Bihar Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.4 Textbook Questions and Answers.

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

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.

Solution:

Let O and O’ be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be them common chord.

We have OP = 5 cm, O’P = 3 cm and OO’ = 4 cm.

Since OP² = PO’² + O’O² [∵ 5² = 3² + 4²]

⇒ OOP is a right ∠d ∆, right angled as O’.

From (1) and (2), we have

2 x PL = 6 ⇒ PL = 3

We know that when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord i.e., 00′ is the perpendicular bisector of AB.

∴ PQ = 2 x PL = (2 x 3) cm = 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.

Solution:

Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.

To prove : (i) AP = PD (ii) PB = CP.

Construction : Draw OM ⊥ AB, ON ⊥ CD.

Join OP.

AM = MB = \(\frac { 1 }{ 2 }\)[Perpendicular from centre bisects the chord]

CN = ND = \(\frac { 1 }{ 2 }\) CD [Perpendicular from centre bisects the chord]

AM = ND and MB = CN s …(1) [∵ AB = CD (given)]

In As OMP and ONP, we have

OM = ON

[Equal chords of a circle are equidistant from the centre]

∠OMP = ∠ONP [∵ Each = 90°]

OP = OP [Common]

By RHS criterion of congruence,

∆ OMP = ∆ ONP

⇒ MP = PN … (2) [C.P.C.T.]

Adding (1) and (2), we have

AM + MP = ND + PN ⇒ AP = PD

Subtracting (2) from (1), we have

MP – MP = CN – PN

PB = CP

Hence, (i) AP = PD and (ii) PB = CP.

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.

Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.

To prove : ∠OPE = ∠OPF.

Construction : Draw OE ⊥ AB and OF ⊥ CD, Join OP.

Solution:

In ∆s OEP and OFP, we have

∠OEP = ∠OFP [∵ Each = 90°]

OP = OP

OE = OF [Equal chords of a circle are equidistant from the centre]

∴ By RHS criterion of congruence

∆ OEP ≅ ∆ OFP

⇒ ∠OPE = ∠OPF

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 figure).

Solution:

Let OM be perpendicular from O on line l. We know that the perpendicular from the centre of a circle to a chord, bisects the chord.

Since BC is a chord of the smaller circle and OM ⊥ BC.

BM = CM … (1)

Again, AD is a chord of the larger circle and OM ⊥ AD.

∴ AM = DM … (2)

Subtracting (1) from (2), we get

AM – BM = DM – CM ⇒ 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 each, what is the distance between Reshma and Mandip?

Solution:

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius 5 cm at points B, A and C respectively.

We know that if AB and AC are two equal chords of a circle, then the centre of the circle lies on the bisector of ∠BAC.

Here, AB = AC = 6 cm. So, the bisector of ∠BAC passes through the centre O i.e., OA is the bisector of ∠BAC.

Since the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio 6 : 6 = 1 : 1 i.e., M is the middle point of BC.

Now, M is the mid-point of BC ⇒ OM ± BC.

In right ∠d A ABM, we have

AB² = AM² + BM²

⇒ 36 = AM² + BM²

⇒ BM² = 36 – AM² … (1)

In the right A OBM, we have

OB² = OM² + BM²

⇒ 25 = (OA – AM)² + BM²

⇒ BM² = 25 – (OA – AM)²

⇒ BM² = 25 – (5 – AM)² … (2)

From (1) and (2), we get

36 – AM² = 25 – (5 – AM)²

⇒ 11 – AM² + (5 – AM)² = 0

⇒ 11 – AM² + 25 – 10AM + AM² = 0

⇒ 10AM = 36

⇒ AM = 3.6

Putting AM = 36 in (1), we get

BM² = 36 – (3.6)² = 36 – 12.96

⇒ BM = \(\sqrt{36-12.96}\) = \(\sqrt{23.04}\) = 4.8 cm

⇒ BC = 2BM = 2 x 4.8 = 9.6 cm

Hence, the distance between Reshma and Mandip = 9.6 cm.

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.

Solution:

Let ABC is an equilateral triangle of side 2x metres.

Clearly, BM = \(\frac { BC }{ 2 }\) = \(\frac { 2x }{ 2 }\) = x metres.

In right ∠d ∆ ABM,

AM² = AB² – BM²

= (2x)² – x² = 4x² – x² = 3x²

⇒ AM = \(\sqrt{3} x\)

Now, OM = AM – OA = (\(\sqrt{3} x\) – 20) m

In right ∠d ∆ OEM, we have

OB² = BM² + OM²

⇒ 20² = x² + (\(\sqrt{3} x\) – 20)²

⇒ 400 = x² + 3x² – 40\(\sqrt{3} x\) + 400

⇒ 4x² – 40\(\sqrt{3} x\) =0

⇒ 4x(x – 10\(\sqrt{3}\)) = 0

Since x ≠ 0, ∴ x – 10\(\sqrt{3}\) = 0 ⇒ x = 10\(\sqrt{3}\)

Now, BC = 2BM = 2x = 20\(\sqrt{3}\)

Hence, the length of each string = 20\(\sqrt{3}\) m.