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The area of the triangular region
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17 Jun 2018, 13:33
17 Jun 2018, 13:33
1
1
Expert Reply
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Question Stats:
77% (00:39) correct
22% (00:53) wrong based on 76 sessions
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Attachment:
triangle.jpg [ 6.88 KiB | Viewed 5881 times ]
Quantity A |
Quantity B |
The area of the triangular region |
25 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: The area of the triangular region
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18 Jun 2018, 20:55
18 Jun 2018, 20:55
2
From the figure we know that this is a 45-45-90 triangle.
In a 45-45-90 triangle sides are distributed with length that it is \(x-x-x\sqrt{2}\)
Given the side with length 10 is opposite to the 90 degree angle.
\(x\sqrt{2} = 10\)
Therefore, \(x = 10/\sqrt{2}\)
\(x = 2*5/\sqrt{2}\)
\(x = 5\sqrt{2}\)
Since this is an isosceles right triangle with height = base = x = \(5\sqrt{2}\)
we can get the area of a triangle as \(\frac{1}{2}* base * height\)
area = \(\frac{1}{2} * 5\sqrt{2}* 5\sqrt{2} = 25\)
In a 45-45-90 triangle sides are distributed with length that it is \(x-x-x\sqrt{2}\)
Given the side with length 10 is opposite to the 90 degree angle.
\(x\sqrt{2} = 10\)
Therefore, \(x = 10/\sqrt{2}\)
\(x = 2*5/\sqrt{2}\)
\(x = 5\sqrt{2}\)
Since this is an isosceles right triangle with height = base = x = \(5\sqrt{2}\)
we can get the area of a triangle as \(\frac{1}{2}* base * height\)
area = \(\frac{1}{2} * 5\sqrt{2}* 5\sqrt{2} = 25\)
Re: The area of the triangular region
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19 Jan 2019, 13:40
19 Jan 2019, 13:40
its 45 45 90
Now area is 1/2 * 10/√2 *10/√2
=25
Now area is 1/2 * 10/√2 *10/√2
=25
