The document provides examples of problems involving tangents and secants of circles and their solutions. It gives real-life examples using bicycle gears, bridges, and other objects to represent circles, tangents, secants, and related geometric concepts. It then presents multiple multi-part problems involving finding measures of arcs, angles, segments and solving equations related to tangents and secants. The problems are solved step-by-step showing the work.

1. Sophia Marie D. Verdeflor Grade 10-1 STE
Activity 8: MyReal 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 representsthe 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 PA
P. Arc IZA and arc IA are the two interceptedarcs 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 theyare
AP IP
congruent.
c. ↔ and ↔ are tangents. The measure of ↔ is 21 cmand the measure of ↔ is x^2+ 5.
IP AP IP AP
Find x.
x^2+ 5 = 21
- 5 -5
√x^2= √16
x = 4

2. d. ↔ and ↔ are tangents. The measure of arc IZA measures 158° and the measure of
IP AP
L 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 IA
arc 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. x
II. ↔
IP
III. ↔
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

3. 2. The picture below shows a bridge in the form of an arc. It also shows how secantis
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 PH
measures 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 suchthat arcs IP = 10, IH = 9,
PS = 2x,and HO = 2x +3.What is the measure of segment SI?
The products of the external segmentand 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 + 108
2x = 8
x = 4
Since SI equals 2x + 10
SI = 2(4) + 10
SI = 18
O
S
P
H
I.
A

4. 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 + 18
0 = 6x2+ 21x - 66
0 = 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 discardthis answersince 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 externalangle P whenformed 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 + 10
x = 40º
Since we were giventhat arc SO= 2x
SO = 2(40º)
SO = 80°

5. 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 interceptedby 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 – 55
3x – 2x = 55 – 20
x = 55 – 20
x = 35
We were given that angle I = x – 10, substituting x = 35
mL I = 35 – 10
mL I = 25°