one vertex of a square and the midpoints of the two sides not containing this vertex are vertices of a triangle. if the area of the square is 16, what is the area of the triangle?
2007-11-07 11:26:05 · 5 answers · asked by kkga_1 1 in Science & Mathematics ➔ Mathematics
one example is to draw the triangle with vertices at:
(0,0), (4,2), and (2,4)
should be fairly straightforward from there
the length of the two equal sides are 2sqrt(5)
and the base is 2sqrt(2), so the height is
3sqrt(2)
area = b*h/2 = 2sqrt(2)* 3sqrt(2) / 2 = 6
of course it's easier to notice that the sides of the triangles divide in half rectangles or squares within the larger square and it is easy to calculate the area outside the triangle
you will have two 2*4 rectangles that are bisected
so both combined is an area of 2*4=8 outside the triangle
a 2 * 2 square is also bisected, so another 2*2/2=2 outside the triangle
subtract from the overall square area
16 - 8 - 2 = 6
wow, same answer from two different methods.
the math must work.
2007-11-07 11:31:25 · answer #1 · answered by Anonymous · 0⤊ 0⤋
16-10 = 6
2007-11-07 19:31:43 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
8
2007-11-07 19:29:22 · answer #3 · answered by OBESE HAS CHANGED HIS NAME 2 · 0⤊ 2⤋
A = 16 - (1/2)4*2 - (1/2)4*2 - (1/2)2*2
A = 16 - 4 - 4 - 2
A = 6
2007-11-07 19:34:24 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋
it's the same distance.
2007-11-07 19:28:05 · answer #5 · answered by Anonymous · 0⤊ 2⤋