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04 Dec 2022, 12:13
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Difficulty:
85%
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Question Stats:
26% (02:38) correct
74%
(03:43)
wrong
based on 19
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In the figure above an equilateral triangle with a perimeter of 18 inches is surrounded by a border \(\sqrt{3}\) inches thick. What is the area of the border?
(A) \(18\sqrt{3}\)
(B) \(21\sqrt{3}\)
(C) \(24\sqrt{3}\)
(D) \(27\sqrt{3}\)
(E) \(36\sqrt{3}\)
Attachment:
Triangle.gif [ 2.43 KiB | Viewed 1275 times ]
05 Dec 2022, 06:23
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Area of the border will be: Area of Big Triangle - Area of Small Triangle
Area of Small Triangle:
Formula for the area of an equilateral triangle is: \(s^2 * \frac{\sqrt{3}}{4}\) where s = side
The small triangle has sides of 6, which means that the area will be: \(6^2 * \frac{\sqrt{3}}{4}\)
\(= 9\sqrt{3}\)
Area of Big Triangle:
Using the sides of the small triangle and the width of the border to create 3 rectangles of \(6*\sqrt{3}\), will leave us with 3 kites at each of the large triangles vertices (I've attached an image of what one of the kites). Solving for the length of the longer sides of the kite, multiplying the answer by 2 and adding it to 6 will give us the length of the sides of the bigger triangle.
The smaller sides of the kite will each be \(\sqrt{3}\), and the angle where the two meet will be \(120°\). The angle equals \(120°\) given that there are two \(90°\), and a \(60°\) from the smaller triangle. All four angles together must sum to \(360°\).
The vertex where the two longer sides of the kite meet will be \(60°\), given that is one of the angles of the larger equilateral triangle.
So when we bisect the kite we get two equal \(90°,60°,30°\) triangles, where the corresponding \(30°\) side has a length of \(\sqrt{3}\). As the longer sides of the kite correspond with the \(60°\), it will have a length of \(\sqrt{3}*\sqrt{3} = 3\)
Which means that the length of the sides of the bigger triangle are \((2*6)+6 = 12\).
Area of the bigger triangle will therefore be: \(12^2 * \frac{\sqrt{3}}{4}\)
\(= 36\sqrt{3}\)
Area of Small Triangle:
Area of border: \(36\sqrt{3} - 9\sqrt{3}\)
\(27\sqrt{3}\)
Answer D
Area of Small Triangle:
Formula for the area of an equilateral triangle is: \(s^2 * \frac{\sqrt{3}}{4}\) where s = side
The small triangle has sides of 6, which means that the area will be: \(6^2 * \frac{\sqrt{3}}{4}\)
\(= 9\sqrt{3}\)
Area of Big Triangle:
Using the sides of the small triangle and the width of the border to create 3 rectangles of \(6*\sqrt{3}\), will leave us with 3 kites at each of the large triangles vertices (I've attached an image of what one of the kites). Solving for the length of the longer sides of the kite, multiplying the answer by 2 and adding it to 6 will give us the length of the sides of the bigger triangle.
The smaller sides of the kite will each be \(\sqrt{3}\), and the angle where the two meet will be \(120°\). The angle equals \(120°\) given that there are two \(90°\), and a \(60°\) from the smaller triangle. All four angles together must sum to \(360°\).
The vertex where the two longer sides of the kite meet will be \(60°\), given that is one of the angles of the larger equilateral triangle.
So when we bisect the kite we get two equal \(90°,60°,30°\) triangles, where the corresponding \(30°\) side has a length of \(\sqrt{3}\). As the longer sides of the kite correspond with the \(60°\), it will have a length of \(\sqrt{3}*\sqrt{3} = 3\)
Which means that the length of the sides of the bigger triangle are \((2*6)+6 = 12\).
Area of the bigger triangle will therefore be: \(12^2 * \frac{\sqrt{3}}{4}\)
\(= 36\sqrt{3}\)
Area of Small Triangle:
Area of border: \(36\sqrt{3} - 9\sqrt{3}\)
\(27\sqrt{3}\)
Answer D
Attachments
kite to use.png [ 17.03 KiB | Viewed 1065 times ]
