Jharkhand Board JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.4 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.4

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Question 1.

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer:

OP = 5 cm, PS = 3 cm and OS = 4 cm.

We know that if two circles intersect at two points then their centres lie on the perpendicular bisector of the common chord.

So, OS is perpendicular bisector of PQ.

⇒ PQ = 2PR and ∠PRO = ∠PRS = 90°

Let RS be x ⇒ OR = 4 – x

In ∆POR,

OP^{2} = OR^{2} + PR^{2}

⇒ 5^{2} = (4 – x)^{2} + PR^{2}

⇒ 25 = 16 + x^{2} – 8x + PR^{2}

⇒ PR^{2} = 9 – x^{2} + 8x …(i)

In ∆PRS,

PS^{2} – PR^{2} + RS^{2}

⇒ 3^{2} = PR^{2} + x^{2}

⇒ PR^{2} = 9 – x^{2} …(ii)

Equating (i) and (ii),

9 – x^{2} + 8x = 9 – x^{2}

⇒ 8x = 0

⇒ x = 0

Putting the value of x in (i), we get

PR^{2} – 9 – 0^{2}

⇒ PR = 3 cm

Length of the Chord PQ = 2PR = 2 × 3 = 6 cm

Question 2.

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer:

Given: AB and CD are chords intersecting at E.

AB = CD

To prove: AE = DE and CE = BE

Construction: OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof: OM bisects AB (OM ⊥ AB)

⇒ AM = MB = \(\frac{1}{2}\) AB

ON bisects Cp (ON ⊥ CD)

⇒ DN = CN = \(\frac{1}{2}\) CD

As AB = CD

thus, AM = ND …….(i)

and MB = CN ……..(ii)

In ∆OME and ∆ONE,

∠OME = ∠ONE = 90° (Perpendiculars)

OE = OE (Common)

OM = ON (AB = CD and thus equidistant from the centre)

∆OME ≅ ∆ONE by RHS congruence criterion.

ME = EN by CPCT …(iii)

Adding (i) and (ii), we get

AM + ME = ND + EN

⇒ AE = ED

Subtracting (iii) from (ii), we get

MB – ME = CN – EN

⇒ EB = CE

Question 3.

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer:

Given: AB and CD are chords intersecting at E.

AB = CD, PQ is the diameter.

To prove: ∠BEQ = ∠CEQ

Construction: OM ⊥ AB and ON ⊥ CD. OE is joined.

In ∆OEM and ∆OEN,

OM = ON (AB = CD and Equal chords are equidistant from the centre)

OE = OE (Common)

∠OME = ∠ONE = 90° (Perpendiculars)

∆OEM = ∆OEN (by RHS congruence criterion)

Thus, ∠OME = ∠ONE (by CPCT)

⇒ ∠BEQ = ∠CEQ

Question 4.

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig).

Answer:

OM ⊥ AD is drawn from O.

So, OM bisects AD (perpendicular from centre to the chord bisects the chord)

⇒ AM = MD …(i)

Also, OM bisects BC as OM ⊥ BC.

⇒ BM = MC …(ii)

Subtracting (ii) from (i), we get

AM – BM = MD – MC

⇒ AB = CD

Question 5.

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m e ach, what is the distance between Reshma and Mandip?

Answer:

Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.

AB = 6 m and BC = 6 m Radius OA = 5 m

BM ⊥ AC is drawn.

ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.

Let, AM = y and OM = x then BM = (5 – x).

Applying Pythagoras theorem in ∆OAM,

OA^{2} = OM^{2} + AM^{2}

⇒ 5^{2} = x^{2} + y^{2} …(i)

Applying Pythagoras theorem in ∆AMB,

AB^{2}= BM^{2} +AM^{2}

⇒ 6^{2} = (5-x)^{2} + y^{2} …(ii)

Subtracting (i) from (ii), we get

36 – 25 = (5 – x)^{2} – x^{2}

⇒ 11 = 25 – 10x

⇒ 10x = 14

⇒ x = \(\frac{7}{5}\)

Substituting the value of x in (i), we get

y^{2} + \(\frac{49}{25}\) =25

⇒ y^{2} = 25 – \(\frac{49}{25}\)

⇒ y^{2} = \(\frac{625-49}{25}\)

⇒ y^{2} = \(\frac{576}{25}\)

⇒ y = \(\frac{24}{5}\)

Thus,

AC = 2 × AM = 2 × y

= 2 × \(\frac{24}{5}\) m = \(\frac{48}{5}\) m = 9.6 m

Distance between Reshma and Mandip is 9.6 m.

Question 6.

A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus, ABC is an equilateral triangle.

AD ⊥ BC is drawn.

Now, AD is median of ∆ABC and it passes through the centre O.

Also, O is the centroid of the ∆ABC.

OA is the radius of the triangle.

OA = \(\frac{2}{3}\) AD

Let the side of the triangle be a metres

then BD = \(\frac{a}{2}\) m.

Applying Pythagoras theorem in ∆ABD,

AB^{2} = BD^{2} + AD^{2}

⇒ AD^{2} = AB^{2} – BD^{2}

⇒ AD^{2} = a^{2} – (\(\frac{a}{2}\))^{2}

⇒ AD^{2} = \(\frac{3 a^2}{4}\)

⇒ AD = \(\frac{\sqrt{3} a}{2}\)

OA = \(\frac{2}{3}\) AD

⇒ 20m = \(\frac{2}{3}\) × \(\frac{\sqrt{3} a}{2}\)

⇒ a = \(20 \sqrt{3} \)m

∴ Length of the string is \(20 \sqrt{3} \)m