Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle. Perimeter of field = 2 x (length + breadth) = 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km = 2.4 x 1000 m = 2400 m Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km https://www.tiwariaRead more
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 x (length + breadth)
= 2 x (0.7 + 0.5)
= 2 x 1.2
= 2.4 km
= 2.4 x 1000 m
= 2400 m
Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km
Length of wooden strip = Perimeter of photograph Perimeter of photograph = 2 x (length + breadth) = 2 (32 + 21) = 2 x 53 cm = 106 cm Thus, the length of the wooden strip required is equal to 106 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Length of wooden strip = Perimeter of photograph
Perimeter of photograph = 2 x (length + breadth)
= 2 (32 + 21)
= 2 x 53 cm
= 106 cm
Thus, the length of the wooden strip required is equal to 106 cm.
Length of table top = 2 m 25 cm = 2.25 m Breadth of table top = 1 m 50 cm = 1.50 m Perimeter of table top = 2 x (length + breadth) = 2 x (2.25 + 1.50) = 2 x 3.75 = 7.50 m Thus, the perimeter of table top is 7.5 m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 x (length + breadth)
= 2 x (2.25 + 1.50)
= 2 x 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.
Total length of tape required = Perimeter of rectangle = 2 (length + breadth) = 2 (40 + 10) = 2 x 50 = 100 cm = 1 m Thus, the total length of tape required is 100 cm or 1 m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Total length of tape required = Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 + 10)
= 2 x 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.
(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects. (b) Hindi. (c) Social Studies. (d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-9/
(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects.
(b) Hindi.
(c) Social Studies.
(d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle. Perimeter of field = 2 x (length + breadth) = 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km = 2.4 x 1000 m = 2400 m Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km https://www.tiwariaRead more
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 x (length + breadth)
= 2 x (0.7 + 0.5)
= 2 x 1.2
= 2.4 km
= 2.4 x 1000 m
= 2400 m
Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
See lessWhat is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Length of wooden strip = Perimeter of photograph Perimeter of photograph = 2 x (length + breadth) = 2 (32 + 21) = 2 x 53 cm = 106 cm Thus, the length of the wooden strip required is equal to 106 cm. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Length of wooden strip = Perimeter of photograph
Perimeter of photograph = 2 x (length + breadth)
= 2 (32 + 21)
= 2 x 53 cm
= 106 cm
Thus, the length of the wooden strip required is equal to 106 cm.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
See lessA table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Length of table top = 2 m 25 cm = 2.25 m Breadth of table top = 1 m 50 cm = 1.50 m Perimeter of table top = 2 x (length + breadth) = 2 x (2.25 + 1.50) = 2 x 3.75 = 7.50 m Thus, the perimeter of table top is 7.5 m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 x (length + breadth)
= 2 x (2.25 + 1.50)
= 2 x 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
See lessThe lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Total length of tape required = Perimeter of rectangle = 2 (length + breadth) = 2 (40 + 10) = 2 x 50 = 100 cm = 1 m Thus, the total length of tape required is 100 cm or 1 m. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
Total length of tape required = Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 + 10)
= 2 x 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-10/
See lessObserve this bar graph which shows the marks obtained by Aziz in half yearly examination in different subjects: Answer the given questions: (a) What information is does the bar graph give? (b) Name the subject in which Aziz scored maximum marks. (c) Name the subject in which he has scored minimum marks. (d) State the name of the subjects and marks obtained in each of them.
(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects. (b) Hindi. (c) Social Studies. (d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-9/
(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects.
(b) Hindi.
(c) Social Studies.
(d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-9/
See less