FINDING LENGTHS OF TWO SECANTS INTERSECTING IN THE EXTERIOR OF CIRCLE

AH and HY are the two secant segments intersecting at point H. SH is the external secant segment of the whole secant segment AH, and LH is the external secant segment of HY. Thus, according to the theorem, we have

SH × AH = HL × HY

secant-secant-theorem.png

Find the value of x in the following.

Problem 1 :

secant-secant-theoremq1.png

Solution :

 (x + 3) = 4 • (8 + 4)

3(x + 3) = 4 (12)

x + 3 = 16

x = 16 - 3

x = 13

Problem 2 :

CD and EF are chords of the circle with centre O. The chords meet at the point P, which is outside of the circle

secant-secant-theoremq2.png

Solution :

PD • PC = PF • PE

x • (x + 12) = 13 • (13 + 15)

x(x + 12) = (13 • 28)

x2 + 12x = 364

x2 + 12x - 364 = 0

(x + 26)(x - 14) = 0

x = -26 and x = 14

Problem 3 :

CD and EF are chords of the circle below. The chords meet at the point P, which is outside of the circle.

secant-secant-theoremq3.png

Solution :

PD • PC = PF • PE

9 • (9 + 11) = 10 • (10 + EF)

Let EF be x.

9 • (20) = 10 • (10 + x)

10 + x = 180/10

10 + x = 18

x = 18 - 10

x = 8

Problem 4 :

secant-secant-theoremq4.png

Solution :

3 • (3 + 5) = 4 • (4 + x)

3 • 8 = 4 • (4 + x)

Dividing by 4 on both sides.

6 = 4 + x

x = 6 - 4

x = 2

Problem 5 :

secant-secant-theoremq5.png

Solution :

x • (x + 3x) = 2 • (2 + 6)

x • 4x = 2 • 8

4x2 = 16

Dividing by 4 on both sides.

x2 = 4

x = 2

Problem 6 :

secant-secant-theoremq7.png

Solution :

6 • (6 + 3) = (12 - x) • 12

6(9) = 12(12 - x)

54 = 12(12 - x)

Dividing by 12 on both sides,

4.5 = 12 - x

x = 12 - 4.5

x = 7.5

Problem 7 :

CD and EF are chords of the circle with centre O. The chords meet at the point P, which is outside of the circle.  Calculate the area of triangle OEF.

secant-secant-theoremq8.png

Solution :

OE = OF (radii)

PD • PC = PF • PE

20 • (20 + 12) = 16 • (16 + EF)

20 (32) = 16 (16 + EF)

Dividing by 16 on both sides.

20(2) = 16 + EF

EF = 40 - 16

EF = 24

Drawing the perpendicular from O.

secant-secant-theoremq8p1.png

In triangle OGF,

OF2 = OG2 + GF2

152 = OG2 + 122

225 - 144 = OG2

OG2 = 81

OG = 9

Area of triangle OEF = (1/2) x base x height

= (1/2) x 24 x 9

= 12 x 9

= 108 cm2

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