Find the perimeter and area of rectangular field of length 96 m and breadth 48 m Also find the cost of levelling the field at a 12 per m²
Answers
Answer:
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Step-by-step explanation:
with solved examples.
If l denotes the length and b denotes the breadth of the rectangle, then the
Perimeter and Area of Rectangle
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● Perimeter of the rectangle = 2(l + b) units
● Length of the rectangle = P2 - b units
● Breadth of the rectangle = P2 - l units
● Area of the rectangle = l × b sq. units.
● Length of the rectangle = Ab units .
● Breadth of the rectangle = Al units
● Diagonal of the rectangle = l2+b2−−−−−√ units
Let us consider a rectangle of length 'a' units and breadth 'b' units.
Perimeter of Rectangle
Therefore, perimeter of the rectangle ABCD
= (AB + BC + CD + DA) units
= (a + b + a + b) units
= (2a + 2b) units
= 2 (a + b) units
Therefore, perimeter of the rectangle = 2 (length + breadth) units
We know that the area of the rectangle is given by
Area = length × breadth
A = a × b square units
⇒ a = Ab, i.e., length of the rectangle = Areabreadth
And b = Aa, i.e., breadth of the rectangle = Arealength
Worked-out problems on Perimeter and Area of Rectangle:
1. Find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm.
Solution:
Given: length = 17 cm, breadth = 13 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (17 + 13) cm
= 2 × 30 cm
= 60 cm
We know that the area of rectangle = length × breadth
= (17 × 13) cm2
= 221 cm2
2. Find the breadth of the rectangular plot of land whose area is 660 m2 and whose length is 33 m. Find its perimeter.
Solution:
We know that the breadth of the rectangular plot = Arealength
= 660m233m
= 20 m
Therefore, the perimeter of the rectangular plot = 2 (length + breadth)
= 2(33 + 20) m
= 2 × 53 m
= 106 m
3. Find the area of the rectangle if its perimeter is 48 cm and its breadth is 6 cm.
Solution:
P = 2 (l + b)
Here, P = 48 cm; b = 6 cm
Therefore, 48 = 2 (l + 6)
⇒ 482 = l + 6
⇒ 24 = l + 6
⇒ 24 - 6 = l
⇒ 18 = l
Therefore, length = 18 cm
Now, area of rectangle = l × b = 18 × 6 cm2 = 108 cm2
4. Find the breadth and perimeter of the rectangle if its area is 96 cm2
and the length is 12 cm.
Solution:
Given, A = 96 cm2 and l = 12 cm
A = l × b
Therefore, 96 = 12 × b
⇒ 9612 = b
⇒ b = 8 cm
Now, P = 2 (l + b)
= 2 (12 + 8)
= 2 × 20
= 40 cm
5. The length and breadth of a rectangular courtyard is 75 m and 32 m. Find the cost of leveling it at the rate of $3 per m2. Also, find the distance covered by a boy to take 4 rounds of the courtyard.
Solution:
Length of the courtyard = 75 m
Breadth of the courtyard = 32 m
Perimeter of the courtyard = 2 (75 + 32) m
= 2 × 107 m
= 214 m
Distance covered by the boy in taking 4 rounds = 4 × perimeter of courtyard
= 4 × 214
= 856 m
We know that area of the courtyard = length × breadth
= 75 × 32 m2
= 2400 m2
For 1 m2, the cost of levelling = $3
For 2400 m2, the cost of levelling = $3 × 2400
= $7200
Solved examples on Perimeter and Area of Rectangle:
6. A floor of the room 8 m long and 6 m wide is to be covered by square tiles. If each square tile is 0.8 m, find the number of tiles required to cover the floor. Also, find the cost of tiling at the rate of $7 per tile.
Solution:
Length of the room = 8 m
Breadth of the room = 6 m
Area of the room = 8 × 6 m2 {Area of room = Area of tiles that are put on the floor of the room.}
= 48 m2
Area of one square tile = 0.8 × 0.8 m2 = 0.64 m2
Number of tiles required = AreaoffloorAreaoftiles
= 480.64
= 48×10064
= 75 tiles
For 1 tile, the cost of tiling is $7
For 7 tiles, the cost of tiling is $(7 × 75) = $525
7. The breadth of the rectangle is 8 cm and A its diagonal is 17 cm. Find the area of the rectangle and its perimeter.
Solution:
Area of the Rectangle
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Using Pythagoras theorem,
BD2 = DC2 + BC2
⇒ 172 = DC2 + 82
⇒ 289 - 64 = DC2
⇒ 225 = DC2
⇒ 15 = DC
Therefore, length of rectangle = 15 cm
So, area of rectangle = l × b
= 15 × 8 cm2
= 120 cm2
Also, perimeter of rectangle = 2 (15 + 8) cm
= 2 × 23 cm
= 46 cm
8. The length and breadth of the rectangle park are in the ratio 5 : 4 and its area is 2420 m2, find the cost of fencing the park at the rate of $10 per metre.
Solution:
Let the common ratio b x,
then length of rectangular park = 5x
Breadth of rectangular park = 4x
Area of rectangular park = 5x × 4x
= 20x2
According to the question,
20x2 = 2420
⇒ x2 = 242020
⇒ x2 = 121
⇒ x = 11
Therefore, 5x = 5 × 11 = 55 and 4x = 4 × 11 = 44
So, the perimeter of the rectangular park = 2 (l + b)
= 2 (55 + 44)
= 2 × 99
= 198 cm
For 1 m, the cost of fencing = $10
For 198 m, the cost of fencing = $198 × 10
= $1980
9. How many envelopes can be made out of a sheet of paper 100 cm by 75
Answer:
area = 96 x 48 = 4608
perimeter = 2(96+48) = 2(144) = 288
Step-by-step explanation:
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