Given:
QW = 12 and WS = 4x + 1
PW = 14 and WR = 3x + 3
To find:
The length of the chord QS.
Solution:
If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
[tex]\Rightarrow QW \cdot WS= PW\cdot WR[/tex]
[tex]\Rightarrow 12 \cdot (4x+1)=14 \cdot (3x+3)[/tex]
[tex]\Rightarrow 48x+12= 42x+42[/tex]
Subtract 12 from both sides.
[tex]\Rightarrow 48x+12-12= 42x+42-12[/tex]
[tex]\Rightarrow 48x= 42x+30[/tex]
Subtract 42x from both sides.
[tex]\Rightarrow 48x-42x= 42x+30-42x[/tex]
[tex]\Rightarrow 6x= 30[/tex]
Divide by 6 on both sides, we get
⇒ x = 5
QS = QW + WS
[tex]=12+4x+1[/tex]
[tex]=12+4(5)+1[/tex]
[tex]=13+20[/tex]
= 43
The length of QS is 43.
