Question

1. which figure shows the rotation image of pentagon a after a 216° clockwise rotation about its ce

pentagon a
figure i
figure ii
figure iii
figure iv
a. figure iii
b. figure i
c. figure iv
d. figure ii

1. quadrilateral p is rotated 90° clockwise about vertex a, then 270° counterclockwise about vert

which quadrilateral shows the final position of quadrilateral p?
a. p
b. s
c. q
d. r
: circle geometry

1. o is the centre of this circle and point m is a point of tangency.

e the value of x°.

Explanation:

Question 23

A regular pentagon has an interior angle of \(108^\circ\), and the central angle between each vertex is \( \frac{360^\circ}{5}=72^\circ\). A \(216^\circ\) clockwise rotation is equivalent to \(216\div72 = 3\) steps of \(72^\circ\) each. Starting from Pentagon A (with vertices P, Q), rotating 3 times \(72^\circ\) clockwise will move P and Q to the positions shown in Figure iii.

First, a \(90^\circ\) clockwise rotation about A, then a \(270^\circ\) counterclockwise rotation about A. The total rotation is \(90^\circ - 270^\circ=- 180^\circ\) (or a \(180^\circ\) clockwise rotation, which is the same as a \(180^\circ\) rotation). A \(180^\circ\) rotation about A will map Quadrilateral P to Quadrilateral Q.

Step1: Recall tangent - radius property

A tangent to a circle is perpendicular to the radius at the point of tangency. So \(OM\perp MN\), \(\angle OMN = 90^\circ\).

Step2: Use angle sum property of triangle

In \(\triangle OMN\), \(\angle O+\angle OMN+\angle x = 180^\circ\). We know \(\angle O = 37^\circ\) and \(\angle OMN = 90^\circ\). So \(37^\circ+90^\circ+x^\circ = 180^\circ\).

Step3: Solve for \(x\)

\(x^\circ=180^\circ-(37^\circ + 90^\circ)=180^\circ - 127^\circ = 53^\circ\).

Answer:

a. Figure iii

Question 24