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

about 3 years ago 0 Answer 482 views

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

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Distance formula

about 2 years ago 0 Answer 261 views

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)

Rightarrowy = 10+4x -4 

Rightarrow y =4x +6

Table of solutions

X 0 1 -1
Y 6 10 2

 

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 

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 

 

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. 

 

 

 

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) 

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.

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.

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}}} = xsqrt{3} .................(2)

Comparing equation (1) and (2)

ie., frac{x + 12}{sqrt{3}} = xsqrt{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 = 1.73times 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?

a. Radius = 30 cm 

Distance the circle rolls = 3.14 cm 

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

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

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

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

b. Consider triangleABC which is an equilateral triangle. also triangle BDC is 300, 600, 900 triangle 

sin 300frac{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 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

Circum radius 

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 =  sqrt{(-3-3)^{2} + (7 + 7)^{2}}

                                        = sqrt{(-6)^{2} + (14)^{2}}

                                        = sqrt{36+196}

                                        = sqrt{232}

                                       =15.23

Distance between B and C

                     = sqrt{(-3+3)^{2} + (-7-7)^{2}}

                     = sqrt{(14)^{2}}

                     = 14

Distance between C and A

                    = sqrt{(3--3)^{2}+ (-7+7)^{2}} 

                    = sqrt{ (6)^{2}}

                    = 6

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 triangle AOB

AB = 10 cm, AO = 8 cm 

Then OB = sqrt{AB^{2} -AO^{2}}

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 = frac{1}{2}, , , d1d2

                                                      = frac{1}{2}times , , , 12 times 16

                                                      = 96 cm2 

c. Distance between the two parallel sides 

Consider triangle AMO

OM = sqrt{AO^{2}-AM^{2}}

       = sqrt{8^{2}-5^{2}} = sqrt{64-25}

       = sqrt{ 39}

Then LM = 2 x sqrt{ 39} = 2sqrt{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 6times , 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 thetriangle 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 BDtimes a : frac{1}{2}times BDtimes b = 30 : 15.

a : b = 30 : 15

a : b = 2 : 1

The table below shows the time taken by 30 students to complete a long distance race.

(Time (Minutes)              Number of children 

10-13                                        2

13-16                                        5

16-19                                       12

19-22                                        8

22-25                                        3

a. How many students have taken below 13 minutes to complete the race?

b. How many students have taken 19 minutes or above to complete the race?

c. If only one student took 16 minutes to complete te race, then how many taken between 16 and 19 minutes to                  complete the race?                               

a. 2 students 

b. 8 + 3 = 11

c. 12 - 1 = 11.

(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 + 100times 0.4, 0.6x - 0.4x = 100 times 0.4

   0.2x = 100times 0.4

       x = frac{100times 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. 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.