Question

a blender has blades as shown. what counterclockwise rotation will map the blade formed by △gfh onto the blade formed by △cjd? a rotation of □° will map the blade formed by △gfh onto the blade formed by △cjd.

Explanation:

Step1: Analyze the figure's symmetry

The blender blades form a symmetric star - like figure (a quadrilateral - based star, likely a square - based rotation - symmetric figure). The center of rotation is the center of the star. To map \(\triangle GFH\) onto \(\triangle CJD\) counter - clockwise, we consider the angle between the corresponding sides or vertices.
In a square - like rotational symmetry (since the figure has 4 - fold symmetry, as it looks like a four - bladed star), the angle between adjacent blades (triangles) when rotating counter - clockwise: The total angle around a point is \(360^{\circ}\), and for 4 equal - spaced blades, the angle between each adjacent pair of blades is \(\frac{360^{\circ}}{4}=90^{\circ}\). But let's check the positions. From \(\triangle GFH\) to \(\triangle CJD\), we can see that we need to rotate by \(90^{\circ}\times2 = 180^{\circ}\)? Wait, no. Wait, let's label the center as \(O\). Let's take a vertex, say \(G\) in \(\triangle GFH\) and its corresponding vertex \(C\) in \(\triangle CJD\). The angle between \(OG\) and \(OC\) when rotating counter - clockwise: If we consider the direction, from \(G\) (left - side) to \(C\) (right - side) in a counter - clockwise rotation, the angle between them is \(180^{\circ}\)? Wait, no, maybe I made a mistake. Wait, let's look at the figure again. The figure is a four - pointed star, so the rotation symmetry: when we rotate a blade (triangle) counter - clockwise to the next non - adjacent blade, the angle is \(180^{\circ}\). Wait, let's count the number of blades. There are 4 blades? Wait, the triangles are \(\triangle GFH\), \(\triangle CJD\), and two others. Wait, the center is the same. Let's consider the rotation that maps \(F\) to \(D\), \(G\) to \(C\), and \(H\) to \(J\). The angle between the vectors from the center to these points. If we rotate counter - clockwise, the angle between \(GFH\) and \(CJD\) is \(180^{\circ}\)? Wait, no, maybe \(90^{\circ}\) per step. Wait, let's think of the standard square rotation. In a square, rotating \(90^{\circ}\) counter - clockwise moves a vertex to the next one. But here, the triangles are arranged such that from \(\triangle GFH\) to \(\triangle CJD\), we need to rotate by \(180^{\circ}\)? Wait, no, let's check the positions. Let's assume the center is \(O\). The vector \(OG\) and \(OC\): if \(G\) is on the left, \(C\) is on the right, \(H\) is on the top - left, \(J\) is on the top - right, \(F\) is on the bottom - left, \(D\) is on the bottom - right. So to map \(G\) (left) to \(C\) (right), \(H\) (top - left) to \(J\) (top - right), \(F\) (bottom - left) to \(D\) (bottom - right), we need a rotation of \(180^{\circ}\) counter - clockwise? Wait, no, \(180^{\circ}\) rotation maps a point \((x,y)\) to \((-x,-y)\). But in terms of the angle, the angle between the initial and final position for a \(180^{\circ}\) rotation is \(180^{\circ}\). Wait, maybe I was wrong earlier. Let's re - evaluate. The figure is a four - bladed star, so the angle between each blade (triangle) when rotating counter - clockwise is \(90^{\circ}\). But \(\triangle GFH\) and \(\triangle CJD\): let's see the number of steps between them. From \(GFH\) to the next blade (say, the one with \(I\) and \(J\)) is \(90^{\circ}\), then to \(CJD\) is another \(90^{\circ}\), so total \(180^{\circ}\). Wait, yes, because \(GFH\) and \(CJD\) are opposite each other (across the center), so the counter - clockwise rotation angle is \(180^{\circ}\).

Step2: Confirm the rotation angle

In a rotational symmetry, when two figures are congruent and are related by a rotation, and they are opposite each other (symmetric with respect to the center), the rotation angle is \(180^{\circ}\). So a counter - clockwise rotation of \(180^{\circ}\) will map \(\triangle GFH\) onto \(\triangle CJD\).

Answer:

\(180\)