Chapter-11 perimeters and area

Solution:-

Here, in the question, length and breadth of the rectangular piece of land is given Length = 500 m

Breadth = 300 m Now,

1. Area of rectangle = Length × Breadth

= 500 × 300

= 150000 m²

So, the area of the land is 150000m².

2. In the question, the cost of the land is also give which is for 1 m² = ₹10000 So, the cost of the land for 150000 m² = 10000 × 150000

= ₹ 1500000000

The cost of the total land is ₹ 1500000000.

Solution:-

Here, in the question, the perimeter of the square park is given. So, perimeter = 320 m

As we know that the formula to find the perimeter, by using that formula finding out the length

4 × Length of the park = 320 m Length of the side park = 320/4

Length of the park = 80 m

So, to find out the are, we simply use the equation for the area. Area of the square park = (Length of the park)²

= 80²

= 6400 m²

So, the area of the square park is 6400 m².

Solution:-

Here, in the question area and length of the rectangular plot is given as 440 m² and 22m respectively.

Now, Area of the rectangle = Length × Breadth 440 = 22 × Breadth

Breadth = 440/22 Breadth = 20 m

Now, to find perimeter we apply the equation of perimeter for rectangle. Perimeter of the rectangle = 2(Length + Breadth)

= 2 (22 + 20)

= 2(42)

= 84 m

Therefore, 84 m is the perimeter of the rectangular plot.

Solution:-

Here, in the question perimeter and length of the rectangular plot is given as 100 m and 35 m respectively.

Now,

Perimeter of the rectangle = 2 (Length + Breadth) 100 = 2 (35 + Breadth)

(100/2) = 35 + Breadth 50 – 35 = Breadth Breadth = 15 cm

Now, the area of the rectangle

Area of the rectangle = Length × Breadth

= 35 × 15

= 525 cm²

Therefore, 525 cm² is the area of the rectangular sheet.

Solution:-

Here, in the question it is given that the area of a square park is the same as that of a rectangular park.

Now,

Length of the square park = 60 m Length of the rectangular park = 90 m

We know how to find the area of square and rectangle, by comparing both the areas we get the breadth of the rectangle.

Area of the square park = ( Length of the square)²

= 60²

= 3600 m²

Both the areas are equal. So, area of the rectangular park = 3600 m²

Length × Breadth = 3600 90 × Breadth = 3600 Breadth = 3600/90 Breadth = 40 m

So, the breadth of the rectangular park is 40 m.

Solution:-

Here, in the question both sides of the rectangle are given. And it is also mentioned that the perimeter of the rectangle is equal to the square.

Length = 40 cm Breadth = 22 cm

Now, comparing perimeter's of both, rectangle and square. Perimeter of the rectangle = Perimeter of the Square

2 (Length + Breadth) = 4 × side 2 (40 + 22) = 4 × side

2 (62) = 4 × side

124 = 4 × side Side = 124/4 Side = 31 cm

Now, to find which shape enclose more area, Area of the rectangle = (Length × Breadth)

= 40 × 22

= 880 cm²

Area of square = side²

= 312

= 31 × 31

= 961 cm2

From the calculation, we can say that square-shaped encloses more area.

Solution:-

Here, in the question perimeter and breadth of the rectangle is given as 130 cm and 30 cm respectively.

Perimeter = 130 cm Breadth = 30

From the perimeter and breadth we can find out the length of the rectangle Perimeter of rectangle = 2 (Length + Breadth)

130 = 2 (length + 30) 130/2 = length + 30 Length + 30 = 65

Length = 65 – 30 Length = 35 cm Now,

Area of the rectangle = Length × Breadth

= 35 × 30

= 1050 cm²

The length and area of the rectangle is 35 cm and 1050 cm² respectively.

Solution:-

Here, in the question it is mentioned that the door is of rectangular size and it's length and breadth are given as 2 m and 2 m respectively.

Length = 2 m Breadth = 1 m

And dimensions of the wall is also given as Length= 4.5 m

Breadth = 3.6 m Bow,

Area of the door = Length × Breadth

= 2 × 1

= 2 m²

Area of the wall = Length × Breadth

= 4.5 × 3.6

= 16.2 m²

Now, to find out the area to be whitewashed subtract the area of door from the area of wall

= 16.2 – 2

= 14.2 m2

Now, cost of whitewashing is also given as 1 m² area = ₹ 20 So, the cost of whitewashing 14.2 m² area = 14.2 × 20

= ₹ 284

Solution:-

(a)

Here, by seeing the table we can say that, Base of parallelogram = 20 cm

Height of parallelogram =?

Area of the parallelogram = 246 cm2

Now, applying the equation for the area of the parallelogram Area of parallelogram = Base × Height

246 = 20 × height Height = 246/20 Height = 12.3 cm

So, the height of the parallelogram is 12.3 cm. (b)

Here, by seeing the table we can say that, Base of parallelogram =?

Height of parallelogram =15 cm

Area of the parallelogram = 154.5 cm2

Now, applying the equation for the area of the parallelogram Area of parallelogram = Base × Height

154.5 = base × 15

Base = 154.5/15 Base = 10.3 cm

So, the base of the parallelogram is 10.3 cm. (c)

Here, by seeing the table we can say that, Base of parallelogram =?

Height of parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm2

Now, applying the equation for the area of the parallelogram Area of parallelogram = Base × Height

48.72 = base × 8.4 Base = 48.72/8.4 Base = 5.8 cm

So, the base of the parallelogram is 5.8 cm. (d)

Here, by seeing the table we can say that, Base of parallelogram = 15.6 cm

Height of parallelogram =?

Area of the parallelogram = 16.38 cm2

Now, applying the equation for the area of the parallelogram Area of parallelogram = Base × Height

16.38 = 15.6 × height Height = 16.38/15.6 Height = 1.05 cm

So, the height of the parallelogram is 1.05 cm.

Solution:-

(a)

Here, by seeing the table we can say that, Height of triangle =?

Base of triangle = 15 cm

Area of the triangle = 16.38 cm2

Now, applying the equation for the area of the triangle Area of triangle = ½ × Base × Height

87 = ½ × 15 × height Height = (87 × 2)/15 Height = 174/15 Height = 11.6 cm

So, the height of the triangle is 12.3 cm. (b)

Here, by seeing the table we can say that, Height of triangle =31.4 mm

Base of triangle =?

Area of the triangle = 1256 mm2

Now, applying the equation for the area of the triangle Area of triangle = ½ × Base × Height

1256 = ½ × base × 31.4 Base = (1256 × 2)/31.4 Base = 2512/31.4

Base = 80 mm = 8 cm

So, the base of the triangle is 8 cm. (c)

Here, by seeing the table we can say that, Height of triangle =?

Base of triangle = 22 cm

Area of the triangle = 170.5 cm2

Now, applying the equation for the area of the triangle Area of triangle = ½ × Base × Height

170.5 = ½ × 22 × height

170.5 = 1 × 11 × height Height = 170.5/11 Height = 15.5 cm

So, the height of the triangle is 15.5 cm

 Solution:-

Here, by seeing the figure we can say the

SR = 12 cm, QM = 7.6 cm

1. Now, applying the equation for the parallelogram Area of the parallelogram = Base × Height

= SR × QM

= 12 × 7.6

= 91.2 cm²

Hence, the area of the parallelogram is 91.2 cm².

2. Now, applying the equation for parallelogram to find out the value of QN Area of the parallelogram = Base × Height

91.2 = PS × QN

91.2 = 8 × QN QN = 91.2/8 QN = 11.4 cm

Hence, the value of QN is 11.4 cm

Solution:-

Here, by seeing the figure we can say that, Area of the parallelogram = 1470 cm2

AB = 35 cm AD = 49 cm

Now, applying the equation for the area of the parallelogram to find out the value of DL Area of the parallelogram = Base × Height

1470 = AB × BM

1470 = 35 × DL DL = 1470/35 DL = 42 cm

Hence, the value of DL is 42 cm.

Now, applying the equation for the area of the parallelogram to find out the value of BM Area of the parallelogram = Base × Height

1470 = AD × BM

1470 = 49 × BM BM = 1470/49 BM = 30 cm

Hence, the value of BM is 30 cm

Solution:-

Here, by seeing the figure we can say that, AB = 5 cm, BC = 13 cm, AC = 12 cm

Now, applying the equation for the area of the triangle Area of the ΔABC = ½ × Base × Height

= ½ × AB × AC

= ½ × 5 × 12

= 1 × 5 × 6

= 30 cm²

Hence, the area of the triangle is 30 cm².

Now, applying the equation for the area of the triangle to find the value of AD

Area of ΔABC = ½ × Base × Height 30 = ½ × AD × BC

30 = ½ × AD × 13 (30 × 2)/13 = AD AD = 60/13

AD = 4.6 cm

Hence, the value of AD is 4.6 cm.

Solution:-

Here, by seeing the figure we can say that, AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm

Now, applying the equation for the area of the triangle Area of ΔABC = ½ × Base × Height

= ½ × BC × AD

= ½ × 9 × 6

= 1 × 9 × 3

= 27 cm²

Hence, the area of the triangle is 27 cm².

Now, applying the equation for the area of the triangle to find out the value of CE Area of ΔABC = ½ × Base × Height

27 = ½ × AB × CE

27 = ½ × 7.5 × CE (27 × 2)/7.5 = CE CE = 54/7.5

CE = 7.2 cm

Hence, the value of CE is 7.2 cm.

Solution:-

Here, the radius of the circle is given as 14 cm.

Now, using the equation for the circumference of the circle = 2πr

= 2 × (22/7) × 14

= 2 × 22 × 2

= 88 cm

Hence, the circumference of the circle with radius 14 cm is 88 cm.

 Solution:-

Here, the radius of the circle is given as 28 mm.

Now, using the equation for the circumference of the circle = 2πr

= 2 × (22/7) × 28

= 2 × 22 × 4

= 176 mm

Hence, the circumference of the circle with radius 28 mm is 176 mm.

Solution:-

Here, the radius of the circle is given as 21 cm.

Now, using the equation for the circumference of the circle = 2πr

= 2 × (22/7) × 21

= 2 × 22 × 3

= 132 cm

Hence, the circumference of the circle with radius 21 cm is 132 cm

1.  Solution:

Here, the radius of the circle is given as 14 mm.

Now, using the equation for the area of the circle = πr² Area of the circle = πr²

= 22/7 × 142

= 22/7 × 196

= 22 × 28

= 616 mm²

Hence, the area of the circle with radius 14 mm is 616 mm².

 Solution:

Here, the diameter of the circle is given as 49 m. So, radius of the circle = d/2=49/2=24.4 m

Now, using the equation for the area of the circle = πr² Area of the circle = πr²

= 22/7 × (24.5)²

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m²

Hence, the area of the circle with diameter 49 m is 1886.5 m²

Solution:

Here, the radius of the circle is given as 5 cm.

Now, using the equation for the area of the circle = πr²

= 22/7 × 52

= 22/7 × 25

= 550/7

= 78.57 cm²

Hence, the area of the circle with radius 5 cm is 78.57 cm².

Solution:-

Here, the circumference of the circle is given as 154 m. Let's suppose the radius of the circle is r.

Now, using the equation for the circumference of the circle Circumference of the circle = 154 m

154 = 2 × (22/7) × r

154 = 44/7 × r

r = (154 × 7)/44 r = (14 × 7)/4

r = (7 × 7)/2

r = 49/2

r = 24.5 m

Hence, the radius of the circle is 24.5 m.

Now, using the equation for the area of the circle Area of the circle = πr²

= 22/7 × (24.5)²

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m²

Hence, the area of the circle with radius 24.5 m is 1886.5 m².

Solution:-

Here, the diameter of the circle is given as 21 m. So, radius of the circle = d/2=21/2=10.5 m

Now, using the equation for the circumference of the circle Circumference of the circle = 2πr

= 2 × (22/7) × 10.5

= 462/7

= 66 m

Gardner wants to make 2 rounds. Hence, the length of rope required = 2 × 66 = 132 m In the question, cost of the rope per m is given as ₹ 4

So, the cost of 132 m rope = ₹ 4 × 132

= ₹ 528

The Gardner need ₹ 528 to fence a circular garden, two times.

Solution:-

Here, the radius of the circle is given as 4 cm and the circle which is going to be removed having the radius 3 cm.

Now, using the equation for the area of the circle to find the remaining area, subtract the area of the circle having radius 3 cm from the circle having radius 4 cm.

So, the remaining area of the sheet = πR² – πr²

= π (R² – r²)

= 3.14 (4² – 3²)

= 3.14 (16 – 9)

= 3.14 × 7

= 21.98 cm²

Hence, the area remaining on the sheet after removing the area of 3 cm is 21.98 cm².

Solution:-

Here, the diameter of the circular table is given as 1.5 m. So, radius of the circular table = d/2=1.5/2= 0.75 m

Now, using the equation for the circumference of the circle. Circumference of the circle = 2πr

= 2 × 3.14 × 0.75

= 4.71 m

So, the length of the lace required is 4.71 m

Now, in the question, the cost for 1 m lace is given as ₹ 15. So, the total cost required for 4.71 m lace = ₹ 15 × 4.71

= ₹ 70.65

Solution:-

Here, the diameter of the semicircle is given as 10 cm. So, radius of the semicircle = d/2=10/2= 5 cm

Now, using the equation for the circumference of the semicircle. Circumference of the semi-circle = πr + 2r

= 3.14(5) + 2(5)

= 5 [3.14+ 2]

= 5 [5.14]

Hence, the perimeter of the semicircle is 25.7 cm

1. Find the cost of polishing a circular table top of diameter 1.6 m, if the rate of polishing is

₹15/m2. (Take π = 3.14) Solution:-

Here, the diameter of the circle is given as 1.6 m. So, radius of the circle = d/2=1.6/2= 0.8 m

Now, using the equation for the area of the circle. Now, here in the question it is mentioned that we have to polish the top area which is nothing but the total area of the table.

Area of the circle = πr²

= 3.14 × 0.82

= 3.14 × 0.8 ×0.8

= 2.0096 m²

Now, in the question the cost for polishing 1 m2 area is given as ₹ 15. So, the cost for polishing 2.0096 m² area = ₹ 15 × 2.0096

= ₹ 30.144

Therefore, the cost for polishing the area 2.0096 m² is ₹ 30.144.

Solution:-

Here, the circumference of the circle is given as 44 cm. Let's suppose the radius of the circle as r.

Now, using the equation for the circumference of the circle. The circumference of the circle = 2πr

44 = 2 × (22/7) × r

44 = 44/7 × r (44 × 7)/44 = r

r = 7 cm

Now, using the equation of area we will find out the area of the circle having radius 7 cm. Area of the circle = πr²

= 22/7 × 7²

= 22/7 × 7 ×7

= 22 × 7

= 154 cm²

Now, we are bending the wire in the form of a square.

So, using the geometry we can say that the side of a square is one-fourth of the perimeter. The length of each side of the square = 44/4

= 11 cm

Now, use the equation of are for the square having side length as 11 cm.

Area of the square = (Length of the side of square)²

= 112

= 121 cm²

Now, we obtained area of the circle as 154 cm² and for the same length of perimeter the area of the square we are getting as 121 cm². Now, it is clear that the area enclosed by the circle is more compare to square.

Solution:-

Here, the radius of the circle card sheet is given as 14 cm.

Now, we have to remove two small circle and a rectangle from this card sheet.

So, the radius of two small circle is 3.5 cm and length and breadth of the rectangle are 3 cm and 1 cm respectively.

Now, find out the total area of the card sheet than subtract the area of two small circle having radius 3.5 cm and a rectangle having length and breadth as 3 cm and 1 cm respectively.

So, the area circular card sheet = πr²

= 22/7 × 14²

= 22/7 × 14 × 14

= 22 × 2 × 14

= 616 cm²

And area of two small circles = 2 × πr²

= 2 × (22/7 × 3.5²)

= 2 × (22/7 × 3.5 × 3.5)

= 2 × ((22/7) × 12.25)

= 2 × 38.5

= 77 cm²

Area of the rectangle = Length × Breadth

= 3 × 1

= 3 cm²

Now, subtract both the area of two small circle and a rectangle from the card sheet.

The area of the remaining part = Card sheet area – (Area of two small circles + Rectangle area)

= 616 – (77 + 3)

= 616 – 80

= 536 cm²

Hence, the remaining area is 536 cm

Solution:-

Here, the side length of the square is given as 6 cm.

And we have to cut a circle from it having radius as 2 cm.

So, first find out the total area of the square and subtract the area of the circle from it. Now, using the equation for the area of the square.

Area of the square = side2

= 6²

= 36 cm²

Now, using the equation for the area of the circle. Area of the circle = πr²

= 3.14 × 2²

= 3.14 × 2 × 2

= 3.14 × 4

= 12.56 cm²

Now, subtracting the area of the circle from the area of the square.

So, the area remaining = Area of the aluminium square sheet – The area of the circle

= 36 – 12.56

= 23.44 cm²

Solution:-

Here, the circumference of the circle is given as 31.4 cm. let's suppose the radius as r.Now, using the equation for the circumference of the circle. Circumference of a circle = 2πr

31.4 = 2 × 3.14 × r

31.4 = 6.28 × r

31.4/6.28 = r

r = 5 cm

Here, the radius of the circle we obtained as 5 cm. Now, using the equation for the area of the circle. Area of the circle = πr²

= 3.14 × 5²

= 3. 14 × 25

= 78.5 cm

Hence, the area of the circle with radius 5 cm is 78.5 cm².

Solution:-

Here, the diameter of the circular flower bed is given as 66 m. So, radius of the circle = d/2=66/2=33 m

Now, using the equation for the area of the circle we can find out the area of the circular flower bed.

Area of the circular flower bed = πr²

= 3.14 × 33²

= 3.14 × 1089

= 3419.46 m

Now, this circular flower bed is surrounded by a 4 m wide path. So, we have to find the area with that path combined.

So, the radius of the flower bed and path combined = 33 + 4 = 37 m Now, the area of the circular flower bed and path combined = πr²

= 3.14 × 37²

= 3.14 × 1369

= 4298.66 m

At last, to find the area of only path, subtract the area of circular flower bed from the total area.

Area of the path = Area of the flower bed and path together – Area of the flower bed

= 4298.66 – 3419.46

= 879.20 m²

Solution:-

Here, the area of the circular flower garden is given as 314m² and there is a sprinkler at the centre of the garden which covers the area of the radius 12 m.

let's suppose that the radius of sprinkler as r.

Now, using the equation for the area of the circle we can find out the area covered by the sprinkler.

Area of the sprinkler = πr²

= 3.14 × 12²

= 3.14 × 144

= 452.16 m

Hence, the area covered by the sprinkler is more than the area of the garder. So, sprinkler can cover the whole circular flower garden.

∴ Radius of the circular flower garden is 10 m.

1. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Solution:-

Here, the radius of the outer circle is given as 19 m.

So, the radius of inner circle = outer circle radius – 10

= 19 – 10

= 9 m

Now, using the equation of the circumference we can calculate the circumference of inner and outer circle.

Circumference of the inner circle = 2πr

= 2 × 3.14 × 9

= 56.52 m

Now, radius of the outer circle is given as 19 m. Circumference of the outer circle = 2πr

= 2 × 3.14 × 19

= 119.32 m

Solution:-

Here, the radius of the wheel is given as 28 cm.

When is rotate one time means it traveled the distance equals to its curving. Circumference of the wheel = 2πr

= 2 × 22/7 × 28

= 2 × 22 × 4

= 176 cm

Now, the total distance the wheel have to travel is given as 352 m. The distance given in meter so first cover if into centimetre.

352 m = 35200 cm.

Now, to find the number of rotation divide the total distance with the circumference of the wheel.

= Total distance to be covered/ Circumference of the wheel

= 35200 cm/ 176 cm

= 200

Hence, the wheel will rotate 200 times to cover the distance of 352 m

Solution:-

Here, the length of the minute have is given as 15 cm. Which means the radius of the circle is given as 15 cm.

Distance travelled by the tip of the minute hand in 1 hour will be equal to the circumferencer covered by the radius of 15 cm.

Distance travelled by the tip of minute hand in 1 hour = Circumference of the clock

= 2πr

= 2 × 3.14 × 15

= 94.2 cm

So, the distance travelled by the tip of the minute hand in 1 hour is 94.2 cm.

Solution:-

Here, in the question, length of the garden is given as 90 m and breadth of given as 75 m. Note, we have to construct a wide path of 5 m outside the garden.

Length and breadth of the garden is given as 90 m and 75 m respectively.

To find the area of the path we have to minus the area of the garden from the area of the garden plus path

Area of the garden = Length × Breadth

= 90 × 75

= 6750 m²

To find the total area of garden plus path, new length and breadth after adding 5 m path it will become 100 m and 85 m respectively.

New area of the garden = 100 × 85

= 8500 m²

To find the area of the path subtract the area of the garden from the total area of garden plus path.

The area of path = Area of the garden plus path – Area of garden

= 8500 – 6750

= 1750 m²

Now, for 1 hectare = 10000 m²

Therefore, the area of the garden in hectares is 6750/10000

= 0.675 hectare

Solution:-

Here, in the question, length of the rectangular park is given as 125 m and breadth of given as 65 m.

Note, a wide path of 3 m runs outside and around the rectangular park.. Length and breadth of the garden is given as 125 m and 65 m respectively.

To find the area of the path we have to minus the area of the rectangular park from the area of the rectangular park plus path

Area of the rectangular park = Length × Breadth

= 125 × 65

= 8125 m²

To find the total area of rectangular park plus path, new length and breadth after adding 3 m path it will become 131 m and 71 m respectively.

New area of the park = 131 × 71

= 9301 m²

To find the area of the path subtract the area of the rectangular park from the total area of rectangular park plus path.

The area of path = Area of the rectangular park plus path – Area of rectangular park

= 9301 – 8125

= 1176 m²

Solution:-

Here, in the question, from the dimensions given it is clear that it is a rectangular cardboard. Now, length and breadth of the cardboard is given as 8 cm long and 5 cm wide.

There is a margin on each side of the picture, that is 1.5 cm.

So, the length and breadth of the picture is 5 cm and 2 cm respectively.

To find the area of the margin we have to subtract the area of the picture from the area of the cardboard.

Area of the cardboard = Length × Breadth

= 8 × 5

= 40 cm²

New area of the picture = 5 × 2

= 10 cm²

The area of margin = Area of the cardboard – Area of the picture

= 40 – 10

= 30 cm²

Hence, the area of the margin is 30 cmcm²

Solution:

(i)Here, the length and the breadth ofroom is given.

Length and breadth of the room is 5.5 m and 4 m respectively. Length (L) = 5.5 m

Breadth (B) = 4 m

Now, using the equation for the area of rectangular we can find the area of the room because by looking the dimension we can say that the room is in the rectangular shape.

Area of the room = Length × Breadth

= 5.5 × 4

= 22 m²

Now, after considering the verandah in the room the dimensions of the room will increase. So, the new length and breadth of the room are 10 m and 8.5 m, respectively.

Now, we can find the area of the new room which include the verandah. So, the area of new room involving verandah

= 10 × 8.5

= 85 m²

Now, to find the area of the verandah we have to subtract the area of the room from the total area of the room including verandah.

Area of the verandah = Area of the room involving verandah – Area of the room

= 85 – 22

= 63 m² (ii)

In the question, it is also given that, the cost for cementing per m² is ₹ 200. So, the total amount required for cementing the verandah of area 63 m² is

= 200 × 63

= ₹ 12600

Solution:-

(i)

In the question, the side length of the square garden is given. Side length of the square garden (s) = 30 m

Now, using the equation of the area for square we can find out the area of the square garden.

Area of the square garden = S²

= 30²

= 30 × 30

= 900 m²

Now, there is 1m wide path inside the garden, so the length of the garden will reduce by 1 m from all the for sides.

So, the new side of the garden after removing the space for the path is 28 m. The area of this new garden which doesn't include the path= 28²

= 28 × 28

= 784 m²

Now, to find the area of the path we gave to subtract the area of the garden which doesn't include the path from the area of the garden which include the path

= 900 – 784

= 116 m² (ii)

In the question, it is also given that to plant a grass in 1 m², the cost required is

= ₹ 40 per m²

So, the cost required for planting the grass in the remaining portion of the garden with the area 784 m²

= 784 × 40

= ₹ 31360

Solution:-

Here, in the question the dimensions of the rectangular park is given. Length and breadth are 700 m and 300 m respectively.

Length (L) = 700 m Breadth (B) = 300 m

Now, using the equation of area for the rectangle we find out the area of the rectangular park Area of the park = Length × Breadth

= 700 × 300

= 210000 m²

Not, let's assume that ABCD is the level of one crossroad and EFGH is the length of another crossroad in the park.

So, the length of ABCD cross road = 700 m and the length of EFGH cross road = 300 m And we know that the width of both the crossroads is same = 10 m

Now, the shape of the cross road ABCD is rectangle, so, by using the formula we can find out the area of the cross road ABCD.

So, the area of the cross road ABCD = Length × Breadth

= 700 × 10

= 7000 m²

Now, the shape of the cross road EFGH is rectangle, so, by using the formula we can find out the area of the cross road ABCD.

So, the area of the cross road EFGH = Length × Breadth

= 300 × 10

= 3000 m²

Now, the intersection of both the road is calculated in both the road calculation so we need to subtract that portion of the road.

So, the area of the intersection IJKL = Length × Breadth

= 10 × 10

= 100 m²

So, the area of the total road = Area of ABCD + Area of EFGH – Area of IJKL

= 7000 + 3000 – 100

= 10000 – 100

= 9900 m²

And we know that for 1 hectare = 10000 m²

Therefore, the area of the cross roads in hectares = 9900/10000

= 0.99 hectare

Now, at last to calculate the area of the park without road subtract the total area covered by the road from the total area of the rectangular park

The area of the rectangular park without roads = Area of the park – Area of the roads

= 210000 – 9900

= 200100 m²

= 200100/10000

= 20.01 hectare

Solution:- (i)

Here, in the question the dimensions of the rectangular park is given. Length and breadth are 90 m and 60 m respectively.

Length (L) = 90 m Breadth (B) = 60 m

Now, using the equation of area for the rectangle we find out the area of the rectangular park Area of the park = Length × Breadth

= 90 × 60

= 5400 m²

Not, let's assume that ABCD is the level of one crossroad and EFGH is the length of another crossroad in the park.

So, the length of ABCD cross road = 90 m and the length of EFGH cross road = 60 m And we know that the width of both the crossroads is same = 3 m

Now, the shape of the cross road ABCD is rectangle, so, by using the formula we can find out the area of the cross road ABCD.

So, the area of the cross road ABCD = Length × Breadth

= 90 × 3

= 270 m²

Now, the shape of the cross road EFGH is rectangle, so, by using the formula we can find out the area of the cross road ABCD.

So, the area of the cross road EFGH = Length × Breadth

= 60 × 3

= 180 m²

Now, the intersection of both the road is calculated in both the road calculation so we need to subtract that portion of the road.

So, the area of the intersection IJKL = Length × Breadth

= 3 × 3

= 9 m²

So, the area of the total road = Area of ABCD + Area of EFGH – Area of IJKL

= 270 + 180 – 9

= 441 m² (ii)

In the question, it is given that to construct on m² road the cost required is ₹ 110. So, the total cost to construct the road of area 441 m2 = 441 × 110

= ₹ 48510

Solution:-

Here, in the question radius of the circular pipe is given to us and also the measurement of the side of square is also given.

So, the radius of a circular pipe = 4 cm and the length of a square = 4 cm Now, for circle we use the formula 2πr to find its perimeter.

So, the perimeter of the circular pipe = 2πr

= 2 × 3.14 × 4

= 25.12 cm

And the perimeter of the square = 4 × Side of the square

= 4 × 4

= 16 cm

Now, to find the cord left we have to subtract the perimeter of the square from the perimeter of Circle according to the question.

So, the length of cord left = Perimeter of the circular pipe – Perimeter of the square

= 25.12 – 16

= 9.12 cm

Yes, pragya is still having a cord of length 9.12 cm

Solution:- 

(i)

Here, in the question length and breadth of the rectangular lawn is given to us. Length and breadth of the rectangular lawn is 10 m and 5 m respectively.

Length (L) = 10 m Breadth (B) = 5 m

So, the area of the rectangular lawn = Length × Breadth

= 10 × 5

= 50 m² (ii)

Here, in the question, radius of the flower bed is also given to us. Radius of the flower bed = 2 m

From radius we can calculate the area of the flower bed by using the equation given for the circle.

Area of the flower bed = πr²

= 3.14 × 2²

= 3.14 × 4

= 12.56 m²

(iii)

Now, to find the area of the lawn without the flower bed, we can find that area by subtracting the area of the flower bed from the area of the rectangle lawn.

= Area of the rectangular lawn – Area of the flower bed

= 50 – 12.56

= 37.44 m²

(iv)

In the question, radius of the flower bed is provided to us.

So, by using the formula of Circle we can find the circumference of the flower bed. So, the circumference of the flower bed = 2πr

= 2 × 3.14 × 2

= 12.56 m

Solution:

Here, AC = 22 cm, BM = 3 cm,

DN = 3 cm, and BM ⊥ AC, DN ⊥ AC

Here, we can divide the quadrilateral into two triangle.

AC = 22 cm, BM = 3 cm DN = 3 cm and BM ⊥ AC, DN ⊥ AC

Now, to find the area of quadrilateral ABCD, we can first find the area of both the triangle and then add then up.

So, the area of the triangles ΔABC and ΔADC ΔABC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm²

Area of ΔADC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm²

Now, by adding the area of both the triangle we can obtain the area of the quadrilateral. Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC

= 33 + 33

= 66 cm².