-2

12-36+22=-2

12 - 36+ 22 = -36+34=-2 So the answer is -2

-2

12-36+22

= -36 + 34

= - 2

As 14 and 28 comes under 14 table: We can write: 14x + 28 = 14x + 14(2) =14(x+2)

14x + 28

= 14x + 14(2)

= 14( x + 2)

X(1/x + y/x)

Mam mera Naam bhi anushka hi hai

4/9 = ×\27

9x = (4 × 27)

x = (4 × 27)/9

x = (4 × 3) 

x = 12

Aryabhatta

Aryabhatta ha

Mayans

Aryabhatta

Aryabhattt

Point intersects x-axis, so y = 0

  3x +2y=12

3x + 2(0) = 12

3x = 12

x = 12/3 = 4

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠DBC = 60° and ∠BAC = 40°. Find

(i) ∠BCD,

(ii) ∠CAD.

(i) We know that the angles in the same segment are equal

So we get

∠BDC = ∠BAC = 40^o

Consider △ BCD

Using the angle sum property

∠BCD + ∠BDC + ∠DBC = 180^o

By substituting the values

∠BCD + 40^o + 60^o = 180^o

On further calculation

∠BCD = 180^o – 40^o – 60^o

By subtraction

∠BCD = 180^o – 100^o

So we get

∠BCD = 80^o

(ii) We know that the angles in the same segment are equal

So we get

∠CAD = ∠CBD

So ∠CAD = 60^o

y=2x−4

⇒y=2x+4

When x=0,y=4

When y=0,x=−2

When x=1,y=6

x	0	−2	1
y=2x+4	4	0	6

Plot the points A(0,4),B(−2,0) and C(1,6) on the graph paper.

Join these points by a straight line BC

0.75=75/100
1.2=12/10
First we have to make the denominators equal.
LCM of 10 and 100 is 100.
The two rational numbers are 120/100 and 75/100. The rational numbers between them can be 76/100, 77/100, 78/100, 79/100, etc. until 120/100.
Therefore, there are many rational numbers between them.

1

Gotta have a fun

STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit.

2: Mark O the mid point of AC

3: Draw a semicircle with O as centre & OA as radius

4: At B draw a perpendicular BD.

5: BD = √9.3 unit

6: Now, B as centre, BD as radius, draw an arc, meeting the numberline at E.

Now, with BD or BE = √9.3 as radius , with 0 ( origin) of the number line as centre, draw an arc on the the number line, intersecting at point ‘k’. And this point ‘k' lies between integers 3 & 4, & represents √9.3

JUSTIFICATION:

BD = √ {(10.3/2)² - (8.3/2)²}

=> BD = √{10.3²- 8.3²)/4 }

=> BD = √{(10.3+8.3)(10.3–8.3)/4}

=> BD = √{18.6*2/4}

=> BF = √{37.2/4}

=> BD = √9.3 = BE
 

CBSE Class 9 Mathematics NCERT Solutions CHAPTER 13 Surface Areas and Volumes(Ex. 13.1) 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1m2 cost Rs.20. Ans. (i) Given: Length l = 1.5 m, Breadth b = 1.25 m and Depth h = 65 cm = 0.65 m Area of the sheet required for making the box open at the top = = = = = 3.575 + 1.875 = 5.45 m2 (ii) Since, Cost of 1 m2 sheet = Rs. 20 Cost of 5.45 m2 sheet = 20 x 5.45 = Rs. 109 2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2. Ans. Given: Length = 5 m, Breadth = 4 m and Height = 3 m Area of the four walls = Lateral surface area = = = 2 x 3 (4 + 5) = 2 x 9 x 3 = 54 m2 Area of ceiling = = 5 x 4 = 20 m2 Total area of walls and ceiling of the room = 54 + 20 = 74 m2 Now Cost of white washing for 1 m2 = Rs. 7.50 Cost of white washing for 74 m2 = 74 x 7.50 = Rs. 555 3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall. Ans. Given: Perimeter of rectangular wall = = 250 m ……….(i) Now Area of the four walls of the room = = = 1500 m2 ……….(ii) Area of the four walls = Lateral surface area = = 1500 [using eq. (i) and (ii) = 6 m Hence required height of the hall is 6 m. 4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container? Ans. Given: Length of the brick = 22.5 cm, Breadth = 10 cm and Height = 7.5 m Surface area of the brick = = 2 (22.5 x 10 + 10 x 7.5 + 7.5 x 22.5) = 2 (225 + 75 + 168.75) = 937.5 cm2 = 0.09375 m2 [1 cm = 0.01 m] Now No. of bricks to be painted = = = 100 Hence 100 bricks can be painted. 5. A cubical box has each edge 10 cm and a cuboidal box is 10 cm wide, 12.5 cm long and 8 cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller total surface area and how much? Ans. (i) Lateral surface area of a cube = 4 (side)2 = 4 x (10)2 = 400 cm2 Lateral surface area of a cuboid = = 2 x 8 (12.5 + 10) = 16 x 22.5 = 360 cm2 Lateral surface area of cubical box is greater by (400 – 360) = 40 cm2 (ii) Total surface area of a cube = 6 (side)2 = 6 x (10)2 = 600 cm2 Total surface area of cuboid = = 2 (12.5 x 10 + 10 x 8 + 8 x 12.5) = 2 (125 + 80 + 100) = 2 x 305 = 610 cm2 Total surface area of cuboid box is greater by (610 – 600) = 10 cm2 6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. (i) What is the surface area of the glass? (ii) How much of tape is needed for all the 12 edges? Ans. (i) Given: Length of glass herbarium = 30 cm, Breadth = 25 cm and Height = 25 m Total surface area of the glass = = 2 (30 x 25 + 25 x 25 + 25 x 30) = 2 (750 + 625 + 750) = 2 x 2125 = 4250 cm2 Hence 4250 cm2 of the glass is required to make a herbarium. (ii) Tape is used at 12 edges. Tape is used at 4 lengths, 4 breadths and 4 heights. Total length of the tape = = 4 (30 + 25 + 25) = 320 cm Hence 320 cm of the tape if needed to fix 12 edges of herbarium. 7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm by 20 cm by 5 cm and the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of the card board is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind. Ans. Given: Length of bigger cardboard box (L) = 25 cm Breadth (B) = 20 cm and Height (H) = 5cm Total surface area of bigger cardboard box = 2 (LB + BH + HL) = 2 (25 x 20 + 20 x 5 + 5 x 25) = 2 (500 + 100 + 125) = 1450 cm2 5% extra surface of total surface area is required for all the overlaps. 5% of 1450 = = 72.5 cm2 Now, total surface area of bigger cardboard box with extra overlaps = 1450 + 72.5 = 1522.5 cm2 Total surface area with extra overlaps of 250 such boxes = 250 x 1522.5 = 380625 cm2 Since, Cost of the cardboard for 1000 cm2 = Rs. 4 Cost of the cardboard for 1cm2 = Rs. Cost of the cardboard for 380625 cm2 = Rs. = Rs. 1522.50 Now length of the smaller box = 15 cm, Breadth = 12 cm and Height = 5 cm Total surface area of the smaller cardboard box = = 2 (15 x 12 + 12 x 5 + 5 x 15) = 2 (180 + 60 +75) = 2 x 315 = 630 cm2 5% of extra surface of total surface area is required for all the overlaps. 5% of 630 = = 31.5 cm2 Total surface area with extra overlaps = 630 + 31.5 = 661.5 cm2 Now Total surface area with extra overlaps of 250 such smaller boxes = 661.5 x 250 = 165375 cm2 Cost of the cardboard for 1000 cm2 = Rs. 4 Cost of the cardboard for 1cm2 = Rs. Cost of the cardboard for 165375 cm2 = Rs. = Rs. 661.50 Total cost of the cardboard required for supplying 250 boxes of each kind = Total cost of bigger boxes + Total cost of smaller boxes = Rs. 1522.50 + Rs. 661.50 = Rs. 2184 8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m with base simensions 4 m x 3 m? Ans. Given: Length of base = 4 m, Breadth = 3 m and Height = 2.5 m Tarpaulin required to make shelter = Surface area of 4 walls + Area of roof = = 2 (4 + 3) 2.5 + 4 x 3 = 35 + 12 = 47 m2 Hence 47 m2 of the tarpaulin is required to make the shelter for the car.

Use the app (all ncert solutions app)

Please refer to video for the figure of different pair of circles. There can be three possibilities.

(i) When two circles are intersecting each other, then there are two common points. So, the maximum number of common points in this case, is 2.

(ii) When two circles are just touching each other, then there is a single common point. So, the maximum number of common points in this case, is 1.

(iii) When two circles are not touching each other, then there is no common point. So, the maximum number of common points in this case, is 0.

 

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a² + b² + c² = ab + bc + ca

 

On multiplying both sides by “2”, it becomes

 

2 ( a² + b² + c² ) = 2 ( ab + bc + ca)

 

2a² + 2b² + 2c² = 2ab + 2bc + 2ca

 

a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0

 

a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0

 

(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0

 

(a – b)² + (b – c)² + (c – a)² = 0

 

=> Since the sum of square is zero then each term should be zero

 

⇒ (a –b)² = 0, (b – c)² = 0, (c – a)² = 0

 

⇒ (a –b) = 0, (b – c) = 0, (c – a) = 0

 

⇒ a = b, b = c, c = a

 

∴ a = b = c.

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1. In which quadrant or on which axis do each of the points (– 2, 4), (3, – 1), (– 1, 0), (1, 2) and (– 3, – 5) lie? Verify your answer by locating them on the Cartesian plane.

Solution:

• (– 2, 4): Second Quadrant (II-Quadrant)
• (3, – 1): Fourth Quadrant (IV-Quadrant)
• (– 1, 0): Negative x-axis
• (1, 2): First Quadrant (I-Quadrant)
• (– 3, – 5): Third Quadrant (III-Quadrant)

Ment ?? Its meant not ment Study hard ?✏??

. Is .which is called .

The answer is 0

8. Praveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5m, with base dimensions 4m×3m?

Solution:

Let l, b and h be the length, breadth and height of the shelter.

Given:

l = 4m

b = 3m

h = 2.5m

Tarpaulin will be required for the top and four wall sides of the shelter.

Using formula, Area of tarpaulin required = 2(lh+bh)+lb

On putting the values of l, b and h, we get

= [2(4×2.5+3×2.5)+4×3] m²

= [2(10+7.5)+12]m²

= 47m²

Therefore, 47 m² tarpaulin will be required.

To find distance we can use various metod but according to me the best method is s=ut +1/2at^2 but if it is given in to find distance in straight line you can just add it.

To solve for distance use the formula for distance d = st, or distance equals speed times time. Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t.

Ex 9.2 no=4

a² - 3a - 1 = 0

 

=> a² - 1 = 3a

 

From the given equation divide by a from both side.

 

a² - 1 = 3a

 

=> (a² - 1 ) / a = 3a / a

=> a - 1/a = 3

 

We know,

 

( a - 1/a)² = a² + 1/a² - 2 { where a ≠ 0}

 

=> (a - 1/a)² + 2 = a² + 1/a²

 

=> 3² + 2 = a² + 1/a² { since, (a - 1/a)² = 3) }

 

=> 9 + 2 = a² + 1/a²

=> 11 = a² + 1/a²