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Eighteen equilateral triangles are joined together as shown in the f
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27 Jan 2025, 23:44
27 Jan 2025, 23:44
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Eighteen equilateral triangles are joined together as shown in the figure below. If the length of the sides of each triangle is 1, what is the perimeter of the whole figure?
GRE polygon.png [ 25.85 KiB | Viewed 107 times ]
A. 10
B. 11
C. 12
D. 13
E. 14
Attachment:
GRE polygon.png [ 25.85 KiB | Viewed 107 times ]
A. 10
B. 11
C. 12
D. 13
E. 14
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Re: Eighteen equilateral triangles are joined together as shown in the f
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03 Mar 2025, 13:12
03 Mar 2025, 13:12
Expert Reply
Perimeter of any figure is nothing but sum of all sides which form the boundary of the figure.
As the figure is formed by joining identical equilateral triangles of side 1 units each, we have the lengths of the outer sides of the figure as $\(1,1,1,1,(4 \times 1)=4 \&(4 \times 1)=4\)$
So, the perimeter of the figure is $\(1+1+1+1+4+4=12\)$
Hence the answer is (C).
As the figure is formed by joining identical equilateral triangles of side 1 units each, we have the lengths of the outer sides of the figure as $\(1,1,1,1,(4 \times 1)=4 \&(4 \times 1)=4\)$
So, the perimeter of the figure is $\(1+1+1+1+4+4=12\)$
Hence the answer is (C).
gmatclubot
Re: Eighteen equilateral triangles are joined together as shown in the f [#permalink]
03 Mar 2025, 13:12
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