Answer:
i think its D.
A reflection, a rotation, and a translation will prove that shape 2 is congruent to shape 1.
Step-by-step explanation:
If these are the missing parts:
What is the quadrilateral?
a.rhombus
b. rectangle
c. trapezoid
<span>d. square
The quadrilateral formed by the </span><span>two equilateral triangles taped together is A. RHOMBUS.
Rhombus has equal sides but its is different from a square because it looks like a diamond.
Rhombus also have parallel opposite sides and equal opposite angles. </span>
Consider the attached figure. If AB has length 1, then BC has length sin(15°) and CD (the altitude of triangle ABC) has length sin(15°)·cos(15°).
By the double angle formula for sin(α), ...
... sin(2α) = 2sin(α)cos(α)
Rearranging, this gives
... sin(α)·cos(α) = sin(2α)/2
We have
... CD = sin(15°)·cos(15°) = sin(2·15°)/2
... CD = sin(30°)/2 = (1/2)/2 = 1/4
That is, the altitude, CD, is 1/4 the hypotenuse, AB, of triangle ABC.