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What is the area of the triangle shown above?
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01 Apr 2020, 09:41
01 Apr 2020, 09:41
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What is the area of the triangle shown above?
A. \(\frac{25\sqrt{2}}{3}\)
B. \(\frac{25\sqrt{3}}{2}\)
C. 25
D. \(25\sqrt{2}\)
E. \(25\sqrt{3}\)
Kudos for the right answer and explanation
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Re: What is the area of the triangle shown above?
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01 Apr 2020, 09:48
01 Apr 2020, 09:48
1
Carcass wrote:
What is the area of the triangle shown above?
A. \(\frac{25\sqrt{2}}{3}\)
B. \(\frac{25\sqrt{3}}{2}\)
C. 25
D. \(25\sqrt{2}\)
E. \(25\sqrt{3}\)
Kudos for the right answer and explanation
Since angles in a triangle must add to 180°, we can see that the missing angle is 60°, which means we have a Special 30-60-90 Special Triangle
So let's compare the given 30-60-90 triangle with the base 30-60-90 triangle
In the base triangle, the side opposite the 90-degree angle has length 2, and in the given triangle, the side opposite the 90-degree angle has length 10
10/2 = 5, which means the given triangle is 5 times the size of the base triangle
Now that we know the Magnification Factor, we can determine the lengths of the remaining sides.
In the base 30-60-90 triangle, the side opposite the 30-degree angle has length 1
So, in the given triangle, x = (5)(1) = 5
Likewise, in the base 30-60-90 triangle, the side opposite the 60-degree angle has length √3
So, in the given triangle, x = (5)(√3) = 5√3
We get:
We now have enough information to find the area of a triangle.
Area of triangle = (base)(height)/2
If we let side AC be the base, and let side CB be the height, then the area = (5)(5√3)/2 = (25√3)/2
Answer: B
Cheers,
Brent
gmatclubot
Re: What is the area of the triangle shown above? [#permalink]
01 Apr 2020, 09:48
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