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02 Oct 2024, 12:26
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
53% (02:03) correct
46% (01:58) wrong based on 13 sessions
History
Date
Time
Result
Not Attempted Yet
Quantity A
The length of a leg of an isosceles right triangle with area R
Quantity B
The length of a side of a square with area R
The length of a leg of an isosceles right triangle with area R
Quantity B
The length of a side of a square with area R
04 Mar 2025, 12:29
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OE
In an isosceles right triangle, both legs have the same length. Te area of an isosceles right triangle with legs of length x is equal to \(\frac{x^2}{2}\). The area of a square with sides of length \(s\) is \(s^2\). Since you are given that an isosceles right triangle and a square have the same area R, it follows that \(\frac{x^2}{2} = s^2, \) and so \(x = \sqrt{2}s\). Since 2 is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A.
In an isosceles right triangle, both legs have the same length. Te area of an isosceles right triangle with legs of length x is equal to \(\frac{x^2}{2}\). The area of a square with sides of length \(s\) is \(s^2\). Since you are given that an isosceles right triangle and a square have the same area R, it follows that \(\frac{x^2}{2} = s^2, \) and so \(x = \sqrt{2}s\). Since 2 is greater than 1, the length of a leg of the triangle is greater than the length of a side of the square. The correct answer is Choice A.
