Question

lesson 14 practice problems

1. draw the image of quadrilateral abcd when rotated 120° counter - clockwise around the point d.
2. there is an equilateral triangle, abc, inscribed in a circle with center d. what is the smallest angle you can rotate triangle abc around d so that the image of a is b?

a. 60°
b. 90°
c. 120°
d. 180°

1. which segment is the image of ab when rotated 90° counter - clockwise around point p?

Explanation:

1.

Step1: Recall rotation rules

To rotate a point $(x,y)$ counter - clockwise about a center point $(a,b)$ by an angle $\theta$, we can use rotation matrices or geometric construction. For a rotation of a quadrilateral about a point $D$, we consider each vertex of the quadrilateral $A$, $B$, $C$, and $D$ separately.

1. First, connect each vertex $A$, $B$, $C$ to the center of rotation $D$.
2. Then, measure an angle of $120^{\circ}$ counter - clockwise from the line segments $DA$, $DB$, $DC$ respectively.
3. Mark the new points $A'$, $B'$, $C'$ such that $DA'=DA$, $DB' = DB$, $DC'=DC$.
4. Connect $A'$, $B'$, $C'$, $D$ to form the rotated quadrilateral.

Since this is a geometric construction problem and not a numerical calculation, we can't provide a numerical answer but the steps for construction are as above.

2.

Step1: Recall properties of an equilateral triangle

An equilateral triangle has all angles equal to $60^{\circ}$ and the central angles subtended by its sides at the center of the circum - circle are equal. The central angle of an equilateral triangle inscribed in a circle is $120^{\circ}$ for each side.
If we want to rotate the triangle $ABC$ around the center $D$ such that the image of $A$ is $B$, we need to rotate it by the central angle subtended by side $AB$ at the center $D$.
The central angle of an equilateral triangle inscribed in a circle is $120^{\circ}$.

Step1: Recall rotation properties

When we rotate a line segment $AB$ counter - clockwise about a point $P$ by $90^{\circ}$, we consider the movement of points $A$ and $B$ around $P$.

1. For point $A$, we draw a line from $P$ to $A$. Then we rotate this line $90^{\circ}$ counter - clockwise about $P$.
2. Similarly for point $B$, we draw a line from $P$ to $B$ and rotate it $90^{\circ}$ counter - clockwise about $P$.
3. By observing the grid, when we rotate segment $AB$ $90^{\circ}$ counter - clockwise about $P$, the image of $AB$ is $FG$.

Answer:

C. $120^{\circ}$

3.