Practice Set 3.5
In figure 3.77, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.…
In figure 3.78, chord MN and chord RS intersect at point D.(1) If RD = 15, DS = 4, MD =…
In figure 3.79, O is the centre of the circle and B is a point of contact. seg OE seg…
In figure 3.80, if PQ = 6, QR = 10, PS = 8 find TS. 8
In figure 3.81, seg EF is a diameter and seg DF is a tangent segment. The radius of the…
Practice Set 3.5
Question 1.
In figure 3.77, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
Answer:
Given PQ = 12, PR = 8
SP × RP = PQ2
This property is known as tangent secant segments theorem.
⇒PS × 8 = 122
RS = PS – RP = 18 – 8 = 10
Question 2.
In figure 3.78, chord MN and chord RS intersect at point D.
(1) If RD = 15, DS = 4, MD = 8 find DN
(2) If RS = 18, MD = 9, DN = 8 find DS
Answer:
(1) Given RD = 15, DS = 4, MD = 8
MD × DN = RD × DS
This property is known as theorem of chords intersecting inside the circle.
⇒ 8 × DN = 15 × 4
(2)Given RS = 18, MD = 9, DN = 8
Here, RS = 18
Let RD = x and DS = 18 – x
MD × DN = RD × DS
This property is known as theorem of chords intersecting inside the circle.
⇒ 8 × 9 = x × (18 – x)
⇒18x – x2 = 72
⇒x2 – 18x + 72 = 0
⇒ (x – 12)(x – 6) = 0
⇒ x = 12 or 6
⇒ DS = 6 or 12
Question 3.
In figure 3.79, O is the centre of the circle and B is a point of contact. seg OE 
seg AD, AB = 12, AC = 8, find
(1) AD
(2) DC
(3) DE.
Answer:
(1)Given: OE AD, AB = 12, AC = 8
⇒ AD × AC = AB2
This property is known as tangent secant segments theorem.
⇒ AD × 8 = 122
(2)DC = AD – AC = 18 – 8 = 10
(3)As we know that a perpendicular from centre divides the chord in two equal parts. Here, OE AD.
⇒ DE = EC
⇒ DE + EC = DC
⇒2DE = DC
Question 4.
In figure 3.80, if PQ = 6, QR = 10, PS = 8 find TS.
Answer:
Given: PQ = 6, QR = 10, PS = 8
PT × PS = PR × PQ
This property is known as theorem of chords intersecting outside the circle.
⇒ PR = PQ + RQ = 6 + 10 = 16
⇒ PT × 8 = 16 × 6
⇒ PT = 12
TS = PT – PS = 12 – 8 = 4
Question 5.
In figure 3.81, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE × GE = 4r2
Answer:
In ∆DEF,
∠DFE = 90° {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}
Given: EF = diameter of the circle.
DE2 = DF2 + EF2 {Using Pythagoras theorem}
⇒ DE2 = DF2 + (2r)2
⇒ DE2 = DF2 + 4r2
⇒ DF2 = DE2- 4r2
Also, DE × DG = DF2
This property is known as tangentsecant segments theorem.
⇒DE × DG = DE2 - 4r2
⇒ DE2- DE × DG = 4r2
⇒DE(DE – DG) = 4r2
⇒ DE × EG = 4r2
Hence, proved.
