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What is the area of the above right triangle, in terms of x?

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What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 02 Sep 2018, 20:59
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What is the area of the above right triangle, in terms of x?


A. x^2/2

B. \(\frac{x^2\sqrt{3}}{2}\)

C. \(x^2\sqrt{3}\)

D. 2x^2

E. \(2x^2\sqrt{3}\)


Attachment:
image005.jpg
image005.jpg [ 1.59 KiB | Viewed 740 times ]

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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 02 Sep 2018, 21:52
Bunuel wrote:
Image
What is the area of the above right triangle, in terms of x?


A. x^2/2

B. \(\frac{x^2\sqrt{3}}{2}\)

C. \(x^2\sqrt{3}\)

D. 2x^2

E. \(2x^2\sqrt{3}\)


Attachment:
1535954058494.jpg
1535954058494.jpg [ 77.24 KiB | Viewed 649 times ]
Answer is B

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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 03 Sep 2018, 07:52
Bunuel wrote:
Image
What is the area of the above right triangle, in terms of x?


A. x^2/2

B. \(\frac{x^2\sqrt{3}}{2}\)

C. \(x^2\sqrt{3}\)

D. 2x^2

E. \(2x^2\sqrt{3}\)


Attachment:
The attachment image005.jpg is no longer available
This is a right-angled isosceles triangle
Attachment:
Distance.jpg
Distance.jpg [ 3.44 KiB | Viewed 618 times ]
So, Area of the Triangle must be \(\frac{x^2}{2}\), Answer must be (A)

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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 03 Sep 2018, 09:02
Abhishek009 wrote:
Bunuel wrote:
Image
What is the area of the above right triangle, in terms of x?


A. x^2/2

B. \(\frac{x^2\sqrt{3}}{2}\)

C. \(x^2\sqrt{3}\)

D. 2x^2

E. \(2x^2\sqrt{3}\)


Attachment:
image005.jpg
This is a right-angled isosceles triangle
Attachment:
Distance.jpg
So, Area of the Triangle must be \(\frac{x^2}{2}\), Answer must be (A)


The given triangle is a 60-90-30 triangle.
So, sides are in the ratio: \(\sqrt{3}x(=base):2x(=hypotenuse):x(=height)\)

Area of triangle=\(\frac{1}{2}*base*height=\frac{1}{2}*\sqrt{3}x*x\)=\(\frac{x^2\sqrt{3}}{2}\)

Ans. (B)
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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 07 Sep 2018, 16:09
Bunuel wrote:
Image
What is the area of the above right triangle, in terms of x?


A. x^2/2

B. \(\frac{x^2\sqrt{3}}{2}\)

C. \(x^2\sqrt{3}\)

D. 2x^2

E. \(2x^2\sqrt{3}\)


Attachment:
image005.jpg


We see that we have a 30-60-90 right triangle, so the base, which is opposite the 60 degree angle, is x√3.

Thus, the area of the triangle is x * x√3 * 1/2 = (x^2√3)/2.

Answer: B
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Re: What is the area of the above right triangle, in terms of x?  [#permalink]

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New post 08 Sep 2018, 08:30
This right triangle is a 30, 60, 90 triangle.
So, its side lengths are \(x\), \(\sqrt{3}x\), \(2x\)
Area =\(\frac{1}{2}*Base*Height\)
Area= \(\frac{1}{2}*x*\sqrt{3}x\)
Area= \(x^2\sqrt{3}x/2\)
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Re: What is the area of the above right triangle, in terms of x? &nbs [#permalink] 08 Sep 2018, 08:30
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