geometry problem?

Question

ABCDEFGH is a cube having a side length of 2. P is the midpoint of EF, as shown. Determine the area of triangle ABP.http://www.e-zgeometry.com/pow/pictures/33.gif

Answer 1

As always, the area of a triangle is half of the base times the height.

Let AB be the base of ABP.
The height is then equal to BF, which is 2*sqrt(2).

The area of ABP is (1/2)*2*2sqrt(2), or 2*sqrt(2).

You can go through a heckuva lot more math than this in order to solve this problem, but why make things harder than they need to be? :-)

Answer 2

Make triangle DEP. This triangle has side lengths of 1 and 2, so the third side length, DP, has length sqrt5. Then, make triangle ADP. It has sidelengths sqrt5 and 2, so the third side length, AP, has length 3. The same goes for BP, where you make triangles CFP and BPC. The lengths of the two sides of the triangle ABP are 3, the other length being two. Split ABP into two right triangles, the base being one and the hypotenuse being three. The length of the height would be 'b' in the pythagorean theorem, so 1+h^2=9. h=sqrt8. A=(1/2)(b)(h). A=1*sqrt8, so the area of the triangle ABP is the square root of 8, or 2 root 2

Answer 3

Cube has six square sides. you have the base, it's AB which is
2. Perpendicular height ABP = BF

You have BG which is 2 and you have GF which is also 2, so you can get BF using pythagerous' theorem.

BF squared = 2 squared + 2 squared
BF = square root(8)

Now, area of triangle ABP = 1/2 AB * BF
= 1/2 of 2 * square root 8
= square root of eight

Answer 4

the hieght of the traingle ABP is also the diagnal of DAHE

since it is a square the diagnal which is also the height of the triangle ABP, is 2sqrt(2)

A = 1/2bh

A = 1/2 * 2 * 2sqrt(2)

A = 2sqrt(2)

Answer 5

144