Question

A rope of length 60 cm is cut into two pieces. One piece is used to form a rectangle of length 12 cm and breadth 6 cm. The other piece is shaped into a hexagon. What is the length of each side of the hexagon?​

Answer 1

Answer:

4 cm each side of hexagon

Step-by-step explanation:

perimeter of rectangle is 2( 12+6 ) = 36

60-36= 24 and hexagon has 6 sides then, 24 ÷ 6 = 4 ,, So , the answer is 4 cm

Answer 2

This is your Answer

Length of wire = 50 m (Given)

Length of wire = 50 m (Given)Let length of one piece for shape of square

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²A = (x ^ 2)/16 + ((50 - x) ^ 2)/(4pi)

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²A = (x ^ 2)/16 + ((50 - x) ^ 2)/(4pi)Differentiating w.r. to x, we get (dA)/(dx) = (2x)/16 + (2(50 - x) * (- 1))/(4pi)

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²A = (x ^ 2)/16 + ((50 - x) ^ 2)/(4pi)Differentiating w.r. to x, we get (dA)/(dx) = (2x)/16 + (2(50 - x) * (- 1))/(4pi)(dA)/(dx) = x/8 + (x - 50)/(2pi)

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²A = (x ^ 2)/16 + ((50 - x) ^ 2)/(4pi)Differentiating w.r. to x, we get (dA)/(dx) = (2x)/16 + (2(50 - x) * (- 1))/(4pi)(dA)/(dx) = x/8 + (x - 50)/(2pi)лx + 4x 200

Length of wire = 50 m (Given)Let length of one piece for shape of square= xm=(50 - x)m...Length of other piece for shape of circle Now perimeter of square= 4a = x X ⇒a=4 and circumference of circle=2str = 50 -X50 - x 2元 ⇒ r =Combined Area = a ^ 2 + pi * r ^ 2x² 16 十介 • (50-x) ²x² 16 +π. (50 - x)² 4л²A = (x ^ 2)/16 + ((50 - x) ^ 2)/(4pi)Differentiating w.r. to x, we get (dA)/(dx) = (2x)/16 + (2(50 - x) * (- 1))/(4pi)(dA)/(dx) = x/8 + (x - 50)/(2pi)лx + 4x 2008元 x(4+ л) - 200

For extremum, (dA)/(dx) = 0

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)(d ^ 2 * A)/(d * x ^ 2) > 0

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)(d ^ 2 * A)/(d * x ^ 2) > 0A is minimum at x = 200/(4 + pi)

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)(d ^ 2 * A)/(d * x ^ 2) > 0A is minimum at x = 200/(4 + pi)...Length of square of square wire, x =

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)(d ^ 2 * A)/(d * x ^ 2) > 0A is minimum at x = 200/(4 + pi)...Length of square of square wire, x =200

For extremum, (dA)/(dx) = 0x(4 + pi) - 200 = 0x = 200/(4 + pi)(d ^ 2 * A)/(d * x ^ 2) > 0A is minimum at x = 200/(4 + pi)...Length of square of square wire, x =200-m 4