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Practice Set 3.5 Circle Class 10th Mathematics Part 2 MHB Solution

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.