Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm. (a) Area of parallelogram = base x height = 12 x 7.6 = 91.2 cm² (b) Area of parallelogram = base x height ⇒ 91.2 = 8 x QN ⇒ QN = 91.2/8 = 11.4 cm Class 7 Maths Chapter 11 Exercise 11.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/cRead more
Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.
(a) Area of parallelogram = base x height
= 12 x 7.6 = 91.2 cm²
(b) Area of parallelogram = base x height
⇒ 91.2 = 8 x QN
⇒ QN = 91.2/8 = 11.4 cm
We know that the area of triangle = 1/2 x base x height In first row, base = 15 cm and area = 87 cm² ∴ 87 = 1/2 x 15 x height ⇒ height = 87x2/15 11.6 cm In second row, height = 31.4 mm and area = 1256 mm² ∴ 1256 = 1/2 x base x 31.4 ⇒ base = 1256 x 2/31.4 = 80 mm In third row, base = 22 cm and area =Read more
We know that the area of triangle = 1/2 x base x height
In first row, base = 15 cm and area = 87 cm²
∴ 87 = 1/2 x 15 x height
⇒ height = 87×2/15 11.6 cm
In second row, height = 31.4 mm and area = 1256 mm²
∴ 1256 = 1/2 x base x 31.4
⇒ base = 1256 x 2/31.4 = 80 mm
In third row, base = 22 cm and area = 170.5 cm²
∴ 170.5 = 1/2 x 22 x height
⇒ height = 170.5 x 2/22 = 15.5 cm
We know that the area of parallelogram = base x height (a) Here, base = 20 cm and area = 246 cm² ∴ Area of parallelogram = base x height ⇒ 246 = 20 x height ⇒ height = 246/20 = 12.3 cm (b) Here, height = 15 cm and area = 154.5 cm² ∴ Area of parallelogram = base x height ⇒ 154.5 = base x 15 ⇒ base =Read more
We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246 cm²
∴ Area of parallelogram = base x height
⇒ 246 = 20 x height
⇒ height = 246/20 = 12.3 cm
(b) Here, height = 15 cm and area = 154.5 cm²
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15
⇒ base = 154.5/15 = 10.3 cm
(c) Here, height = 8.4 cm and area = 48.72 cm²
∴ Area of parallelogram = base x height
⇒ 48.72 = base x 8.4
⇒ base = 48.72/8.4 = 5.8 cm
(d) Here, base = 15.6 cm and area = 16.38 cm²
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height
⇒ height = 16.38/15.6 = 1.05 cm
We know that the area of triangle = 1/2 x base x height (a) Here, base = 4 cm and height = 3 cm ∴ Area of triangle = 1/3 x 4 x 3 = 6 cm² (b) Here, base = 5 cm and height = 3.2 cm ∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm² (c) Here, base = 3 cm and height = 4 cm ∴ Area of triangle = 1/2 x 3 x 4 = 6 cRead more
We know that the area of triangle = 1/2 x base x height
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/3 x 4 x 3 = 6 cm²
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm²
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 x 3 x 4 = 6 cm²
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 x 3 x 2 = 3 cm²
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m² Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m² Now, Area of wall excluding door = Area of wall including door – Area of door = 16.2 – 2 = 14.2 m² Since, The rate of white washing of 1 m2 the wall = ₹20 ThereforeRead more
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m²
Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m²
Now, Area of wall excluding door
= Area of wall including door – Area of door
= 16.2 – 2 = 14.2 m²
Since, The rate of white washing of 1 m2 the wall = ₹20
Therefore, the rate of white washing of 14.2 m2 the wall = 20 x 14.2 = ₹284
Thus, the cost of white washing the wall excluding the door is ₹284.
Perimeter of rectangle = 130 cm ⇒ 2 (length + breadth) = 130 cm ⇒ 2 (length + 30) = 130 ⇒ length + 30 = 130/2 ⇒ length + 30 = 65 ⇒ length = 65 – 30 = 35 cm ⇒ Now area of rectangle = length x breadth = 35 x 30 = 1050 cm² Thus, the area of rectangle is 1050 cm². Class 7 Maths Chapter 11 Exercise 11.1Read more
Perimeter of rectangle = 130 cm
⇒ 2 (length + breadth) = 130 cm
⇒ 2 (length + 30) = 130
⇒ length + 30 = 130/2
⇒ length + 30 = 65
⇒ length = 65 – 30 = 35 cm
⇒ Now area of rectangle = length x breadth = 35 x 30 = 1050 cm²
Thus, the area of rectangle is 1050 cm².
According to the question, Perimeter of square = Perimeter of rectangle ⇒ 4 x side = 2 (length + breadth) ⇒ 4 x side = 2 (40 + 22) ⇒ 4 x side = 2 x 62 ⇒ side = 2x62 = 31 Thus, the side of the square is 31 cm. Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm² And Area of square = side x sRead more
According to the question,
Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (length + breadth)
⇒ 4 x side = 2 (40 + 22)
⇒ 4 x side = 2 x 62
⇒ side = 2×62 = 31
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm²
And Area of square = side x side = 31 x 31 = 961 cm²
Therefore, on comparing, the area of square is greater than that of rectangle.
Given: The side of the square park = 60 m The length of the rectangular park = 90 m According to the question, Area of square park = Area of rectangular park ⇒ side x side = length x breadth ⇒ 60 x 60 = 90 x breadth ⇒ breadth = 60x60 = 40m Thus, the breadth of the rectangular park is 40 m. Class 7 MRead more
Given: The side of the square park = 60 m
The length of the rectangular park = 90 m
According to the question,
Area of square park = Area of rectangular park
⇒ side x side = length x breadth
⇒ 60 x 60 = 90 x breadth
⇒ breadth = 60×60 = 40m
Thus, the breadth of the rectangular park is 40 m.
Area of rectangular park = 440 m² ⇒ length x breadth = 440 m² ⇒ 22 x breadth = 440 ⇒ breadth = 440/20 = 20 m Now, Perimeter of rectangular park = 2 (length + breadth) = 2 (22 + 20) = 2 x 42 = 84 m Thus, the perimeter of rectangular park is 84 m. Class 7 Maths Chapter 11 Exercise 11.1 for more answerRead more
Area of rectangular park = 440 m²
⇒ length x breadth = 440 m²
⇒ 22 x breadth = 440
⇒ breadth = 440/20 = 20 m
Now, Perimeter of rectangular park = 2 (length + breadth)
= 2 (22 + 20)
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find: (a) the area of the parallelogram PRS (b) QN, if PS = 8 cm
Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm. (a) Area of parallelogram = base x height = 12 x 7.6 = 91.2 cm² (b) Area of parallelogram = base x height ⇒ 91.2 = 8 x QN ⇒ QN = 91.2/8 = 11.4 cm Class 7 Maths Chapter 11 Exercise 11.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/cRead more
Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.
(a) Area of parallelogram = base x height
= 12 x 7.6 = 91.2 cm²
(b) Area of parallelogram = base x height
⇒ 91.2 = 8 x QN
⇒ QN = 91.2/8 = 11.4 cm
Class 7 Maths Chapter 11 Exercise 11.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Find the missing values:
We know that the area of triangle = 1/2 x base x height In first row, base = 15 cm and area = 87 cm² ∴ 87 = 1/2 x 15 x height ⇒ height = 87x2/15 11.6 cm In second row, height = 31.4 mm and area = 1256 mm² ∴ 1256 = 1/2 x base x 31.4 ⇒ base = 1256 x 2/31.4 = 80 mm In third row, base = 22 cm and area =Read more
We know that the area of triangle = 1/2 x base x height
In first row, base = 15 cm and area = 87 cm²
∴ 87 = 1/2 x 15 x height
⇒ height = 87×2/15 11.6 cm
In second row, height = 31.4 mm and area = 1256 mm²
∴ 1256 = 1/2 x base x 31.4
⇒ base = 1256 x 2/31.4 = 80 mm
In third row, base = 22 cm and area = 170.5 cm²
∴ 170.5 = 1/2 x 22 x height
⇒ height = 170.5 x 2/22 = 15.5 cm
Class 7 Maths Chapter 11 Exercise 11.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/
Find the missing values:
We know that the area of parallelogram = base x height (a) Here, base = 20 cm and area = 246 cm² ∴ Area of parallelogram = base x height ⇒ 246 = 20 x height ⇒ height = 246/20 = 12.3 cm (b) Here, height = 15 cm and area = 154.5 cm² ∴ Area of parallelogram = base x height ⇒ 154.5 = base x 15 ⇒ base =Read more
We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246 cm²
∴ Area of parallelogram = base x height
⇒ 246 = 20 x height
⇒ height = 246/20 = 12.3 cm
(b) Here, height = 15 cm and area = 154.5 cm²
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15
⇒ base = 154.5/15 = 10.3 cm
(c) Here, height = 8.4 cm and area = 48.72 cm²
∴ Area of parallelogram = base x height
⇒ 48.72 = base x 8.4
⇒ base = 48.72/8.4 = 5.8 cm
(d) Here, base = 15.6 cm and area = 16.38 cm²
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height
⇒ height = 16.38/15.6 = 1.05 cm
Class 7 Maths Chapter 11 Exercise 11.2
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Find the area of each of the following triangles:
We know that the area of triangle = 1/2 x base x height (a) Here, base = 4 cm and height = 3 cm ∴ Area of triangle = 1/3 x 4 x 3 = 6 cm² (b) Here, base = 5 cm and height = 3.2 cm ∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm² (c) Here, base = 3 cm and height = 4 cm ∴ Area of triangle = 1/2 x 3 x 4 = 6 cRead more
We know that the area of triangle = 1/2 x base x height
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/3 x 4 x 3 = 6 cm²
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm²
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 x 3 x 4 = 6 cm²
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 x 3 x 2 = 3 cm²
Class 7 Maths Chapter 11 Exercise 11.2
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A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m².
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m² Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m² Now, Area of wall excluding door = Area of wall including door – Area of door = 16.2 – 2 = 14.2 m² Since, The rate of white washing of 1 m2 the wall = ₹20 ThereforeRead more
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m²
Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m²
Now, Area of wall excluding door
= Area of wall including door – Area of door
= 16.2 – 2 = 14.2 m²
Since, The rate of white washing of 1 m2 the wall = ₹20
Therefore, the rate of white washing of 14.2 m2 the wall = 20 x 14.2 = ₹284
Thus, the cost of white washing the wall excluding the door is ₹284.
Class 7 Maths Chapter 11 Exercise 11.1
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The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.
Perimeter of rectangle = 130 cm ⇒ 2 (length + breadth) = 130 cm ⇒ 2 (length + 30) = 130 ⇒ length + 30 = 130/2 ⇒ length + 30 = 65 ⇒ length = 65 – 30 = 35 cm ⇒ Now area of rectangle = length x breadth = 35 x 30 = 1050 cm² Thus, the area of rectangle is 1050 cm². Class 7 Maths Chapter 11 Exercise 11.1Read more
Perimeter of rectangle = 130 cm
⇒ 2 (length + breadth) = 130 cm
⇒ 2 (length + 30) = 130
⇒ length + 30 = 130/2
⇒ length + 30 = 65
⇒ length = 65 – 30 = 35 cm
⇒ Now area of rectangle = length x breadth = 35 x 30 = 1050 cm²
Thus, the area of rectangle is 1050 cm².
Class 7 Maths Chapter 11 Exercise 11.1
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A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
According to the question, Perimeter of square = Perimeter of rectangle ⇒ 4 x side = 2 (length + breadth) ⇒ 4 x side = 2 (40 + 22) ⇒ 4 x side = 2 x 62 ⇒ side = 2x62 = 31 Thus, the side of the square is 31 cm. Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm² And Area of square = side x sRead more
According to the question,
Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (length + breadth)
⇒ 4 x side = 2 (40 + 22)
⇒ 4 x side = 2 x 62
⇒ side = 2×62 = 31
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm²
And Area of square = side x side = 31 x 31 = 961 cm²
Therefore, on comparing, the area of square is greater than that of rectangle.
Class 7 Maths Chapter 11 Exercise 11.1
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The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 cm, find the breadth of the rectangular park.
Given: The side of the square park = 60 m The length of the rectangular park = 90 m According to the question, Area of square park = Area of rectangular park ⇒ side x side = length x breadth ⇒ 60 x 60 = 90 x breadth ⇒ breadth = 60x60 = 40m Thus, the breadth of the rectangular park is 40 m. Class 7 MRead more
Given: The side of the square park = 60 m
The length of the rectangular park = 90 m
According to the question,
Area of square park = Area of rectangular park
⇒ side x side = length x breadth
⇒ 60 x 60 = 90 x breadth
⇒ breadth = 60×60 = 40m
Thus, the breadth of the rectangular park is 40 m.
Class 7 Maths Chapter 11 Exercise 11.1
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The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Perimeter of the rectangular sheet = 100 cm ⇒ 2 (length + breadth) = 100 cm ⇒ 2 (35 + breadth) = 100 ⇒ 35 + breadth = 100/2 ⇒ 35 + breadth = 50 ⇒ breadth = 50 – 35 ⇒ breadth = 15 cm Now, Area of rectangular sheet = length x breadth = 35 x 15 = 525 cm² Thus, breadth and area of rectangular sheet areRead more
Perimeter of the rectangular sheet = 100 cm
⇒ 2 (length + breadth) = 100 cm
⇒ 2 (35 + breadth) = 100
⇒ 35 + breadth = 100/2
⇒ 35 + breadth = 50
⇒ breadth = 50 – 35
⇒ breadth = 15 cm
Now, Area of rectangular sheet = length x breadth
= 35 x 15 = 525 cm²
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm² respectively.
Class 7 Maths Chapter 11 Exercise 11.1
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Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Area of rectangular park = 440 m² ⇒ length x breadth = 440 m² ⇒ 22 x breadth = 440 ⇒ breadth = 440/20 = 20 m Now, Perimeter of rectangular park = 2 (length + breadth) = 2 (22 + 20) = 2 x 42 = 84 m Thus, the perimeter of rectangular park is 84 m. Class 7 Maths Chapter 11 Exercise 11.1 for more answerRead more
Area of rectangular park = 440 m²
⇒ length x breadth = 440 m²
⇒ 22 x breadth = 440
⇒ breadth = 440/20 = 20 m
Now, Perimeter of rectangular park = 2 (length + breadth)
= 2 (22 + 20)
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
Class 7 Maths Chapter 11 Exercise 11.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-11/