Exercise 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.

Ans.

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

∴ AC = CB

It is given that, OO' = 4 cm

Let OC be x. Therefore, O'C will be 4 − x.

In ΔOAC,

OA² = AC² + OC²

⇒ 5² = AC² + x²

⇒ 25 − x² = AC² ... (1)

In ΔO'AC,

O'A² = AC² + O'C²

⇒ 3² = AC² + (4 − x)²

⇒ 9 = AC² + 16 + x² − 8x

⇒ AC² = − x² − 7 + 8x ... (2)

From equations (1) and (2), we obtain

25 − x² = − x² − 7 + 8x

8x = 32

x = 4

Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC² = 25 − x² = 25 − 4² = 25 − 16 = 9

∴ AC = 3 m

Length of the common chord AB = 2 AC = (2 × 3) m = 6 m

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.

Ans.

Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.

In ΔOVT and ΔOUT,

OV = OU (Equal chords of a circle are equidistant from the centre)

∠OVT = ∠OUT (Each 90°)

OT = OT (Common)

∴ ΔOVT ≅ ΔOUT (RHS congruence rule)

∴ VT = UT (By CPCT) ... (1)

It is given that,

PQ = RS ... (2)

⇒ 

⇒ PV = RU ... (3)

On adding equations (1) and (3), we obtain

PV + VT = RU + UT

⇒ PT = RT ... (4)

On subtracting equation (4) from equation (2), we obtain

PQ − PT = RS − RT

⇒ QT = ST ... (5)

Equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.

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.

Ans.

Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.

In ΔOVT and ΔOUT,

OV = OU (Equal chords of a circle are equidistant from the centre)

∠OVT = ∠OUT (Each 90°)

OT = OT (Common)

∴ ΔOVT ≅ ΔOUT (RHS congruence rule)

∴ ∠OTV = ∠OTU (By CPCT)

Therefore, it is proved that the line joining the point of intersection 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 figure 10.25).

Ans.

Let us draw a perpendicular OM on line AD.

It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.

We know that perpendicular drawn from the centre of the circle bisects the chord.

∴ BM = MC ... (1)

And, AM = MD ... (2)

On subtracting equation (2) from (1), we obtain

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 each, what is the distance between Reshma and Mandip?

Ans.

Draw perpendiculars OA and OB on RS and SM respectively.

OR = OS = OM = 5 m. (Radii of the circle)

In ΔOAR,

OA² + AR² = OR²

OA² + (3 m)² = (5 m)²

OA² = (25 − 9) m² = 16 m²

OA = 4 m

ORSM will be a kite (OR = OM and RS = SM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

∴∠RCS will be of 90° and RC = CM

Area of ΔORS =

Therefore, 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 each other. Find the length of the string of each phone.

Ans.

It is given that AS = SD = DA

Therefore, ΔASD is an equilateral triangle.

OA (radius) = 20 m

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write

∴ AB = OA + OB = (20 + 10) m = 30 m

In ΔABD,

AD² = AB² + BD²

AD² = (30)² + 

Therefore, the length of the string of each phone will be m.

Exercise 10.3

Question 1:

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Ans.

Consider the following pair of circles.

The above circles do not intersect each other at any point. Therefore, they do not have any point in common.

The above circles touch each other only at one point Y. Therefore, there is 1 point in common.

The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.

These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.

Question 2:

Suppose you are given a circle. Give a construction to find its centre.

Ans.

The below given steps will be followed to find the centre of the given circle.

Step1. Take the given circle.

Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords.

Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.

Question 3:

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Ans.

Consider two circles centered at point O and O’, intersecting each other at point A and B respectively.

Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O.

Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will also pass through O’.

Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.

Exercise 10.2

Question 1:

Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans.

A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius.

Consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths.

In ΔAOB and ΔCO'D,

AB = CD (Chords of same length)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

∴ ΔAOB ≅ ΔCO'D (SSS congruence rule)

⇒ ∠AOB = ∠CO'D (By CPCT)

Hence, equal chords of congruent circles subtend equal angles at their centres.

Question 2:

Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Ans.

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

In ΔAOB and ΔCO'D,

∠AOB = ∠CO'D (Given)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

∴ ΔAOB ≅ ΔCO'D (SSS congruence rule)

⇒ AB = CD (By CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Exercise 4.4

Question 1:

Give the geometric representation of y = 3 as an equation

(I) in one variable

(II) in two variables

Ans.

n one variable, y = 3 represents a point as shown in following figure.

In two variables, y = 3 represents a straight line passing through point (0, 3) and parallel to x-axis. It is a collection of all points of the plane, having their y-coordinate as 3.

Question 2:

Give the geometric representations of 2x + 9 = 0 as an equation

(1) in one variable

(2) in two variables

Ans.

(1) In one variable, 2x + 9 = 0 represents a point as shown in the following figure.

(2) In two variables, 2x + 9 = 0 represents a straight line passing through point 
(−4.5, 0) and parallel to y-axis. It is a collection of all points of the plane, having their x-coordinate as 4.5.

Exercise 4.3

Question 1:

Draw the graph of each of the following linear equations in two variables:

(i) (ii) (iii) y = 3x (iv) 3 = 2x + y

Ans.

(i)

It can be observed that x = 0, y = 4 and x = 4, y = 0 are solutions of the above equation. Therefore, the solution table is as follows.

x	0	4
y	4	0

The graph of this equation is constructed as follows.

(ii)

It can be observed that x = 4, y = 2 and x = 2, y = 0 are solutions of the above equation. Therefore, the solution table is as follows.

x	4	2
y	2	0

The graph of the above equation is constructed as follows.

(iii) y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.

x	− 1	1
y	− 3	3

The graph of the above equation is constructed as follows.

(iv) 3 = 2x + y

It can be observed that x = 0, y = 3 and x = 1, y = 1 are solutions of the above equation. Therefore, the solution table is as follows.

x	0	1
y	3	1

The graph of this equation is constructed as follows.

Question 2:

Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Ans.

It can be observed that point (2, 14) satisfies the equation 7x − y = 0 and

x − y + 12 = 0.

Therefore, 7x − y = 0 and x − y + 12 = 0 are two lines passing through point (2, 14).

As it is known that through one point, infinite number of lines can pass through, therefore, there are infinite lines of such type passing through the given point.

Question 3:

If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Ans.

Putting x = 3 and y = 4 in the given equation,

3y = ax + 7

3 (4) = a (3) + 7

5 = 3a

Question 4:

The taxi fare in a city is as follows: For the first kilometre, the fares is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Ans.

Total distance covered = x km

Fare for 1^st kilometre = Rs 8

Fare for the rest of the distance = Rs (x − 1) 5

Total fare = Rs [8 + (x − 1) 5]

y = 8 + 5x − 5

y = 5x + 3

5x − y + 3 = 0

It can be observed that point (0, 3) and satisfies the above equation. Therefore, these are the solutions of this equation.

x	0
y	3	0

The graph of this equation is constructed as follows.

Here, it can be seen that variable x and y are representing the distance covered and the fare paid for that distance respectively and these quantities may not be negative. Hence, only those values of x and y which are lying in the 1^st quadrant will be considered.

Question 5:

From the choices given below, choose the equation whose graphs are given in the given figures.

	For the first figure		For the second figure
(i)	y = x	(i)	y = x +2
(ii)	x + y = 0	(ii)	y = x − 2
(iii)	y = 2x	(iii)	y = − x + 2
(iv)	2 + 3y = 7x	(iv)	x + 2y = 6

Ans.

Points on the given line are (−1, 1), (0, 0), and (1, −1).

It can be observed that the coordinates of the points of the graph satisfy the equation x + y = 0. Therefore, x + y = 0 is the equation corresponding to the graph as shown in the first figure.

Hence, (ii) is the correct answer.

Points on the given line are (−1, 3), (0, 2), and (2, 0). It can be observed that the coordinates of the points of the graph satisfy the equation y = − x + 2.

Therefore, y = − x + 2 is the equation corresponding to the graph shown in the second figure.

Hence, (iii) is the correct answer

Question 6:

If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is

(i) 2 units (ii) 0 units

Ans.

Let the distance travelled and the work done by the body be x and y respectively.

Work done ∝ distance travelled

y ∝ x

y = kx

Where, k is a constant

If constant force is 5 units, then work done y = 5x

It can be observed that point (1, 5) and (−1, −5) satisfy the above equation. Therefore, these are the solutions of this equation. The graph of this equation is constructed as follows.

(i)From the graphs, it can be observed that the value of y corresponding to x = 2 is 10. This implies that the work done by the body is 10 units when the distance travelled by it is 2 units.

(ii) From the graphs, it can be observed that the value of y corresponding to x = 0 is 0. This implies that the work done by the body is 0 units when the distance travelled by it is 0 unit.

Question 7:

Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Ans.

Let the amount that Yamini and Fatima contributed be x and y respectively towards the Prime Minister’s Relief fund.

Amount contributed by Yamini + Amount contributed by Fatima = 100

x + y = 100

It can be observed that (100, 0) and (0, 100) satisfy the above equation. Therefore, these are the solutions of the above equation. The graph is constructed as follows.

Here, it can be seen that variable x and y are representing the amount contributed by Yamini and Fatima respectively and these quantities cannot be negative. Hence, only those values of x and y which are lying in the 1^st quadrant will be considered.

Question 8:

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Ans.

(i)

It can be observed that points (0, 32) and (−40, −40) satisfy the given equation. Therefore, these points are the solutions of this equation.

The graph of the above equation is constructed as follows.

(ii) Temperature = 30°C

Therefore, the temperature in Fahrenheit is 86°F.

(iii) Temperature = 95°F

Therefore, the temperature in Celsius is 35°C.

(iv)

If C = 0°C, then

Therefore, if C = 0°C, then F = 32°F

If F = 0°F, then

Therefore, if F = 0°F, then C = −17.8°C

(v)

Here, F = C

Yes, there is a temperature, −40°, which is numerically the same in both Fahrenheit and Celsius.