High School

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1. What is the length segment LP?

A.13.75

B. 1.75

C. 35.2

D. 23.2


2. Line ET is tangent to circle A at T, and the measure of Arc TG is 70°.


What is the measure of ∠GET?


A. 55

B. 20

C. 35

D. 110


3. Given the following: mAB=mBC=mCD

In circle F what is the measure of

A. 36

B.18

C.9

D.72


4. Given that measure of arc AF = 100º, arc = BD = 154º and arc AD = 40º.


What is the measure of ∠ AED?


A.13

B.53

C.26

D.40


5. What is DC if AC = 10 cm, BC = 6 cm, and EC = 12 cm?


A. 5

B. 20

C. 12

D. 4


6. A 48 inch by 36 inch poster is reduced to an image that is 8 inch by 6 inch in size to be used for a flyer. What is the scale factor?


A. 3/2

B. 1/6

C. 1/8

D. 6

1 What is the length segment LP A 13 75 B 1 75 C 35 2 D 23 2 2 Line ET is tangent to

Answer :

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]


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Rewritten by : Barada

its
1.B
2.C
3.A
4.D
5.A
6.B