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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.

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

(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

(2) DC

(3) DE.

(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.

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

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.