Sophia Marie D. Verdeflor Grade 10-1 STE

Activity 8: My Real World

Answer the following. Use the rubric provided to rate your work.

1. The chain and gears of bicycles or motorcycles or belt around two pulleys are some

real-life illustrations of tangents and circles. Using these real-life objects or similar ones, formulate problems involving tangents, then solve.

The big gear represents the circle while the smaller one represents the exterior point of

intersection of the tangent lines.

a. In the figure above, ↔ and ↔ are two tangents intersecting outside the circle at point

PI PAP. Arc IZA and arc IA are the two intercepted arcs of L IPA. If arc IZA measures 220°

and arc IA measures 140°, what is the measure of L IPA?

mL IPA = ½ (arc IZA – arc IA)mL IPA = ½ (220° - 140°)mL IPA = ½ (80°)mL IPA = 40 °

b. ↔ and ↔ are tangents. What is the measure of ↔ if ↔ measures 15 inches? IP AP AP IP

AP = 15 inches; the measure of ↔ is the same with the measure of ↔ because they are

AP IPcongruent.

c. ↔ and ↔ are tangents. The measure of ↔ is 21 cm and the measure of ↔ is x^2 + 5.

IP AP IP APFind x.

x^2 + 5 = 21 - 5 -5

√x^2 = √16 x = 4

d. ↔ and ↔ are tangents. The measure of arc IZA measures 158° and the measure of

IP APL IPA is 68°, what is the measure of the other arc which is arc IA?

68° = ½ (158° - arc IA)2 · 68° = 2 · ½ (158° - arc IA) 136° = 158° – arc IAarc IA = 158° – 136°arc IA = 22 °

e. Suppose the measure of ↔ is 2x + 10 and the measure of ↔ is 3x + 7. Find: IP AP

I. xII. ↔ IPIII. ↔ AP

I. 2x + 10 = 3x + 7 - 7 - 7 2x + 3 = 3x -2x -2x 3 = x

II. 2x + 10 = ↔ IP

2(3) + 10 = ↔ IP

6 + 10 = ↔ IP

16 = ↔ IP

III. 3x + 7 = ↔

AP 3(3) + 7 = ↔

AP 9 + 7 = ↔

AP 16 = ↔

AP

2. The picture below shows a bridge in the form of an arc. It also shows how secant is

illustrated in real life. Using the bridge in the picture and other real-life objects, formulate problems involving secants, then solve them.

The picture of the bridge above shows the real-life application of secant of the circle.

a. In the figure above, ↔ and ↔ are two secants intersecting outside the circle at point A.

SP OH arc SO and arc PH are the two intercepted arcs of L SIO. If arc SO measures 140°

and arc PHmeasures 50°, what is the measure of L SIO?

mL SIO = ½ (arc SO – arc PH)mL SIO = ½ (140° - 50°)mL SIO = ½ (90°)mL SIO = 45 °

b. In circle A above, two secants from point I intersect circle A such that arcs IP = 10, IH = 9,

PS = 2x, and HO = 2x +3. What is the measure of segment SI?

The products of the external segment and the entire secant must be equal for both secants.

We have:

IP (IP + PS) = IH (IH + HO)10 (2x + 10) = 9 (2x + 12)

Solving this equation for x we get

20x + 100 = 18x + 1082x = 8x = 4

O

SP

HI.A

Since SI equals 2x + 10

SI = 2(4) + 10SI = 18

c. In circle A above, secants IS and IO are drawn from point A. We have the following measurements given: IP = 6, PS = 8, IH = x + 2, and HO = 5x +7. What is the measure of

chord HO?

IP (IS) = IH (IO)6 (14) = (x + 2)(6x + 9)

Solving this equation for x gives us

6(14) = (x + 2)(6x + 9)84 = 6x2 + 21x + 180 = 6x2 + 21x - 660 = 3(2x2 + 7x - 22)0 = 3(x - 2)(2x + 11)x = 2 and x = -11/2

Using x = 2, we have

HO = 5(2) + 7 = 17

Using x = -11/2, we have

HO = 5(-11/2) + 7 = -41/2

We must discard this answer since the length of a segment cannot be a

negative number.

d. In circle A shown above, two secants from point I intercept arcs PH = x – 10 and SO = 2x. What is the measure of arc SO if angle I is 25°?

We know that the measure of an external angle P when formed by two secants is equal

to one half the difference of the measures of the intercepted arcs. mL I = ½ (arc SO – arc PH)

mL I = ½ [2x – ( x – 10) ]mL I = ½ (x + 10)25 = (1/2)(x + 10)50 = x + 10x = 40º

Since we were given that arc SO = 2x

SO = 2(40º)SO = 80°

e. In circle A above, angle I is x – 10, arc PH is 55, and arc SO is 3x. What is the measure

of angle I?

SO and PH are arcs intercepted by the secants IS and IO. The measure of angle I must be

½ the difference of the measures of arcs IS and IO.

mL I = ½ (arc SO – arc PH)x – 10 = ½ (3x – 55)

Solving for x:

2x –20 = 3x – 553x – 2x = 55 – 20x = 55 – 20x = 35

We were given that angle I = x – 10, substituting x = 35

mL I = 35 – 10mL I = 25 °