle, and a semicircle—are attached to one another, as shown in the figure above, to cover the floor of a room. If the area of the square is 144 feet and the perimeter of the triangle is 28 feet, what is the perimeter of the room’s floor, in feet?

(A) \(32+12π\)

(B) \(40+6π\)

(C) \(34+12π\)

(D) \(52+6π\)

(E) \(52+12π\)



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Attachment:
The attachment 1.jpg is no longer available
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Solution:

Firstly, the area if the given square = 144.

Let us assume the side of the square be 'a'. Then we can write \(a^2=144\)

\(⇒ a=\sqrt{144}\)

\(⇒ a=12\)

Attachment:
gc3.png
gc3.png [ 22.47 KiB | Viewed 1668 times ]

Now lets us concentrate on the triangle.

The perimeter of the triangle is 28. This includes side 12 as well. so the sum of the other 2 sides (say a and b) is \(28-12=16\).

Attachment:
gc4.png
gc4.png [ 22.33 KiB | Viewed 1653 times ]

Now lets us concentrate on the Semi Circle.

The diameter of semi-circle = 12. Thus radius \(= \frac{12}{2}=6\).

So, the length of boundary or circumference \(= \pi \times r = \pi \times 6=6\pi\)


Ultimately the perimeter of whole figure \(= 12 + b + a + 12 + 6\pi \)

\(= 12 + 16 +12+ 6\pi\)

\(= 40+6\pi \)

Hence the right answer is Option B.
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Three carpet piecesin the shapes of a square, a triangle, and a semic 29 Sep 2021, 02:28
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GMATWhizTeam
Bunuel

Three carpet pieces—in the shapes of a square, a triangle, and a semicircle—are attached to one another, as shown in the figure above, to cover the floor of a room. If the area of the square is 144 feet and the perimeter of the triangle is 28 feet, what is the perimeter of the room’s floor, in feet?

(A) \(32+12π\)

(B) \(40+6π\)

(C) \(34+12π\)

(D) \(52+6π\)

(E) \(52+12π\)



Project PS Butler


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Show Spoiler
Attachment:
1.jpg

Solution:

Firstly, the area if the given square = 144.

Let us assume the side of the square be 'a'. Then we can write \(a^2=144\)

\(⇒ a=\sqrt{144}\)

\(⇒ a=12\)

Attachment:
gc3.png

Now lets us concentrate on the triangle.

The perimeter of the triangle is 28. This includes side 12 as well. so the sum of the other 2 sides (say a and b) is \(28-12=16\).

Attachment:
gc4.png

Now lets us concentrate on the Semi Circle.

The diameter of semi-circle = 12. Thus radius \(= \frac{12}{2}=6\).

So, the length of boundary or circumference \(= \pi \times r = \pi \times 6=6\pi\)


Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \)

\(= 12 + 16 +12+ 6\pi\)

\(= 40+6\pi \)

Hence the right answer is Option B.
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Dear Sir,
Should we not consider both sides of the square to get the entire perimeter?

Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \)

Here, 12+a+b = Triangle's Perimeter
\(6\pi\) = Half of the Circle's Circumference
12 = Just 1 side of the square-shaped part of the carpet.

Please correct me if I am missing sth!
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Three carpet piecesin the shapes of a square, a triangle, and a semic 29 Sep 2021, 03:46
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ShahadatHJ
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Bunuel

Three carpet pieces—in the shapes of a square, a triangle, and a semicircle—are attached to one another, as shown in the figure above, to cover the floor of a room. If the area of the square is 144 feet and the perimeter of the triangle is 28 feet, what is the perimeter of the room’s floor, in feet?

(A) \(32+12π\)

(B) \(40+6π\)

(C) \(34+12π\)

(D) \(52+6π\)

(E) \(52+12π\)



Project PS Butler


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Show Spoiler
Attachment:
The attachment 1.jpg is no longer available

Solution:

Firstly, the area if the given square = 144.

Let us assume the side of the square be 'a'. Then we can write \(a^2=144\)

\(⇒ a=\sqrt{144}\)

\(⇒ a=12\)

Attachment:
The attachment gc3.png is no longer available

Now lets us concentrate on the triangle.

The perimeter of the triangle is 28. This includes side 12 as well. so the sum of the other 2 sides (say a and b) is \(28-12=16\).

Attachment:
The attachment gc4.png is no longer available

Now lets us concentrate on the Semi Circle.

The diameter of semi-circle = 12. Thus radius \(= \frac{12}{2}=6\).

So, the length of boundary or circumference \(= \pi \times r = \pi \times 6=6\pi\)


Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \)

\(= 12 + 16 +12+ 6\pi\)

\(= 40+6\pi \)

Hence the right answer is Option B.
Dear Sir,
Should we not consider both sides of the square to get the entire perimeter?

Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \)

Here, 12+a+b = Triangle's Perimeter
\(6\pi\) = Half of the Circle's Circumference
12 = Just 1 side of the square-shaped part of the carpet.

Please correct me if I am missing sth!
Show more

Yes, you are correct that we need to consider both sides of the square andwe are doing exactly that.

Consider this diagram with labels.

Attachment:
gc5.png
gc5.png [ 23.57 KiB | Viewed 1562 times ]

When I wrote: Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \), I actually moved round the figure clockwise starting from point A.

So the perimeter \(= AB+BC+CD+DE+arc(EA)=12+b+a+12+6\pi\).

Hope this helps!
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Three carpet piecesin the shapes of a square, a triangle, and a semic 29 Sep 2021, 05:47
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GMATWhizTeam
Yes, you are correct that we need to consider both sides of the square andwe are doing exactly that.

Consider this diagram with labels.

Attachment:
gc5.png

When I wrote: Ultimately the perimeter of whole figure \(= 12 + a + b + 12 + 6\pi \), I actually moved round the figure clockwise starting from point A.

So the perimeter \(= AB+BC+CD+DE+arc(EA)=12+b+a+12+6\pi\).

Hope this helps!
Show more
Helped, thanks a lot. :)
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Three carpet piecesin the shapes of a square, a triangle, and a semic 29 Sep 2021, 18:09
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Area of square = 144
=> side of square = 12

=> base of triangle = 12 = diameter of circle

Perimeter of rooms floor
= Sum of other two sides of triangle + 2*side of square + (perimeter of circle)/2
= (28 - 12) + 2*12 + 2π*6/2
= 40 + 6π

Answer: D
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