Question #1
Since [tex] OP||MN [/tex], [tex] \frac{LO}{OM}=\frac{LP}{PN} [/tex]
[tex] \frac{22}{8}= \frac{LP}{5} [/tex]
[tex] \therefore LP=\frac{22\times 5}{8}=13.75 [/tex]
Thus, Option A is the correct answer.
Question #2
It is given that the measure of Arc TG is 70°.
Therefore, from the the [tex] m\angle TNE=\frac{1}{2}\times 70^o=35^o [/tex] (The angle subtended by an arc at the center is twice the angle subtended at the circumference.)
Now, in [tex] \Delta TNE [/tex], [tex] m\angle ETN=90^o [/tex] and [tex] m\angle TNE=35^o [/tex]. Thus, [tex] m\angle GET=m\angle NET=180^o-90^o-35^o=55^o [/tex] (Since the sum of the angles of a triangle is always 180 degrees).
Thus Option A is the correct option.
Question #3
It is given that the measure of angle AFB=36 degrees.
Therefore, since, mAB=mBC=mCD (given), [tex] \angle CFD=36^o [/tex]
Therefore, [tex] \angle CFD=18^o [/tex] because we know that The angle subtended by an arc at the center is twice the angle subtended at the circumference.
Question #4
To solve this question we will use the Angle of Intersecting Chords Theorem, which states that: "If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle."
Arc FB can be found as:
[tex] \overarc{FB}=360^o-154^o-100^o-40^o=66^0 [/tex]
Therefore, [tex] m\angle AED=\frac{1}{2}(m\overarc{FB}+m\overarc{AD})=\frac{1}{2}(66^0+40^0)=53^0 [/tex].
Thus, Option B is the correct option.
Question #5
We know that since these are secants from a point outside the circle, therefore:
[tex] BC\times AC=DC\times EC[/tex]
[tex] 6\times 10=DC\times 12[/tex]
[tex] \therefore DC=\frac{6\times 10}{12}=5[/tex]
Question #6
Scale factor is the ratio of the length of the linear sides of the new figure to the old figure.
Thus, Scale Factor is: [tex] \frac{8}{48}=\frac{6}{36}=\frac{1}{6}[/tex]
