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The area of the triangular region

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The area of the triangular region [#permalink] New post   17 Jun 2018, 13:33
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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.
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C
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Re: The area of the triangular region [#permalink] New post   18 Jun 2018, 20:55
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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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Re: The area of the triangular region [#permalink] New post   19 Jan 2019, 13:40
its 45 45 90
Now area is 1/2 * 10/√2 *10/√2
=25
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Re: The area of the triangular region [#permalink] New post   21 Jul 2022, 22:06
The 45 : 45: 90 triangle theorem states that 45 : 45: 90 special right triangles that have sides of which the lengths are in a special ratio of 1: 1: √2

Now, in this particular question, we have the hypotenuse given as 10.
So, we take a ratio constant "x", which is basically the length of the other two sides of the triangle and hence, we can write,
x * √2 = 10.

So, the other two sides are, x = 10/(√2)

Let us calculate Quantity A, i.e., "Area of the Triangle" = 1/2 * Base * Height

Here, Base = 10/(√2) and Height = 10/(√2)

This implies, "Area of the Triangle" = 1/2 * 10/(√2) * 10/(√2) = 25

So, Quantity A is 25.

On the other hand, Quantity B is also 25 (As Given)

Hence, Quantity A = Quantity B.
So, Answer Choice is C.
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Re: The area of the triangular region [#permalink]
21 Jul 2022, 22:06
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