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In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t

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Kritisood
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In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink] New post   01 Apr 2020, 08:01
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In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is the area of the triangle?

A. √3/2
B. 1
C. √3
D. 2
E. 2√3
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C

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BrentGMATPrepNow
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In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink] New post   01 Apr 2020, 09:40
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Kritisood wrote:
In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is the area of the triangle?

A. √3/2
B. 1
C. √3
D. 2
E. 2√3


Aside: Although we can, indeed, solve this question using trigonometric ratios, the GMAT test makers don't expect students to know about trigonometric ratios. So another approach is to apply what we know about special 30-60-90 right triangles

Since we're told that AB = BC, we know that the triangle is an isosceles triangle, which means ∠BCA = 30°
So if we draw in the altitude from point B, we get two identical right triangles.


Also notice that these right triangles are both special 30-60-90 right triangles
So let's compare these triangles with our base 30-60-90 right triangle


When we do this, we see that the two red right triangles are identical to the base triangle
So, we can add the following measurements to our diagram:


At this point we're ready to find the area of the triangle.

Area of triangle = (base)(height)/2

So, the area \(= \frac{(\sqrt{3}+\sqrt{3})(1)}{2}= \frac{(2\sqrt{3})(1)}{2}=\sqrt{3}\)

Answer: C

Cheers,
Brent
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Re: In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink] New post   01 Apr 2020, 09:49
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Kritisood wrote:
In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is the area of the triangle?

A. √3/2
B. 1
C. √3
D. 2
E. 2√3


Drop a perpendicular BD from point B on AC

Now we have two 30-60-90 right triangles with hypotenuse 2 unit long

the length of BD = 1
AD = √3

i.e. AC = 2√3

Area = (1/2)*AC*BD = (1/2)*2√3*1 = = √3

Answer: Option C
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Re: In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink] New post   01 Apr 2020, 09:17
Kritisood wrote:
In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is the area of the triangle?

A. √3/2
B. 1
C. √3
D. 2
E. 2√3


Attachment:
Img.png
Img.png [ 11.91 KiB | Viewed 2922 times ]

let the side AC be x.
\(\frac{2}{sin(30)}=\frac{x}{sin(120)}\)
\(x = \frac{2*sin(120)}{sin(30)}\)
\(x = \frac{2*cos(30)}{sin(30)}\) = \(\frac{2*2*\sqrt{3}}{2}\); sin(90+x)=cos(x)
\(x = 2*\sqrt{3}\)
To find area we can use formula
\(A =\frac{1*a*b*sin(x)}{2} \); where a,b are two adjacent sides with angle x between them
\(A =\frac{1*2*2*\sqrt{3}*sin(30)}{2} \)
\(A =\sqrt{3}\)
Ans: C
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Re: In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink] New post   15 Apr 2023, 12:03
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Re: In the given isosceles triangle, if AB = BC = 2 and x = 30°, what is t [#permalink]
15 Apr 2023, 12:03
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