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Re: In the figure, what is the area of ∆ABC ?
[#permalink]
02 Oct 2018, 09:25
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Carcass wrote:
Attachment:
triangle.jpg
In the figure, what is the area of ∆ABC ?
(A) 2
(B) \(\sqrt{2}\)
(C) 1
(D) \(\frac{1}{\sqrt{2}}\)
(E) \(\frac{1}{2}\)
Since vertically opposite angles are equal, we can conclude that y° = z°
If y° = z°, then we can take the angle y° (on the bottom right of the triangle) and replace it with z°
Since ∆ABC has two EQUAL angles, we know that it also has two EQUAL SIDES
Since vertically opposite angles are equal, we can conclude that the missing angle (shown below) is 2z°
At this point, the 3 angles are labeled, z°, z° and 2z° Since angles in a triangle must add to 180°, we can write: z + z + 2z = 180 Simplify: 4z = 180 Solve, z = 45° This also means 2z = 90°
When we add these 3 angles to our diagram . . .
. . . we see that we actually have a RIGHT TRIANGLE
ASIDE: The diagrams in GRE math questions are not necessarily drawn to scale. So, even though ∆ABC doesn't LOOK like a right triangle, it IS actually a right triangle. In fact, we could actually redraw the triangle as follows:
At this point, it's easy to calculate the area of ∆ABC
Re: In the figure, what is the area of ABC ?
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22 May 2024, 11:40
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Re: In the figure, what is the area of ABC ? [#permalink]