Hi, Can you please let me know where I am making mistake in this question.
My analysis is as follows::
For Area of Triangle, BCE BC = 1 and ce = 1
I thought the base is 1 and the height is also 1 so I calculated as 1/2.1.1 and calculated the area as 1/2.
Can you please point me on where I am making mistake.
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kamalakarthi
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Hi kamalakarthi,
When calculating the area of a triangle, any side can be the "base", as long as the side is "flat on the ground." Once you choose your base, the "height" will be a line (the math term is an "altitude") that touches the highest point on the triangle AND is perpendicular with "the ground." Sometimes the height is inside the triangle and sometimes it's outside of the triangle.
In your example, segment CE is NOT the height (because it's not perpendicular with the ground). The height of this triangle is actually OUTSIDE of it. To help you determine the height, I'm going to give you a couple of clues....
1) Look at point C. The 3 angles around it will form the center of a circle, so those 3 angles sum to 360 degrees.
2) Both BCE and DCE are the SAME isosceles triangle
3) All the angles in the square are 90 degrees
4) Try drawing a line "out" from BC (to the right) and another line "down" from point E to the base
GMAT assassins aren't born, they're made,
Rich
When calculating the area of a triangle, any side can be the "base", as long as the side is "flat on the ground." Once you choose your base, the "height" will be a line (the math term is an "altitude") that touches the highest point on the triangle AND is perpendicular with "the ground." Sometimes the height is inside the triangle and sometimes it's outside of the triangle.
In your example, segment CE is NOT the height (because it's not perpendicular with the ground). The height of this triangle is actually OUTSIDE of it. To help you determine the height, I'm going to give you a couple of clues....
1) Look at point C. The 3 angles around it will form the center of a circle, so those 3 angles sum to 360 degrees.
2) Both BCE and DCE are the SAME isosceles triangle
3) All the angles in the square are 90 degrees
4) Try drawing a line "out" from BC (to the right) and another line "down" from point E to the base
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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I think your error is in assuming the height of that triangle is 1. Let's see what it really is.
Start with this picture.
Notice that ∆AEG and ∆CEF are both 45-45-90 ∆'s. Since ∆CEF has a hypotenuse of 1, it has legs of √2/2. (The legs of a 45-45-90 are the hypotenuse divided by √2.)
Since EF is the height of ∆BCE, we have the height and the base.
Base = 1
Height = √2/2
Area = 1 * √2/2 * 1/2 = √2/4.
Start with this picture.
Notice that ∆AEG and ∆CEF are both 45-45-90 ∆'s. Since ∆CEF has a hypotenuse of 1, it has legs of √2/2. (The legs of a 45-45-90 are the hypotenuse divided by √2.)
Since EF is the height of ∆BCE, we have the height and the base.
Base = 1
Height = √2/2
Area = 1 * √2/2 * 1/2 = √2/4.
