How to answer the following question?rnAmir starts walking from his house to office. Instead of going to the office directly, he goes to the bank first, from there to his daughter's school and then reaches the office. What

How to find whether the relation is Direct or Indirect? Can anyone help me to solve it?

a) The time taken by a train to cover a fixed distance and the speed of the train.

b) The distance travelled by CN

Distance formula

## Find the co-ordinate of the point which lies on y-axis at a distance of 4 units in negative direction of y-axis. (A) (-4,0)              (B) (4,0) (c) (0,-4)              (D) (0,4)

The point on y- axis  has x-axis has x- coordinate 0.

Since it lies at a distance of 4 units in the negative direction of y-axis.

$\therefore$The point  is (0, -4)

Parallel, 6

## Give the geometrical representation of the equation 3x+15 = 0  as an equation  (i) One variable  (ii) In two variable

3x + 15 = 0

X = -5

(i) Equay=tion in one variable (Number line ): A point P at a distance of 5 units to left of O on the number line.

(ii) In two variables (Cartesian plane): A line AB parallelto y-axis at a distance of 5 units to the left of y-axis.

## The auto rikshaw fare in a city is charged RS . 10 for first kilometer and @Rs.4 per kilometer for subsequent distance covered. Write the linear equation to express the above statment. Draw the graph of the linear  equation.

Total distance covered =  x km.

Total fare = y km.

Fare for the first kilometer

Subsequent distance  = (x-1) km

Fare for the subsequent distance = Rs. 4(x-1)

According to the question,

y = 10+4 (x-1)

$\Rightarrow$y = 10+4x -4

$\Rightarrow y =4x +6$

Table of solutions

 X 0 1 -1 Y 6 10 2

## Ramesh is driving his car with a uniform speed of  90 km per hour. Draw the time- distance graph on the graph paper. From the graph, find the distance travelled by him in: (i)  $\frac{1}{2}$ hour                            (ii) $2$ hours

Graph of equation y = 90x,

where x be time and y be distance from the graph

(i) Distance travelled in $\frac{1}{2}$ hour = 45 km

(ii) Distance travelled in 2 hours  = 180 km.

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

The fare of first km is Rs. 8.00

Let the total distance to be covered is x km.

Fare for (x-1) kilometer at the rate of Rs.5 per km is 5 (x-1)

$\therefore$ Total fare   y = 5 (x-1)+8

y =5x-5+8

y = 5x+3

 X 0 -1 -2 Y 3 -2 -7

## Solve the equation 3x+4 = 5x+8  and represent the solution on (i) the number line (ii) the cartesian plane. What do you get as the representation of the solution on the cartesian plane ? In cartesian planes, how many solutions this equation has?

$3x+4 = 5x+8$

$\Rightarrow 2x = -4$

$\Rightarrow x = -2$

(i) On the number line the point P (-2,0)

represent the solution

(ii) On the Cartesian plane x = -2 is a line parallel to the y-axis at a distance of 2 units to left of it.

It has infinite solutions.

## AB and CD are two parallel chords on the same side of the circle. AB = 6 cm, CD= 8 cm. The small chords is at a distance of 4 cm from the centre. At what distance from the centre is the other chord

In $\Delta OMB,$  OM = 4 cm, MB = 3 cm

OB= OM2+MB2

= 16+9

= 25

$OB =\sqrt{25}$

$= 5 cm$

In  $\Delta OND$

$ON^{2}= OD^{2}-DN^{2}$

$DN = 4 cm$

$OD = OB = 5 cm$    (Radii)

$ON^{2}=5^{2}-4^{2}$

= 25-16

= 9

$ON = 3 cm.$

$\therefore$ The other chord is at a distance of 3 cm from the centre.

## What you mean by abscissa of a point

The distance of a point from the y-axis is called its x- coordinates, or abscissa.

## Point p is on a x axis and is at a distance of 4 units  from y- axis to its left. Write the co-ordintates of the point p.

The p is on x-axis

y = 0

P is at a distance 4 units from y axis to its lft

In second quadrant, the coordinates of the point

p = (-4, 0)

## Find the perpendicular distance of A  (5,12) from the y-axis

The point on the y-axis is (0,12)

Distance between  (5,12) and (0,12)

$d = \sqrt{(0-5)^{2}+(12-12)^{2}}$

$=\sqrt{25+0} =5$ units

## Prove that the points  (2, -2), (-2,1) and (5,2) are the vertices of a right angled triangle. Also, find the area of this triangle.

Let the points be A  (2, -2), B (-2,1) and C(5,2)

Applying distance formula

$AB^{2}= (2+2)^{2}+(-2-1)^{2}$

$= 16+9$

$AB^{2}= 25\Rightarrow AB = 5$

Similarly  $BC^{2} = (-2-5)^{2}+(1-2)^{2}$

$=49+1 =50$

$\Rightarrow BC^{2}= 50 \Rightarrow BC = 5\sqrt{2}$

Also,

$AC^{2}= (2-5)^{2}+(-2-2)^{2}$

$= 9+16$

$= 25$

$\Rightarrow AC^{2} = 25$ and  $\Rightarrow AC = 5$

Clearly  $AB^{2}+AC^{2} =BC^{2}$

25+25 = 50

Hence the triangle is right angled

Area of $\Delta ABC = \frac{1}{2} \times$ Base  x  height

$= \frac{1}{2}\times 5\times 5 =\frac{25}{2}$ sq units

## A man steadily goes 10 m due east and then 24 m due north. (i) Find the distance from the starting point  (ii) Which mathematical concept is used in this problem

(i) Let the initial position of the man be at O and his final position be B. Since the man goes to 10 m due east  and then 24 m due north. Therefore,  $\triangle AOB$ is a right angled triangle right angled triangle  at A  such that OA = 10m and AB  = 24m.

By Pythagoras theorem

$\Rightarrow OB^{2} = OA^{2}+AB^{2}$

$\Rightarrow OB^{2} = (10)^{2}+(24)^{2}$

$= 100+576=676$

$\Rightarrow OB = \sqrt{676} = 26 m$

Hence the man is at a distance of  26 m from the starting point.

(ii) Right angled triangle

## Point A(-1, y) and  B (5,7) lie on a  circle with centre  O (2,-3y). Find the values of y. Hence find the radius of the circle.

Since A (-1,y) and  B(5,7) lie on a circle with centre O (2,-3y)

Distance between OA and OB are equal

OA =OB

$\sqrt{(-1-2)^{2}+(y+3y)^{2}} = \sqrt{(5-2)^{2}+(7+3y)^{2}}$

$\Rightarrow \sqrt{9+(4y)^{2}} = \sqrt{3^{2}(7+3y)^{2}}$

Squaring on both sides, we get

$9+16y^{2}= 9y^{2}+42y+58$

$\Rightarrow y^{2}-6y-7=0$

$\Rightarrow (y+1)(y-7)=0$

$\therefore y=-1,7$

Radius of the circle, when y = -1

(x-2)2+(y-3)2= r2

$\because (5,7)$ lie on the circle

$\therefore (5-2)^{2}+(7-3)^{2}= r^{2}$

$\Rightarrow 9+16 = r^{2}$

$\Rightarrow r = 5 cm .$

## Draw tangents to a circle of radius 6 cm  from a point P at a distance of  10 cm from its centre.

Given: A circle with centre O and a point P outside it

Construction  :  We have to  construct the two tangents from P to the circle

Steps:

1. Draw a line segment PO = 10 cm

2.  From the point O draw  a circle of radius = 6  cm

3.  Draw a perpendicular bisector  of PO. Let M  be the mid- point of PO

4. Taking M as centre and OM  as radius draw a circle .

5. Let this circle intersects  the given circle at the points Q and R.

6. Join PQ and PR.

Then PQ and PR are the required two tangents

## As observed from the top of a 100 m high lighthouses from the sea-level, the angles of depression of two sips are  $30^{\circ}$  and  $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouses, find the distance between two ships. $(Use \sqrt{3} = 1.732)$

Let ships are at distance x from each other

In $\triangle APO$

$tan 45^{\circ} = \frac{100}{y} = 1 \, \, \, \, \, \, \, \, \, \therefore y = 100m$ ............(i)

$In \triangle POB$

$tan \, \, 30^{\circ} = \frac{OP}{OB} = \frac{100}{x+y} = \frac{1}{\sqrt{3}}$

$\sqrt{3} = \frac{x+y}{100}$

$x+y = 100 \sqrt{3}$          ...........................(ii)

$x= 100 \sqrt{3} - y = 100\sqrt{3}-100 (\sqrt{3}-1)$

$\therefore x = 100 (1.732 -1)$

$= 100 \times 0.732$

$= 73.2 m$

$\therefore$ ships are 73.2 meters apart.

## The angle of depression of a car parked on the road from the top of a  150m high tower is  300. Find the distance of the car from the tower.

Let the distance of the car from the above tower be d.

$\angle BAX = 30^{0}= \angle ABC$

(Alternate angles )

tan 300$\frac{150}{d}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{d}$

$\therefore d = 150 \times \sqrt{3} m$

## A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle  300 with it. The distance between the foot of the tree to the point where the top touches the ground is  8m. Find the height of the tree.

In right $\Delta BCD$

$Cos 30^{0} = \frac{CD}{BD}$

$\Rightarrow \frac{\sqrt{3} }{2} = \frac{8}{BD}$

$\Rightarrow BD = \frac{16}{\sqrt{3}}$

and  $tan 30^{0}= \frac{BC}{CD}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{8}$

$\therefore$ Height of tree = BC+BD

=  $\frac{ 8+16}{\sqrt{3}}$

$= \frac{24}{3}\sqrt{3}$

$= 8\sqrt{3} m.$

## From the top of a building100 m high, the angle of depression of two objects are on the same side observed to be 450 and 600. Find the distance between the objects.

Let A be a point and B, C be two objects.

In $\triangle ACO,$               $\frac{AO}{CO} = tan 45^{0}$

$\Rightarrow \frac{100}{CO} = 1$

$\Rightarrow CO = 100 m$

Also in $\triangle ABO ,$      $\frac{AO}{OB} = tan 60^{0}$

$\Rightarrow \frac{ 100}{OB} = \sqrt{3}$

$\Rightarrow OB = \frac{ 100}{ \sqrt{3}}$

$\therefore BC = CO - OB = 100 - \frac{100}{\sqrt{3}}$

$= 100\left ( 1-\frac{1}{\sqrt{3}} \right )$  m.

## The horizontal distance between two poles is  15m. The angle of depression of the top of the first  pole as seen from the top of the second pole is  300. If the height of the first pole is 24m, find the height of the second pole. [use $\sqrt{3} = 1. 732$]

In  $\Delta PTR$

$tan 30^{\circ} = \frac{PT}{TR}$

$\frac{1}{\sqrt{3}} = \frac{h}{15}$

$h = \frac{15}{\sqrt{3}}$

$= 5\sqrt{3}$

$= 5$ x $1.732= 8.66$

PQ = PT +TQ

= 8.66 +24

= 32.66 m

Height of the pole is  32.66 m.

## A vertical tower stands on a horizontal plane and is surmounted by a  flagstaff of height 5m. From a  point on the ground the angles of elevation of top and bottom of the flagstaff are $60^{\circ}$ and  $30^{\circ}$ respectively. Find the height of the tower and the distance of the point from the tower (take $\sqrt{3} = 1.732$)

(i) $\frac{x}{y} = tan30^{0} = \frac{1}{\sqrt{3}}$

$\rightarrow {\sqrt{3}}$

(ii) $\frac{x+5}{\sqrt{3}x} = \sqrt{3}$

Height of tower  = 2.5 m

Distance of P  from tower   = (2.5 x 1.732) or  4.33 m

## A man on the top of vertical tower observes a car moving at a uniform speed towards him. If it takes 12 min for the angle of depression to change from $30^{\circ}$ to $45^{\circ}$, how soon after this, the car will reach the tower ?

Let AB be the tower of height h.

$\angle AQB = 45 ^{0}$

Now in  $\triangle ABQ$

$tan 45^{0} = \frac{AB}{BQ}$

$1 = \frac{h}{BQ }$

$\Rightarrow 1 = BQ = h$

$tan 30^{0} = \frac{AB}{PB}$

$\Rightarrow x+ h = h \sqrt{3}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x+h}$

ie., $x = h (\sqrt{3}-1)$

Thus, Speed  = $\frac{h(\sqrt{3}-1)}{12}$

$Spped = \frac{Distance }{Time }$

Time for remaining distance,

$= h \frac{(\sqrt{3}-1)}{12}$

$= \frac{12(\sqrt{3}+1)}{3-1}$

$= \frac{12}{2} (\sqrt{3}-1)$

$= 6 (\sqrt{3}+1)$

Time  = 16.39 minutes

## Fig. depicts a racing track whose left and right ends are semi-circular. The distance between the two inner parallel line segments is  60 m and they are each 106 m long. If the track is  10 m wide everywhere find the area of the track.

We have,  OB = O'C = 30 m

AB = CD = 10 m

OA = O'D  = (30+10) m

= 40 m

ar  (track) = ar (rect. ABCD) +ar (rect. EFGH)+2 (ar semi-circle of radius 40) -2

ar  (semi circle of radius 30.)

$= \left [ (10\times 106)+ (10\times \times 106)\right ]$

$=+2 \left [ \frac{1}{2}\times \frac{22}{7}\times (40)^{2}-\frac{1}{2}\times \frac{22}{7} \times (30)^{2}\right ]$

$= (2120+2200)m^{2}$

$=4320 m^{2}.$

## Two circular beads of different size are joined together such that the distance between their centres is 14 cm. The sum of their areas is 130 cm2. Find the radius of each bead.

Let the radii of the circles are  r1 cm and r2 cm

$\therefore r_{1}+r_{2} = 14$

And, sum of their areas  $\pi r_{1}^{2}+\pi r_{2}^{2}$

$\Rightarrow 130 = \pi (r_{1}^{2}+r_{2}^{2})$

$\therefore r_{1}^{2}+r_{2}^{2} =130$

$(r_{1}+r_{2})^{2}=r_{1}^{2}+r^{2}_{2}+2r_{1}r_{2}$

$(14)^{2}=130+2_{1}r_{2}$

$\Rightarrow2_{1}r_{2}=196-130$

$=66$

$(r_{1}+r_{2})^{2}=r_{1}^{2}+r^{2}_{2}+2r_{1}r_{2}$

$130-66=64$

$\Rightarrow r_{1}-r_{2} = 8$

From (i) and (ii)

$2 r_{1} = 22$

$r_{1} = 11 cm$

$r_{2} = 14-11$

$= 3 cm$

## In the figure, AB and CD are two chords of a circle with centre O at a distance of 6 cm and 8 cm from O. If the radius of the circle is 10 cm find the length of chords.

Join OA and OC.

Since, Perpendicular from the centre bisects the chord,

$\therefore AP = BP = \frac{1}{2}AB$

$and \, \, CQ = QD = \frac{1}{2} CD$

In  $\Delta OAP,$ By Pythagoras theorem,

$AP^{2}=OA^{2}-OP^{2}$

$= 10^{2}-6 ^{2} = 64$

$\therefore$ AP = 8 cm $\Rightarrow$AB  = 16 cm

In  $\Delta OQC$

$CQ^{2}=OC^{2}-OQ^{2}$

= $10^{2}-8^{2}$

$CQ = 6 cm , CD = 12 cm$

## What do you mean  by abscissa of a point ?

The distance of a point from the y- axis is called its x -co-ordinate or abcissa.

## Point P is on the x-axis and is at a  distance 4 units from y-axis  to its left. write the co-ordinates of the point P.

The P is on x-axis

y= 0

P is at distance of 4 units from y- axis to its left

In second quadrant the coordinates of the point

p = (-4,0)

7

## A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius of the circle.

Given  AB = 10 cm

ON  = 12 cm

Also,    $ON \perp AB$

and  AN = BN

($\therefore$ Perpendicular drawn from the centre of the circle bisects the chord)

In $\Delta ONB ,$

$OB^{2}= ON^{2}+NB^{2}$     ( By pythagoras theorem)

$OB^{2}= 12^{2}+5^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\because BN = 5 cm)$

= 144+25

= 169

$OB = 13 cm$

Hence the radious of the circle is 13 cm.

## From a point P, which is at a distance  of 13 cm from the centre O of a circle radius 5 cm, the pair of tangents PQ and PR are drawn to the circle, find the area of the quadrilateral PQOR (in cm2)

Length of a tangent,

$PQ = \sqrt{OP^{2} - OQ^{2}}$

$= \sqrt{(13)^{2}- (5)^{2}}$

$= \sqrt{ 169-25}$

$= \sqrt{ 144}= 12 cm$

= Base  x-height

= 12 x  5 = 60 cm2

## To divide a line segment AB in the ratio 5 : 7  first AX is drawn, so that $\angle BAX$ is an acute angle and then at equal distance, points are marked on the ray AX, find the minimum number of these points

Minimum number of points marked on

AX= 5+7 = 12

## Draw a circle of radius 3 cm. From a point P, 7 cm away from its centre its centre draw to tangents to the circle. Measures the length of each tangent.

Given: A circle of radius 3 cm with centre O and a point P at a distance of 7 cm from O.

Construction: We have to construct the two tangents from P to the circle.

Step of construction:

1. Draw a line segment PO = 7 cm

2. From the point O, draw a circle of radius =3 cm.

3. Draw a perpendicular bisector of PO. Let M be the midpoint be the midpoint of PO.

4. Taking M as the centre and OM as the radius a circle.

5. Let this circle intersects the given circle at the point Q and R.

6. Join PQ and PR

7. Thus PQ  and PR are the required two tangents length of the tangent  PQ =PR = $\sqrt{PQ^{2}-OQ^{2}}$

$PQ = PR = \sqrt{(7)^{2}-(3)^{2}}$

$= \sqrt{49-9} = \sqrt{40}$ = 6.3 cm.

## The diameter of a wheel is 1.26 m. What the distance covered in 500 revolutions ?

Distance covered in 1 revolutions = Circumference of wheel

$= \pi d$

$= \pi \times 1.26 m$

Distance covered in 500 revolutions

$=500 \times \pi \times 1.26$

$=500 \times \frac{22}{7}\times 1.26$

$= 1980 m.$

## Find the distance of points  C(-3, -2) and D (5,2)  from x-axis and y-axis

C (-3, -2) distance from x-axis = 2

distance from y-axis = 3

D (5,2)      Distance from x-axis = 2

Distance  from y axis = 5

## Find distance of points C (-3,-2) and D (5,2) from x-axis and y-axis

C (-3,-2)  distance from x-axis  = 2

distance from y-axis  = 3

D (5,2)     distance from x-axis  = 2

distance from x-axis  = 5

## A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates what will be the coordinates of a point if it lies on y-axis at a distance of 9 units from x axis

(i) Since the point lies on x-axis at a distance of 9 units from y-axis. Hence its coordinates are (9,0)

(ii) According to the question, the required coordinates are (0,-9)

## Find the distance of points C(-3,-2) and D (5,2) from the x-axis and y-axis.

C(-3,-2) distance from x-axis = 2

distance from y-axis = 3

D (5,2), distance from x-axis = 2

distance from y-axis= 5

(A) P (0,5)

(B) Q (0,-3)

(C) R (5,0)

(D) S (-2,0)

## A point lies on the x-axis at a distance of 9 units from the y-axis. What are its coordinates? what will be the coordinates of a point if it lies on the y-axis at a distance of -9 units from x-axis?

(i) Since the point lies on the x-axis at a distance of 9 units from the y-axis. Hence its coordinates are (9,0).

(ii) According to the question, the required coordinates are (0,-9)

## Plot the points (5,-3), (-6,0), (-2,-3), and (-4,3) on the graph.

For plotting a point A (5,-3) we will take a distance o f-3 units in the negative direction of y-axis and a distance of 5 unit in the positive direction of x-axis, which is shown in the figure given below. similarly, we plot all the points ie., B(-6,0), C(-2,-3) and D(-4,3)

pic

## Plot the points (x,y) given in the following table on the Cartesian plane, choosing suitable units of distance on the axes. x     -1.25        0        3         -1.75         4       -2.25 y         2         2.25    1.5          -2          -3          0

Scale 1 unit = 2 cm

## Rahul walks 12 meters north from his house and turns west to walk 35 meters to reach his friend's house. While returning, he walks diagonally from his friend's house to reach back to his house. what distance he did cover while returning?

Applying pythagoras theorem here

$x^{2} = 12^{2} + 35^{2}$

$x^{2} = 1 44 + 1225$

$x^{2} = 1369$

$x = 37 m$

## Draw the graph for the following table of values, with suitable scales on the axes. Distance travelled by car  Time (in hours)           6 a.m          7 a.m       8 a.m            9 a.m  Distance (in Km)           40               80           120               160      (ii) What was the time when the car had covered a distance of 100 km since its start?(i) How much distance did the car cover during the period 7.30 am to 8 cm?

b. The graph for given data is as shown below:

Here, the x-axis represents the time and scale taken is 2 units = 1 hour, Whereas y-axis represents distance (in km)and scale has taken is 2 units = 40 km.

(i) The car covered the distance of 20 km during the period 7.30 am to 8 am.

(ii) The time was 7.30 am when the car had covered a distance of 100 km since it start.

## How many times will the wheel of a car rotate in a journey of 88 km if it is known that the diameter of the wheel is 56 cm  ( Take  $\pi = \frac{22}{7}$)

change the 88 km into cm.
which will 88 km = 8800000 cm
Diameter of given wheel = 56 cm
then, radius of the wheel = 28 cm
We know that,
circumference of the wheel = 2πr
2×22/7×28 =176 cm
then we just divide it out.
8800000/176.
= 50000
SO WHEEL HAVE TO BE ROTATED 50000 TIMES TO REACH THE DISTANCE OF 88 KM.

20

23

19

## ABCD is a parallelogram. AB = 8 cm, AD = 4cm, $\angle B = 120^{\circ}$ a. What is $\angle A$? b. What is the perpendicular distance from D to AB? c. What is the area of ABCD?

Given ABCD is a parallelogram,

AB = 8 cm , AD = 4 cm, $\angle B = 120^{\circ}$

a. $\angle B = \angle D = 120^{\circ}$

(Opposite angles equal)

$\angle A + \angle C = 360 - (120+120)$

$= 360 - 240 = 120$

$\therefore \angle A = \angle C \, \, \, \frac{120}{2} = 60$

$\angle A = 60^{\circ}$

b. Draw DE as perpendicular from D to AB.

From the figure DE = $2 \sqrt{3}$ cm. [ 1 : $\sqrt{3}$ : 2]

C. The area of ABCD  =  bh (formula)

$= 8 \times 2 \sqrt{3} = 16 \sqrt{3}$ cm2

## Draw a circle of radius 3 cm. Mark a point 7 cm away from its centre. Draw the tangents to the circle from this point?

Given radius  = 3 cm, Distance from the centre  = 7 cm

Construction.

Draw a circle 3 cm radius and O as its centre. from center mark OP = 7 cm. Draw perpendicular bisector of the line OP and meet M. Draw a circle with radius OM and cut the circle at Q and R. Joint PR and PQ. Hence PR and PQ are the tangents.

## A circle with centre (3,2) passes through the point (6,3) a. What is the radius of the circle? b. Check whether each of the points with coordinates (0,2) (3,6) (0,3) inside, outside or on the circle.

a. Given points (3,2) and (6,3)  (using the distance formula)

$r = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

$= \sqrt{(6-3)^{2} + (3-2)^{2}} = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$

b. (1) (3,2)  and (0,2) (Using distance formula)

$= \sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2}} = \sqrt{(0-3)^{2} + (2-2)^{2}} = \sqrt{9} = 3$

Here 3 is less than the radius, so the point is inside the circle

(2) (3,2) and (3,6)      (using the distance formula)

$\sqrt{(3-3)^{2}+(6-2)^{2}} = \sqrt{0^{2} + 4^{2} } = \sqrt{16} = 4$

Here 4 is greater than the radius, so the point be outside the circle.

(3) (3,2) and (0,2)      (Using the distance formula)

$\sqrt{(0-3)^{2} + (3-2)^{2}} = \sqrt{9+1} = \sqrt{10}$

Here $\sqrt{10}$ is equal to the radius, so the point be on the circle.

## The coordinates of the vertices of a triangle are A(1,1) B(5,5) C(2,5) a. Write the coordinates of the midpoint D of AB. b. What is the length of CD? c. What are the coordinates of the point dividing the line CD in the ratio 2:1?

Given, the vertices of the,$\triangle ABC$

A (1,1) B (5,5) C(2,5)

a. Mid-point of D of AB

$\left ( \frac{x_{1} + x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$  (Formula)

Points A (1,1) B (5,5)

ie., D  =  $\left ( \frac{1 + 5 }{2}, \frac{1+5}{2} \right ) = (3,3)$

b. Length of CD points C(2,5) D (3,3) [Using distance formula]

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} = \sqrt{(3-2)^{2}+(3-5)^{2}}$

$= \sqrt{1+4} = \sqrt{5}$

c. Given CD = 2:1            (ie., m:n)

$\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n} \right )$   (Formula)

$= \left ( \frac{2\times 3+1\times 2}{2+1}, \frac{2\times 3+1\times 5}{2+1} \right ) = \left ( \frac{6+2}{3}, \frac{6+5}{3} \right ) = \left ( \frac{8}{3}, \frac{11}{3} \right )$

## A boy saw the top of a building under construction at an elevation of 300. The completed building was 12 m higher and the boy saw its top at an elevation of 600 from the same spot. a. Draw a rough figure based on the given details? b. What is the height of the building? c. What is the distance between the building and the boy?

Consider AC = x, AD = x + 12.

In triangle ABD, $\frac{AD}{AB} = tan\, \, 60^{\circ}$.

$AB = \frac{AD}{tan\, \, 60^{\circ}} \Rightarrow AB = \frac{x + 12}{\sqrt{3}}$.....................(1)

In triangle ABC,

$\frac{AC}{AB} = tan \, \, 30^{\circ} \Rightarrow AB = \frac{AC}{tan \, \, 30^{\circ}}$ $\Rightarrow AB = \frac{x}{\frac{1}{\sqrt{3}}} = x\sqrt{3}$ .................(2)

Comparing equation (1) and (2)

ie., $\frac{x + 12}{\sqrt{3}} = x\sqrt{3} \Rightarrow 3x=x + 12, \Rightarrow 3x - x = 12, \Rightarrow 2x = 12, x = 6$

Hence, the height of the building = 6 + 12 = 18 m

c. Consider the equation (2)

ie., AB = $\sqrt{3}x$

$= \sqrt{3} \times 6 = 1.73\times 6 = 10.38 \, \, m$

Distance between the building and the boy = 10.38 m

## There is a mark on the outermost part of a wheel of radius 30 centimetres. Now the mark is close to the ground as shown in the figure. If the wheel rolls 31.4 centimetres on a straight line, then a. Find the angle by which the wheel rotates (use $\pi$ = 3.14 as an approximation) b. What will be the height of the mark from the ground?

Distance the circle rolls = 3.14 cm

Length of the arc of a circle = $\frac{\Theta }{360} \times 2\pi R$

$3.14 = \frac{\Theta }{360} \times 2 \times 3.14 \times 30$

$\Theta = \frac{31.4\times 360 }{2\times 3.14 \times 30}$

$= \frac{31.4\times 360 }{2\times 3.14 \times 3} = \frac{360}{2\times 3} = 60^{\circ}$

b. Consider $\triangle$ABC which is an equilateral triangle. also $\triangle$ BDC is 300, 600, 900 triangle

sin 300$\frac{BD}{BC} = \frac{BD}{30}$

$\frac{1}{2}= \frac{BD}{30}$

$BD = \frac{30}{2} = 15 cm$

The mark is 15 cm above the ground

## In the figure, the radius of the smaller circle is 3 centimetres, that of the bigger circle is 6 centimetres and the distance between the centres of the circles is 15 centimetres. PQ  is a tangent to both the circles. Find its length.

Here $EB\left \| AR$

$\angle ABC = 90^{\circ}$

$\angle DEC = 90^{\circ}$

AB = ER = 3 cm

PQ = AR

Using pythagoras theorem.

Length of tangents $=\sqrt{2^{2}-(r_{1}+r_{2})^{2}}$

$= \sqrt{ (15)^{2}-(6+3)^{2}}$

$= \sqrt{ 225-(9){2}}$

$= \sqrt{ 225-81}=\sqrt{144}=12 cm$

## In triangle ABC, AB = 5 cm, $\angle A = 80^{\circ}$ and $\angle B = 70^{\circ}$ . Calculate the radius of the circumcircle and length of the other two sides. (Necessary values can be taken from the following table)   Angle        sin       cos         tan 700          0.94     0.34       2.75 800          0.98     0.17       5.67 OR  Gopi and Gautham stand on opposite sides of a tower. The children and the tower are on a straight line also. Gopi sees the top of the tower at an angle of elevation of 360 and Gautham sees it at an angle elevation of 520. The distance between the children is 60 metres. a. Draw a rough figure according to the given information. b. Find the height of the tower. The height of children can be neglected. Necessary values can be taken the following table. Angle        Sin         Cos        tan 360          0.59       0.81      0.72 520          0.79       0.62      1.28

$tan \, \, \, 80^{\circ} = \frac{BC}{AB}$

$BC = tan \, \, 80^{\circ}\times AB$

$= tan \, \, 80^{\circ}\times 5 = 5.67 \times 5 = 28.35\, \, cm$

$tan \, \,70^{\circ} \, \, \, \frac{AC}{AB}$

$AC = 2.75 \times 5$

=13.75 cm

Cos 60= 2.5/R

R = $\frac{2.5}{1/2}= 5 cm$

OR

a.

b. $tan \, \, \, 36^{\circ} = \frac{AB}{BC}$

$0.72 = \frac{AB}{30}$

$AB = 0.72 \times 30$

Height of tower =21.6 cm

## Calculate the distance between each pair of points given below on the number line  a. 3,-7 b. -3,7   c. -3,-7

Distance between A and B =  $\dpi{120} \sqrt{(-3-3)^{2} + (7 + 7)^{2}}$

= $\dpi{120} \sqrt{(-6)^{2} + (14)^{2}}$

$\dpi{120} = \sqrt{36+196}$

$\dpi{120} = \sqrt{232}$

=15.23

Distance between B and C

$\dpi{120} = \sqrt{(-3+3)^{2} + (-7-7)^{2}}$

$\dpi{120} = \sqrt{(14)^{2}}$

$\dpi{120} = 14$

Distance between C and A

= $\dpi{120} \sqrt{(3--3)^{2}+ (-7+7)^{2}}$

= $\dpi{120} \sqrt{ (6)^{2}}$

= 6

## In the figure a line is drawn inside an isosceles triangle. Parallel to the base. What is its length? a. What is the length of such a parallel line 6 centimeters dwn from the to[ of the triangle? b. Prove that the length of such a line varies proportionally as its downward distance.

$\dpi{120} \frac{PQ}{BC} = \frac{AO} {AM}$

$\dpi{120} \frac{x}{6} = \frac{3} {9}$

$\dpi{120} x = 2 cm$

a.

$\dpi{120} \frac{PQ}{BC} = \frac{AO}{AM}$

$\dpi{120} \frac{x}{6}=\frac{6}{9}$

x = 4 cm

b. Downward distance is 1 cm

$\dpi{120} \frac{PQ}{1} \, \, \, \, \frac{x}{6} = \frac{1}{9}$

$\dpi{120} x = \frac{6}{9} = \frac{2}{3}$

Downward distance is 2 cm

$\dpi{120} = \frac{x}{6} = \frac{2}{9}$

$\dpi{120} x = \frac{2 \times 6}{9} = \frac{4}{3}$

Downward distance is 'h'

Length of the line = $\dpi{120} x = \frac{6}{4} \times h \, \, \, \, \, \, \, \, \, \, ie., x \propto h$

## In the semicircle shown, the top chord is parallel to the diameter. What is its length? a. What is the length of such a chord drawn 2 centimetres down from the top of the semicircle? b. Is the length of such a chord proportional to the distance from the top? write the reason

$\dpi{120} \frac{AB}{CD} = \frac{ER}{EG}$

$\dpi{120} \frac{x}{10} = \frac{1}{5}$

$\dpi{120} x= \frac{1\times 10}{5} = 2 cm$

a.

$\dpi{120} \frac{AB}{CD} = \frac{ER}{EG}$

$\dpi{120} \frac{x}{10} = \frac{2}{5}$

$\dpi{120} x = \frac{2\times 10^{2}}{5} = 4 cm$

b.

If the chord is 3 cm down from top of the semicircle.

$\dpi{120} \frac{x}{10}> \frac{3}{5}$

$\dpi{120} x = \frac{3\times 10^{2}}{5} = 6 \, \, cm$

If the chord is 'h' cm down from top of the semicircle.

$\dpi{120} x =\frac{3}{5}\, \, \, h$ ie., $\dpi{120} x \propto h$ and $\dpi{120} \frac{3}{5}$ is the constant of proportionality.

## a. What is the number which gives the midpoint of the points denoted by the numbers x and y on the number line? b. If the numbers x and y are thought of as points on a number line. What is the geometrical meaning of $\left | x-y \right |$ ? c. In each of the equations below. find the number x satisfying it: (i) $\left | x-y \right | = \left | x-3 \right |$   (ii) $\left | x-1 \right | = \left | x+3 \right |$  (iii) $\left | x+1 \right | = \left | x-3 \right |$

a. Midpoints denoted by $\dpi{120} \frac{x+y}{2}$

b. $\dpi{120} \left | x-y \right |$ indicate the distance between the points x and y

c. (i) $\dpi{120} \left | x-y \right | = (x-3)$

$\dpi{120} 2x = y + 3$

$\dpi{120} x = \frac{y + 3}{2}$

(ii) $\dpi{120} \left | x-1 \right | = \left | x+3 \right |$

2x = 3 + 1

$\dpi{120} x = \frac{4}{2} = 2$

(iii) $\dpi{120} \left | x+1 \right | = \left | x-3 \right |$

2x = 4

$\dpi{120} x = \frac{4}{2} = 2$

## All sides of a parallelogram are of length 10 cm and one of its diagonals is of length 16 cm. a. What is the length of the other diagonal. b. What is the area of the parallelogram? c. What is the distance between two of its parallel sides.

Since all sides of ABCD are equal its a Rhombus.

Hence consider $\dpi{120} \triangle AOB$

AB = 10 cm, AO = 8 cm

Then OB = $\dpi{120} \sqrt{AB^{2} -AO^{2}}$

$\dpi{120} \sqrt{10^{2} -8^{2}} = \sqrt{100-64} = \sqrt{36} = 6 cm$

The length of diagonal = 6 x2 = 12 cm

b) Area of the parallelogram ABCD = $\dpi{120} \frac{1}{2}\, \, \, d1d2$

= $\dpi{120} \frac{1}{2}\times \, \, \, 12 \times 16$

= 96 cm2

c. Distance between the two parallel sides

Consider $\dpi{120} \triangle AMO$

$\dpi{120} OM = \sqrt{AO^{2}-AM^{2}}$

$\dpi{120} = \sqrt{8^{2}-5^{2}} = \sqrt{64-25}$

$\dpi{120} = \sqrt{ 39}$

Then LM = 2 x $\dpi{120} \sqrt{ 39} = 2\sqrt{39}$ cm

## ABCD is a trapezium. P is the midpoint of BC. AB = 12 cm, DC = 8 cm. The distance between AB and CD is 6 cm. a. What is the area of the trapezium? b. What is the area of the triangle AQD?

Given, Length of the parallel side of the trapezium

AB and DC = 12 cm and 8 cm.

Distance between AB and DC (h) = 6 cm

a. Area of the trapezium = $\frac{1}{2}\, \, \, h (a+b)$ (formula)

= $\frac{1}{2}\times 6\, \, \, (12+8)$

= $\frac{1}{2}\times 6\times \, 20$

= 60 cm2

b. The area of the $\triangle AQD$ = Area of the quadrilateral ABPD + Area of the $\triangle BQP$.

Consider the $\triangle PCD$ and the $\triangle BQP$

PC = Pb (P is the midpoint of BC)

$\angle DPC = \angle BPQ$ (Opposite angles)

$\angle CDP= \angle BQP$ (alternate angles)

Hence the $\triangle PCD = \triangle BQP$.

Here from the figure we can see that the area of the $\triangle AQD$ is equal to the areas of the trapezium ABCD.

$\therefore$ The area of the $\triangle AQD = 60cm ^{2}$

## In the figure DP is perpendicular to AB.  AB = 10 cm, DP = 6 cm  a. Find the area of triangle ABD. b. If area of triangle BCD is half the area of triangle ABD. What is the area of the quadtilateral ABCD? c. If the distance from A to BD is 'a' and that from C to BD is 'b' then what is a:b?

Given, AB = 10 cm, DP = 6 cm

a. The area of the $\triangle ABD = \frac{1}{2}\, \, bh$ (formula)

$= \frac{1}{2} \times 10\, \, \times 6 = 30 cm^{2}$

b. Given that the area of the $\triangle BCD$ is the half of the area of the $\triangle ABD$

ie., the area  of the $\triangle BCD = \frac{30}{2} = 15 cm^{2}$.

The area of the quadrilateral ABCD = Area of the$\triangle ABD$  +  Area of the $\triangle BCD$

$= 30 \, cm^{2} + 15 cm ^{2} = 45 cm^{2}$

c. Area of the $\triangle ABD$ : Area of the $\triangle BCD$ = 30 : 15.

ie., $\frac{1}{2}\times BD\times a : \frac{1}{2}\times BD\times b = 30 : 15$.

a : b = 30 : 15

a : b = 2 : 1

a. 2 students

b. 8 + 3 = 11

c. 12 - 1 = 11.

## Calculate the distance between each pair of points given below on the number line.  a. 3, -7       b. -3,7          c. -3,-7

a. Distance between the number 3 and -7.

$= \left | 3-(-7) \right | = \left | 3+7 \right | = 10$

[Or distance between 3 and 0 + distance between 0 and -7 = 3+7 =10]

b. Distance between -3 and 7  $= \left | 7-(-3) \right |$

$= \left | 7+3 \right | = 10$

c. Distance between -3 and -7 $= \left | - 3 - (-7)\right |$

$= \left | - 3 +7 \right | = 4$

## (a)  What is the number which gives the midpoint of the points denoted by the numbers x and y         on the number line? (b) If the numbers x and y are thought of as points on a number line, What is the geometrical meaning of $\left | x-y \right |$  ? (c) In each of the equations below, find the number x satisfying it: (i) $\left | x-1 \right |= \left | x -3 \right |$           (ii) $\left | x-1 \right |= \left | x+3 \right |$ (iii)  $\left | x+1 \right |= \left | x-3 \right |$

(a) The number denoting the midpoint of the points denoted by the numbers x and y on the number line = $\frac{x+y}{2}$

(b) $\left | x-y \right |$ Means the distance between the points denoting the number x and y.

(c) (i) $\left | x-1 \right | = \left | x-3 \right |$ means distance between x and 1 and distance between x and 3 are equal.

So the position of x is in the middle of 1 and 3.

$x = \frac{1+3}{2}= 2$

(ii)  $\left | x-1 \right | = \left | x+3 \right |, \left | x-1 \right | = \left | x- (-3) \right |$

$x = \frac{1+(-3)}{2}=\frac{-2}{2}=-1$

(iii) $\left | x+1 \right | = \left | x-3 \right |, \left | x-(-1) \right | = \left | x-3 \right |$

$x = \frac{-1+3}{2} = \frac{2}{2} = 1$

## (a) What are the numbers x which satisfy the equation $\left | x-2 \right | + \left | x-6 \right | = 4$ ? (b) What are the numbers x satisfying the equation $\left | x-2 \right | + \left | x-6 \right | = 5$ (c) Are there numbers x satisfying the equation $\left | x-2 \right | + \left | x-6 \right | = 3$ Write the reason?

(a) $\left | x-2 \right | + \left | x-6 \right | = 4$

The distance between x and 2+ the distance between x and 6 should be 4.

Distance between 2 and 6 $\left | 6-2 \right | = \left | 4 \right | = 4$

So x can be anywhere between 2 and 6 including 2 and 6.

So x can be any number 2 or greater than 2 and 6 or less than 6.

(b) $\left |x -2 \right | + \left | x-6 \right | = 5$

The distance between the x and 2+ the distance between x and 6 should be 5.

Distance between 2 and 6= $\left |6 -2 \right |= \left | 4 \right |$ = 4

For this distance to be 5, x should move 1/2 unit to the left of 2 and 1/2 unit to the right of 6.

Then the distance = $4 + \frac{1}{2} + \frac{1}{2} = 5$

When x is moved $\frac{1}{2}$ unit to the left of 2, it is $1 \, \, \frac{1}{2}$.

When x is moved  $\frac{1}{2}$  unit to the right of 6, it is  $6 \, \, \frac{1}{2}$ .

So values of x are $1 \, \, \frac{1}{2}$ and $6 \, \, \frac{1}{2}$.

(C)

The sum of the distance between x and 2 and x and 6 is equal to or greater than the distance between 2 and 6. The distance between 2 and 6 is 4. So this distance cannot be less than 4.

$\therefore$ There are no numbers for x satisfying

$\left | x-2 \right | + \left | x-6 \right | = 3$

## A man standing on the top of a light houses sees a  ship approching the seashore at an angle of depression of 220. After the ship has travelled 100 metres more, towards the seashore, he sees it at an angle of depression of 310. The ship stops there. a. Draw a rough sketch  b. How far is the ship from the light house? c. Find the area of the triangle ABC $(sin\, \, 70^{\circ} = 0.93)$

a. PQ =  Light houses

A  =  First position of the ship

B  = Last position of the ship

Q  =  Position of the man

b. Let PB = x

From right triangle PBQ  =  tan 31= $\frac{PQ}{x}$$\frac{PQ}{x}$

$PA = x \times tan 31...................(1)$

From right triangle PAQ =  PA =  x + 100

tan 22 = $\frac{PQ}{x + 100}$

$PQ = (x + 100) \times tan 22 ..............(2)$

From (1) and (2), $x \times tan 31 = (x + 100)\times tan 22$

$x \times 0.6 = (x + 100) \times 0.4$

$0.6 x = 0.4x + 100\times 0.4, 0.6x - 0.4x = 100 \times 0.4$

$0.2x = 100\times 0.4$

$x = \frac{100\times 0.4}{0.2} =\frac{40}{0.2}=\frac{400}{2} = 200 m$

Distance to the ship from the light house = 200 m

c. Height of the light house =  PQ =  $x \times tan 31$

$= 200 \times 0.6 = 120 m$

## In the picture midpoint of the sides of the quadrilateral ABCD are joined to draw PQRS. a. Find coordinates of all vertices of R. b. Write coordinates of all vertices of quadrilateral ABCD.

a. The quadrilateral got by joining the midpoint of the sides of a quadrilateral is a parallelogram.PE is drawn parallel to      the x axis and QE parallel to Y axis.

PE =8-4=4

QE=8-3=5

If SF is drawn parallel to the x axis and RF parallel to the y axis.Then SF=4,RF=5.

b. P is the midpoint of AB. Distance between A and the horizontal line through P=3-1=2

Coordiantes of B=(4+3,3+2)=(7,5)

Coordinates of C=(8+1,8+3)=(9,11)

Coordinates of D=(6-3,9-2)=(3,7)

Coordinates of A =(1,1)

## A circle with centre (3,2) passes through the point (6,3). a. What is the radius of the circle? b. Check whether each of the points with coordinates (0,2),(3,6),(0,3) is inside ,outside or on the circle.

a. If the centre of the circle is A(3,2) and B(6,3), a point on the circle.

Radius of the circle =$\sqrt{(6-3)^2 +(3-2)^2 }$

=$\sqrt{9+1}=\sqrt{10}units$

b. Distance between the centre of the circle and (0,2)=$\sqrt{(3-0)^2 +(2-2)^2 }=\sqrt{9+0}=\sqrt{9}=3$

Since this is less than the radius, the point (0,2) is inside the circle.

Distance between the centre of the circle and (3,6) =$\sqrt{(3-3)^2 +(6-2)^2 }=\sqrt{0+16}=\sqrt{16}=4$

Since this is greater than the radius, the point (3,6) in outside the circle.

Distance between the centre of the circle and (0,3) =$\sqrt{(3-0)^2 +(3-2)^2 }=\sqrt{9+1}=\sqrt{10}$

Since this is equal to the radius, the point (0,3) is on the circle.

## A circle with centre (3,4) passes through the origin. a. What is the radius of the circle? b. If a point in the circle is (x,y),write the relation between x,y. c. Check whether the point (-2,1) lies  on this circle.

a. Radius of the circle = distance between (3,4) and (0,0) =$\sqrt{3^2 +4^2}=\sqrt{9+16}=\sqrt{25}=5$

b. equation of a circle : $(x-a)^2 +(y-b)^2 =r^2$

Equation of the circle with centre (3,4) and radius 5 is $(x-3)^2 +(y-4)^2 =5^2$

$x^2 -6x+9+y^2 -8y+16=25$

$x^2 +y^2 -6x-8y=0$

c. If x=-2 and y=1

$x^2 +y^2 -6x-8y=(-2)^2 +1^2 -6\times -2-8\times 1$

$=4+1+12-8=9$

This is not equal to 0. So (-2,1) is not a point on this circle.

## a. Find the centre of the circle with the line joining te points (3,-1), (13,-9) as diameter. Find the equation of this circle. b. Show taht there is no point on the circle whose x and y coordiantes are equal.

a. Centre of the circle = midpoint of the line joining the points (3,-1) and (13,-9)=$\left ( \frac{3+13}{2},\frac{-1+-9}{2} \right )=(8,-5)$

Radius of teh circle = distance between the point (3,-1) and (8,-5)=$\sqrt{(3-8)^2 +(-1--5)^2 }=\sqrt{5^2 +4^2 }=\sqrt{25+16}=\sqrt{41}$

Eqaution of the circle with centre (8,-5) and radius $\sqrt{41}$ is $(x-8)^2 +(y--5)^2 =\left ( \sqrt{41} \right )^2$

$(x-8)^2 (y+5)^2 =41$

b. Ket x=y

$(x-8)^2 +(x+5)^2 =41$

$x^2 -16x+64+x^2 +10x+25-41=0$

$2x^2 -6x+48=0,x^2 -3x+24=0$

$\sqrt{b^2 -4ac}=\sqrt{9-96}=\sqrt{-87}$

Since negative numbers has no squareroot, there is no solution to the equation $x^2 -3x+24=0$.

So there cannot be any point on the circle where x and y coordinates are equal.