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The legs of a right triangle are in the ratio of 3 to 1. If
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17 Nov 2019, 04:38
17 Nov 2019, 04:38
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Question Stats:
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The legs of a right triangle are in the ratio of 3 to 1. If the length of the hypotenuse of the triangle is \(\sqrt{40}\), then the perimeter of the triangle is between
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: The legs of a right triangle are in the ratio of 3 to 1. If
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18 Nov 2019, 06:47
18 Nov 2019, 06:47
2
Carcass wrote:
The legs of a right triangle are in the ratio of 3 to 1. If the length of the hypotenuse of the triangle is \(\sqrt{40}\), then the perimeter of the triangle is between
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
Let x and 3x be the lengths of the two legs.
Attachment:
The legs of a right triangle are in the ratio of 3 to 1.png [ 2.8 KiB | Viewed 7470 times ]
Applying the Pythagorean Theorem, we get: x² + (3x)² = (√40)²
Simplify to get: x² + 9x² = 40
Simplify to get: 10x² = 40
Divide both sides by 10 to get: x² = 4
Solve: x = 2 or -2
Since the lengths must be POSITIVE, it must be the case that x = 2
As for √40, we should recognize that, since √36 = 6, and since √49 = 7, it must be the case that √40 is BETWEEN 6 and 7
So, we can say √40 = 6.something
The PERIMETER = x + 3x + √40
= 2 + 3(2) + 6.something
= 8 + 6.something
= 14.something
Answer: A
Cheers,
Brent
Re: The legs of a right triangle are in the ratio of 3 to 1. If
[#permalink]
18 Nov 2019, 21:28
18 Nov 2019, 21:28
Carcass wrote:
The legs of a right triangle are in the ratio of 3 to 1. If the length of the hypotenuse of the triangle is \(\sqrt{40}\), then the perimeter of the triangle is between
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
A. 14 and 15
B. 13 and 14
C. 12 and 13
D. 11 and 12
E. 10 and 11
Kudos for the right answer and solution.
Applying the Pythagorean Theorem, we get: k² + (3k)² = (√40)²
k² + 9k² = 40
10k² = 40
Divide both sides by 10 to get: k² = 4
k = 2 or -2
Since the lengths can not be NEGATIVE, k = 2
√40 = 2√10.
Now √10 is a little greater than √9 so √10 is approximately 3.1
so √40 = 2(3.1) = 6.2
Perimeter = k + 3k + √40
= 2 + 3(2) + 6.2
= 8 + 6.2
= 14.2
Answer: A
Attachments
triangle.jpg [ 10.15 KiB | Viewed 7308 times ]
