PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

PSEB 9th Class 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.
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 = 1/2AB and CN = DN = 1/2CD.
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.

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

Follow on Facebook page – Click Here

Google News join in – Click Here

Read More Asia News – Click Here

Read More Sports News – Click Here

Read More Crypto News – Click Here