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The perimeter of a square playground

= 4 × 30

= 120 mt

Yes. The area of the pathway can be found by dividing it into rectangles and adding the areas of these rectangles. 

Length of rectangle 1 = 30 + 1.5 + 1.5 

= 33 m 

Breadth of rectangle 1 = 1.5 m 

∴ Area of rectangle 1 = 33 x 1.5 

= 49.5 sq.m 

Area of rectangle 4 = Area of rectangle 1 

= 49.5 sq. m. 

Length of rectangle 2 = 65 m 

breadth of rectangle 2 = 1.5 m 

∴ Area of rectangle 2 = 65 x 1.5 

= 97.5 sq. m. 

Area of rectangle 3 = area of rectangle 2 

= 97.5 sq. m. 

∴ Area of pathway = Sum of area of the 4 rectangles = 49.5 + 49.5 + 97.5 + 97.5 

= 294 sq. m.

First let us calculate the area of the wall to be painted.

Area of the wall = length of the wall × breadth of the wall = 4 × 3 = 12

Thus, the area of the wall is 12 sq m.

Labour cost of 1 sq m is 25 rupees.

So the labour cost for 12 sq m will be = 12 × 25 = 300

The cost of labour for painting the wall will be 300 rupees.

Area of a rectangle = length × breadth 

∴ 102 = 17 × breadth 

∴ breadth = 102/17 = 6 cm 

Perimeter of rectangle = 2 (length + breadth) 

= 2 (17 + 6) 

= 2 × 23 

= 46 cm 

∴ The perimeter of rectangle is 46 cm.

Area of a square = length of side × length of side 

= 15 × 15 = 225 

The area of the square is 225 sq cm.

Given Area = 220 cm², h = 11 cm

Area of a triangle = 1/2b.h

= 1/2 × b × 11 = 220cm² (given)

b × 11 = 220 × 2

b = \(\frac {220\times2}{11} = 40\) 

∴ base of the triangle = 40 cm.

Given : Area = 108 cm²

Let the base of the triangle be 3x and height of the triangle be 2x.

Area of the triangle = 1/2 × base × height = 1/2 x (3x) (2x) = 108 cm² (given)

3x² = 108

x² = 108/3 = 36

x = √36 = 6

∴ base 3x = 3(6) = 18 cm

height = 2x = 2(6) = 12cm

Area of page of a calendar = length × breadth 

= 45 × 26 

= 1170 sq. cm. 

∴ The area of the page of the calendar is 1170 sq. cm.

Length of the matchbox (l) = 4 cm, breadth (b) = 2.5 cm, height (h) = 1.5 cm 

∴ Total surface area of the matchbox = 2 (lb + bh + lh) 

= 2 (4 × 2.5 + 2.5 × 1.5 + 4 × 1.5) 

= 2 (10 + 3.75 + 6) 

= 2 × 19.75 

= 39.5 sq. cm. 

∴ 39.5 sq. cm paper will be required.

We know that ∠ROS + ∠ROQ + ∠QOP + ∠POS = 360°

3x° + 2x° + x° + 2x° = 360°

8x° = 360°

x° = 360°/8

x° = 45°

First we must find the area of the square room.

Area of the square room = length of side × length of side = 4 × 4 = 16

Therefore, the area of the square room is 16 sq m.

The labour cost of laying 1 sq m of flooring is 35 rupees. 

Therefore, the cost of laying 16 sq m of flooring is 16 × 35 = 560 rupees.

The playground is rectangular. 

Rs.ABCD is the playground. Around it is a pathway 1.5 m wide. 

Around Rs.ABCD we get the rectangle 

Rs.PQRS 

Length of new rectangle PQRS = 65 + 1.5 + 1.5 = 68 m 

Breadth of new rectangle PQRS = 30 + 1.5 + 1.5 = 33m 

Area of path = Area of rectangle PQRS – Area of rectangle ABCD 

= 68 x 33 – 65 x 30 

= 2244 – 1950 

= 294 sq m

Let the side of the square be a. 

∴ Area of a square = (side)² = a² New side of the square = 3 × a = 3a 

∴ New area of the square = (3a)² 

= 9a²

= 9 × area of original square 

∴ If the side of a square is tripled, its area will become 9 times the area of the original square.

Area of triangle = 1/2 × base × height 

= 1/2 × 4.8 × 3.6 

= 1/2 × 17.28 

= 8.64 sq. cm. 

∴ The area of the triangle is 8.64 sq. cm.

We know that, only closed shapes have perimeters. Here shape (b) and (c) are not closed, so we cannot find their perimeter. These type of shapes are known as open shapes.

(a) We know that Perimeter of regular shapes = number of sides × length of a side

∴ Perimeter = 3 × 10 = 30 cm.

(b) Perimeter = 4 × 12 = 48 cm.

(c) Perimeter = 5 × 7 = 35 cm.

 Linear pair: The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°

Therefore 3x° = 105°

x° = 105°/3

x° = 45°

Given that, ∠1 = 40°

∠1 and ∠2 is a linear pair [from the figure]

∠1 + ∠2 = 180°

∠2 = 180° – 40°

∠2 = 140°

Again from the figure we can say that

∠2 and ∠6 is a corresponding angle pair

So, ∠6 = 140°

∠6 and ∠5 is a linear pair [from the figure]

∠6 + ∠5 = 180°

∠5 = 180° – 140°

∠5 = 40°

From the figure we can write as

∠3 and ∠5 are alternative interior angles

So, ∠5 = ∠3 = 40°

∠3 and ∠4 is a linear pair

∠3 + ∠4 = 180°

∠4 = 180° – 40°

∠4 = 140°

Now, ∠4 and ∠6 are a pair interior angles

So, ∠4 = ∠6 = 140°

∠3 and ∠7 are pair of corresponding angles

So, ∠3 = ∠7 = 40°

Therefore, ∠7 = 40°

∠4 and ∠8 are a pair corresponding angles

So, ∠4 = ∠8 = 140°

Therefore, ∠8 = 140°

Therefore, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40° and ∠8 = 140°

External measurement of triangle = Sum of three sides of triangle

= 4 cm. + 5 cm. + 6 cm. 

= 15 cm.

(a) Perimeter of the shape = 10 + 7 + 10 + 8 + 7 + 7 + 9 + 7 = 65 cm.

(b) Perimeter of the shape = 10 + 5 + 4 + 5 + 5 + 4 + 5 = 38 cm.

(c) Perimeter of the shape = 4 × side = 4 × 7 = 28 cm.

Area of the rectangular plot

= length x breadth

= 25 x 20 

= 500 sq.m.

Cost of the plot of land 

= Area of the plot x rate 

= 500 x 900 

= 4,50,000 rupees

No, rectangular is not a regular shape.
Because it’s all sides are not equal measurement.
We know that opposite sides of a rectangle are equal.

i.e. perimeter = 2 × length + 2 × breadth

Therefore formula for the perimeter of rectangle

perimeter = 2 × (length + breadth)

This is the perimeter of rectangle.

Now perimeter of given rectangular shape = 2 × (15 + 5)

= 2 × 20 = 40 m.

Area of rectangular field = Length × Width

= 25 × 30

= 750 square meter.

(a) Perimeter of triangle = 3 × side = 3 × 14 = 42 cm.

(b) Perimeter of square = 4 × side = 4 × 6 = 24 cm.

(c) Perimeter of Hexagon = 6 × side = 6 × 6 = 36 cm.

(d) Perimeter of Pentagon = 5 × side = 5 × 9 = 45 cm.

1. Length

2. Area

3. Perimeter

4. No. of sides

5. Length

Given from the figure, ∠1 = ∠3 are the vertically opposite angles

Therefore, ∠3 = 65°

Here, ∠1 + ∠2 = 180° are the linear pair [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°]

Therefore, ∠2 = 180° – 65°

= 115°

∠2 = ∠4 are the vertically opposite angles [from the figure]

Therefore, ∠2 = ∠4 = 115°

And ∠3 = 65°

(i) Shape is a picture of rectangle.

If each side of square is of 1 cm on grid paper then
length of rectangle is = 5 cm
and width = 2 cm.

Number of squares covered by rectangle ABCD = 10

Therefore, area of rectangle ABCD = 10 square cm.

Here, there is a rule been between length, width and area of rectangle

Length × Width = Area

= 5 cm. × 2 cm. 

= 10 square cm.

(ii) Length of square = 4 cm.

Width of square = 4 cm.

Area of square = (side)² = (4)² = 16 square cm.

Number of squares covered by square = 16.

Therefore area of square = 16 square cm.

(iii) Length of rectangle = 7 cm.

Width of rectangle = 3 cm.

Area of rectangle = Length × Width = 7 × 3 = 21 square cm.

Number of squares covered by rectangle ABCD = 21.

Therefore area = 21 square cm.

Area of the rectangular plot = length × breadth 

= 75.5 × 30.5 

= 2302.75 sq. m. 

Value of the plot = area of the plot × rate per square metre = 2302.75 × 1000 

= Rs 230275 

∴ The value of the plot is Rs 23,02,750.

Perimeter of square stool = 4 × side

= 4 × 60

= 240 centimeter

Therefore perimeter of stool = 240 centimeter.

Perimeter of square = 4 x side

∠AOC + ∠BOC = 180° [The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180°]

2y + 5° + 3x = 180°

3x + 2y = 175°

Given If x = 25°, then

3(25°) + 2y = 175°

75° + 2y = 175°

2y = 175° – 75°

2y = 100°

y = 100°/2

y = 50°

(c) 25 square meter

(i) Given length of the rectangular field = 125 m

Breadth of the rectangular field = 400 m

We know that the area of the rectangular field = Length x Breadth

= 125 m x 400 m

= 50000 m² 

= 5 hectares [Since 10000 m² = 1 hectare]

(ii) Given length of the rectangular field =75 m 5 dm

= (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

We know that the area of the rectangular field = Length x Breadth

= 75.5 m x 120 m

= 9060 m² 

= 0.906 hectares [Since 10000 m² = 1 hectare]

(c) Space occupied by object in plane

If the radius is r, then its circumference of the circle is 2πr. 

If radius is halved means r = \(\frac{r}{2}\) (new) 

Write r = \(\frac{r}{2}\) in 2πr. 

Then, the circumference = 2π x \(\frac{r}{2} = \frac{1}{2}\)

2πr 

Circumference is halved. 

So if the radius of a circle is halved, then the circumference is also halved.

(i) Given length of the rectangular field = 200 m

Breadth of the rectangular field = 125 m

We know that area of the rectangular field = Length x Breadth

= 200 m x 125 m

= 25000 m² 

= 250 acres [Since 100 m² = 1 acre]

(ii) Given length of the rectangular field =75 m 5 dm

= (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

We know that area of the rectangular field = Length x Breadth

= 75.5 m x 120 m

= 9060 m² 

= 90.6 acres [Since 100 m² = 1 acre]

Distance walked in two round = 40 meter

∴ Distance walked one round = 40/2 = 20 meter

Distance walked in one round is called perimeter

Therefore perimeter of square field = 20 meter

∵ 4 × side = Perimeter

Side = Perimeter/4 = 20/4 = 5 meter

Therefore side of square field = 5 meter.

(i) Given base =15 dm, altitude = 6.4 dm

By converting these to standard form we get,

Base = 15 dm = (15 x 10) cm 

= 150 cm 

= 1.5 m

Altitude = 6.4 dm = (6.4 x 10) cm 

= 64 cm 

= 0.64 m

We know that area of the parallelogram = Base x Altitude

= 1.5 m x 0.64 m

Area of parallelogram = 0.96 m²

(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]

Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]

We know that area of the parallelogram = Base x Altitude

= 1.4 m x 0.6 m

= 0.84 m² 

Given side of the square (s) = 16 cm 

Area of the square A = s x s 

= 16 x 16 

= 256 sq.cm

Given side of the square = 16.5

dam = 16.5 x 10 m = 165 m [Since 1 dam/dm (decametre) = 10 m]

Area of the square = (Side)² 

= (165 m)² 

= 27225 m²

Measurement of roof, l = 4 m.

b = 3 m.

h = 2.5 m.

∴ Area of Tarpaulin = 2 (lb + bh) + lh 

= 2 [4 × 2.5 + 3 × 2.5] +4 × 3 

= 2 [10 + 7.5] + 12 

= 2 × 17.5 + 12 

= 35 + 12 = 47 m² .

(i) Length of Herbarium, l = 30 cm, 

breadth, b = 25 cm. 

height, h = 25 cm. 

∴ TSA of Herbarium = 2 (lb + bh + Hi) 

= 2 [30 × 25 + 25 × 25 + 30 × 25] 

= 2 [750 + 625 + 750] 

= 2 (2125) = 4250 cm² . 

(ii) Tape required for all 12 edges means, it has 4 length, 4 breadth and 4 heights. 

∴ Length of tape required 

= 4 (l + b + h) 

= 4 (30 + 25 + 25) 

= 4 × 80 

= 320 cm.

Each edge of cubical box, a = 10 cm. 

(i) Curved Surface Area = 4a² = 4(10)² 

= 4 × 100 

= 400 cm² . 

Length of cuboid box, l = 12.5 cm.

 b = 10 cm h = 8 cm. 

Curved Surface area of Cuboid, A = 2h (l+ b) 

A = 2 × 8(12.5 + 10) 

= 16 × 22.5 = 360 cm² . 

∴ 400 – 360 = 40 cm² . 

L.S.A. of cuboid box is 40 cm² greater than rectangular cuboid. 

(ii) Total surface area of cube (T.S.A.) = 6a² 

= 6(10)² = 6 × 100 = 600 cm² . 

T.S.A. of cuboid = 2(Ib + lh + bh) 

= 2[ 12.5 × 10 + 12.5 × 8 + 10 × 8] 

= 2 (125 + 100 + 80) 

= 2 × 305 = 610 cm² . 

Here, cuboid box’s T.S.A. is more than T.S.A. of cubical box, 

i.e, 610 – 600 = 10 cm² more.

(i) Edge of cube = 10 cm

= 4(10 cm)²

= 400 cm²

Lateral surface area of cuboidal box = 2[lh + bh]

= [2(12.5 × 8 + 10 × 8)] cm²

= (2 × 180) cm²

= 360 cm²

Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm² − 360 cm² = 40 cm²

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm².

(ii) Total surface area of cubical box = 6(edge)² = 6(10 cm)² = 600 cm² Total surface area of cuboidal box

= 2[lh + bh + lb]

= [2(12.5 × 8 + 10 × 8 + 12.5 × 100] cm²

= 610 cm²

Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.

Total surface area of cuboidal box − Total surface area of cubical box = 610 cm² − 600 cm² = 10 cm²

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm².

The type and thickness of vegetation changes from place to place, because of the variation in temperature, moisture, altitude, rainfall and thickness of soil.

Correct option is (B) greater than 1

Let there are x number of girls and y number of boys in the class.

The ratio of boys to girls in the class is B.

\(\therefore\) \(\frac xy\) = B         _________(1)

The ratio of girls to boys in the class is G.

\(\therefore\) \(\frac yx\) = G        _________(2)

Now, B+G \(=\frac xy+\frac yx\)

\(=\frac{x^2+y^2}{xy}\)

\(\because(x-y)^2\geq0\)

\(\Rightarrow x^2+y^2-2xy\geq0\)

\(\Rightarrow x^2+y^2\geq2xy\)

\(\Rightarrow\frac{x^2+y^2}{xy}\geq2>1\)

\(\Rightarrow\frac{x^2+y^2}{xy}>1\)

Hence, B+G > 1

Correct option is(B) greater than 1

The correct answer is-

Scientific methods of cultivation and the use of tractors, harvesters and combines have made North America a surplus food producer. The temperature and fertile soil with high humus content make the region suitable for crop production. There is huge production of wheat in the region. Hence, the Prairies are called the 'Granaries of the world'.