PSEB 9th Class Maths Solutions Chapter 10 Circles Ex 10.4

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths 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.
Answer:

The circle with centre O and radius 5 cm intersects the circle with centre P and radius 3 cm at points A and B.
Hence, AB is their common chord.
Then, OP = 4 cm (Given),
OA = 5 cm and PA = 3 cm.
In ∆ OAP, OA² = 5² = 25 and
OP² + AP² = 4² + 3² = 16 + 9 = 25
Thus, in ∆ OAP, OA² = OP² + AP²
∴ ∆ OAP is a right triangle in which ∠OPA is a right angle and OA is the hypotenuse.
Thus, in the circle with centre O, OP is perpendicular from centre O to chord AB.
∴ OP bisects AB.
AB = 2PA = 2 × 3 = 6 cm
Thus, the length of the common chord is 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:

In the circle with centre O, equal chords AB and CD intersect at E.
Draw OM ⊥ AB and ON ⊥ CD.
∴ AM = BM = \(\frac{1}{2}\)AB and CN = DN = \(\frac{1}{2}\)CD.
But, AB = CD
∴AM = BM = CN = DN …………….. (1)
Chords AB and CD, being equal, are equidistant from the centre.
∴ OM = ON
In ∆ OME and ∆ ONE,
∠OME = ∠ONE (Right angles)
OE = OE (Common)
OM = ON
By RHS rule, ∆ OME ≅ ∆ ONE
∴ME = EN (CPCT) ……………… (2)
From (1) and (2),
AM + ME = CN + NE
∴ AE = CE
Similarly, BM – ME = DN – NE
∴ BE = DE
Thus, if two equal chords of a circle intersect within the circle, the segments of one chord are equal to corresponding segments of the other chord.

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:
As the data of example 2 and example 3 are same, we use the proof of example 2 up to the required stage and do not repeat it here.
In example 2, we proved that,
∆ OME ≅ ∆ ONE ,
∴ ∠ OEM = ∠ OEN
∴ ∠ OEA = ∠ OEC
Thus, the line joining the point of intersection of two equal chords of a circle to the centre makes equal angles with the chords.

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 the given figure).

Answer:

From centre O, draw perpendicular OM to line AD.
In the outer circle, OM is the perpendicular drawn from centre O to chord AD.
Hence, M is the midpoint of AD.
∴ MA = MD …………… (1)
In the inner circle, OM is the perpendicular drawn from centre O to chord BC.
Hence, M is the midpoint of BC.
∴ MB = MC ………….. (2)
Subtracting (2) from (1),
MA – MB = 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 each, what is the distance between Reshma and Mandip?
Answer:

Here, OR = OM = OS = 5 m (Radius of the circle) and RS = SM = 6 m.
In quadrilateral ORSM, OR = OM = 5 m and RS = SM = 6 m.
∴ Quadrilateral ORSM is a kite.
∴ It diagonal OS bisects the diagonal RM at right angles.
∴ ∠RKO = 90° ………………. (1)
OK is perpendicular from centre O to chord RM.
Hence, K is the midpoint of RM.
∴ RM = 2RK ………………… (2)
From centre O, draw perpendicular OL to chord RS.
∴ RL = \(\frac{1}{2}\)RS = \(\frac{1}{2}\) × 6 = 3 m
In ∆ RLO, ∠ L = 90°
∴ RO² = OL² + RL²
∴ 5² = OL² + 3²
∴ 25 = OL² + 9
∴ OL² = 16
∴ OL = 4 m
Now, area of ∆ ROS = \(\frac{1}{2}\) × RS × OL
= \(\frac{1}{2}\) × OS × RK [by (1)]
∴RS × OL = OS × RK
∴ 6 × 4 = 5 × RK
∴ 24 = 5 × RK
∴ RK = \(\frac{24}{5}\) = 4.8 m
Then, RM = 2RK [by (2)]
∴ RM = 2 × 4.8
∴ RM = 9.6 m
Thus, the 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 with each other. Find the length of the string of each phone.
Answer:

Here, the circle with centre O represents the park and the points A, S and D represent the positions of Ankur, Syed and David respectively. Since Ankur, Syed and David are sitting at equal distances from the others, ∆ ASD is an equilateral triangle.

Then, drawing the perpendicular bisector of SD from its midpoint M, it will pass through O as well as A.
Suppose, SM = x m
∴ SD = 2SM = 2xm
Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) (side)²
∴ Area of equilateral ∆ ASD = \(\frac{\sqrt{3}}{4}\) × (2x)²
∴ Area of equilateral ∆ ASD = √3x² …………. (1)
In ∆ OMS, ∠M = 90°
∴ OM² = OS² – SM² = (20)² – (x)² = 400 – x²
∴ OM = \(\sqrt{400-x^{2}}\)
Now, area of ∆ OSD = \(\frac{1}{2}\) × SD × OM
∴ Area of ∆ OSD = \(\frac{1}{2}\) × 2x × \(\sqrt{400-x^{2}}\)
∴ Area of ∆ OSD = x\(\sqrt{400-x^{2}}\) …………….. (2)
Here, ∆ OAS, ∆ OSD and ∆ ODA are congruent triangles.
Area of ∆ ASD = Area of ∆ OAS + Area of ∆ OSD + Area of ∆ ODA
∴ Area of ∆ ASD = 3 × Area of ∆ OSD
∴ √3 ∙ x<sup2 = 3 × x\(\sqrt{400-x^{2}}\)
∴x = √3 ∙ \(\sqrt{400-x^{2}}\)
∴ x² = 3(400 – x²2)
∴ x²= 1200 – 3x²
∴ 4x² = 1200
∴x² = 300
∴x= 10 √3
SD = 2x = 2 × 10 √3 = 20 √3 m
Thus, the length of the string of each phone is 20 √3m.