Given: Area of parallelogram = 1470 cm² Base (AB) = 35 cm and base (AD) = 49 cm Since Area of parallelogram = base x height ⇒ 1470 = 35 x DL ⇒ DL = 1470/35 ⇒ DL = 42 cm Again, Area of parallelogram = base x height ⇒ 1470 = 49 x BM ⇒ BM = 1470/49 ⇒ BM = 30 cm Thus, the lengths of DL and BM are 42 cmRead more
Given: Area of parallelogram = 1470 cm²
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ DL = 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ BM = 1470/49
⇒ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
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.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Given: Area of parallelogram = 1470 cm² Base (AB) = 35 cm and base (AD) = 49 cm Since Area of parallelogram = base x height ⇒ 1470 = 35 x DL ⇒ DL = 1470/35 ⇒ DL = 42 cm Again, Area of parallelogram = base x height ⇒ 1470 = 49 x BM ⇒ BM = 1470/49 ⇒ BM = 30 cm Thus, the lengths of DL and BM are 42 cmRead more
Given: Area of parallelogram = 1470 cm²
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ DL = 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ BM = 1470/49
⇒ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Class 7 Maths Chapter 11 Exercise 11.2
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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
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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:
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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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