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01 Apr 2020, 08:01
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C
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Difficulty:
35%
(medium)
Question Stats:
70% (01:36) correct
30%
(01:47)
wrong
based on 152
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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
A. √3/2
B. 1
C. √3
D. 2
E. 2√3
Attachments
Screenshot 2020-04-01 at 8.31.03 PM.png [ 11.4 KiB | Viewed 3957 times ]
01 Apr 2020, 09:40
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Kritisood
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
General Discussion
ArunSharma12
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01 Apr 2020, 09:17
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Kritisood
Attachment:
Img.png [ 11.91 KiB | Viewed 3448 times ]
\(\frac{2}{sin(30)}=\frac{x}{sin(120)}\)
\(x = \frac{2*sin(120)}{sin(30)}\)
\(x = \frac{2*cos(30)}{sin(30)}\) = \(\frac{2*2*\sqrt{3}}{2}\); 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
