Q: Can a regular pentagon with a side measuring 5 cm be congruent to another regular pentagon with a of perimeter 25cm?

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Yes, it definately can and always will be. Since you wrote that both pentagons are regular, all sides (and angles) are equivalent to each other. On the first pentagon, one side measured 5cm, so therefore the rest of the sides total up to 25cm. Since the second pentagon has perimeter 25cm, the pentagons must be congruent.

No, a parallelogram is 4 sided with each 2 sides being parallel to one another. The opposite angles in a parallelogram are congruent. A pentagon is a 5 sided figure with each angle being the same.

The perimeter of a rectangle is the distance all the way round it.That's lenght (4.5m) + a breadth (1.5m) + another length (4.5m) + another breadth (1.5m) = 12m.Algebraically, the perimeter is 2(l + b), where l is the length and b is the breadth, = 2 x (4.5m + 1.5m) = 2 x 6m = 12m.

No. You can have two triangles that are congruent to one another, and two quadrilaterals that are congruent to one another. But the triangle cannot be similar to the quadrilateral!

If you're looking for the length of each of the 6 sides it's 16. To find the perimeter of an object you just add the length of each side to one another. In this case when you already know the perimeter you divide 96 by 6(which represents the six sides of the hexagon).

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Yes, it definately can and always will be. Since you wrote that both pentagons are regular, all sides (and angles) are equivalent to each other. On the first pentagon, one side measured 5cm, so therefore the rest of the sides total up to 25cm. Since the second pentagon has perimeter 25cm, the pentagons must be congruent.

No, a parallelogram is 4 sided with each 2 sides being parallel to one another. The opposite angles in a parallelogram are congruent. A pentagon is a 5 sided figure with each angle being the same.

Answer: Yes. A polygon can have the same perimeter length but smaller area than another polygon. Answer: For congruent or similar shapes, no. For different shapes, yes. Consider, for example, a rectangle 3 x 1, and another rectangle 2 x 2. They have different areas, but the same perimeter.

A pentagon is a 5 sided polygon

The perimeter of a rectangle is the distance all the way round it.That's lenght (4.5m) + a breadth (1.5m) + another length (4.5m) + another breadth (1.5m) = 12m.Algebraically, the perimeter is 2(l + b), where l is the length and b is the breadth, = 2 x (4.5m + 1.5m) = 2 x 6m = 12m.

Another word for the adjective congruent is agreeable. Additional synonyms for the word congruent are harmonious, coinciding, compatible, and in agreement.

If two angles in a triangle are congruent to two angles in another triangle, then the ______________ angles are also congruent.

No. You can have two triangles that are congruent to one another, and two quadrilaterals that are congruent to one another. But the triangle cannot be similar to the quadrilateral!

Another trapezoid that has the same dimensions and angles will be congruent to it.

If you're looking for the length of each of the 6 sides it's 16. To find the perimeter of an object you just add the length of each side to one another. In this case when you already know the perimeter you divide 96 by 6(which represents the six sides of the hexagon).

Since the question does not say so, you may not assume that the pentagon is regular. One way to find the area is to select any point inside the pentagon and join it to each of the vertices. This will divide the pentagon into 5 triangles. You can then measure the sides of each triangle and thereby calculate its area. Then sum the areas of the triangles. You could also select one side in each triangle as the base and then draw and measure the perpendicular distance to the opposite vertex. That is another way to find the area of each triangle. There are other methods, too. To find the perimeter you will need to measure the length of each side of the pentagon and add these lengths together.

When measuring one point to another point you are measuring distance.