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# The area of the triangular region

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Joined: 18 Apr 2015
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Kudos [?]: 1029 [0], given: 4636

The area of the triangular region [#permalink]  17 Jun 2018, 13:33
Expert's post
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Question Stats:

75% (01:06) correct 25% (00:48) wrong based on 12 sessions
Attachment:

triangle.jpg [ 6.88 KiB | Viewed 389 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.
[Reveal] Spoiler: OA

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Joined: 07 Jan 2018
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Kudos [?]: 477 [2] , given: 84

Re: The area of the triangular region [#permalink]  18 Jun 2018, 20:55
2
KUDOS
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 = $$\frac{1}{2} * 5\sqrt{2}* 5\sqrt{2} = 25$$
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: The area of the triangular region   [#permalink] 18 Jun 2018, 20:55
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