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505-555 (Easy)|   Geometry|      
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Bunuel
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What is the area of the above right triangle, in terms of x? 02 Sep 2018, 21:59
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B
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90% (01:02) correct 10% (00:55) wrong based on 54 sessions

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


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B
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What is the area of the above right triangle, in terms of x? 02 Sep 2018, 22:52
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Bunuel

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


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Answer is B

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What is the area of the above right triangle, in terms of x? 03 Sep 2018, 08:52
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Bunuel

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


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This is a right-angled isosceles triangle
Attachment:
Distance.jpg
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So, Area of the Triangle must be \(\frac{x^2}{2}\), Answer must be (A)
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What is the area of the above right triangle, in terms of x? 03 Sep 2018, 10:02
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Abhishek009
Bunuel

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


Show Spoiler
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)
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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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What is the area of the above right triangle, in terms of x? 07 Sep 2018, 17:09
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Bunuel

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


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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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What is the area of the above right triangle, in terms of x? 08 Sep 2018, 09:30
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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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