Answer:You can make a square have any area you want it to, by making its sideslonger or shorter. You just have to keep all the sides the same length andone of its angles 90 degrees, otherwise you won't have a square any more.-- You can make a trapezoid have any area you want it to, by making its sideslonger or shorter. You just have to keep one pair of sides parallel, otherwiseyou won't have a trapezoid any more.-- Since you can make a square with any area you want, and you can make atrapezoid with any area you want, there's no reason why you can't make oneof each, that both have the same area.-- And while you're at it, you could also make a circle that has the same areaas the square and the trapezoid.
Step-by-step explanation:
Yes, a trapezoid and a square can have the same area. Area is the measure of the space inside a shape and two different shapes can occupy the same amount of space. An example is provided to illustrate this.
Yes, a trapezoid and a square can have the same area. Area is simply a measure of the space inside a shape, and it's entirely possible for two different shapes to occupy the same amount of space. For instance, consider a square of side length 4 units. Its area would be 16 square units. Now consider a trapezoid with base lengths of 3 units and 5 units, and a height of 4 units. The area of a trapezoid is calculated as ((base1+base2)/2)*height, which would also give us 16 square units in this case. So, these two shapes - the square and the trapezoid - have the same area.
Learn more about Area of Shapes here:
#SPJ11
Answer:
5 x 5
Step-by-step explanation:
Hope this helps.
From a list of ten books, 210 ways to select a group of 4 books.
A combination is a selection of all or a portion of a group of items, regardless of the sequence in which the items are chosen.
The number of ways to select a group of 4 books from a list of 10 books can be calculated using the combination formula:
nCr = n! / (r! x (n-r)!)
Where n is the total number of items (in this case, 10 books) and r is the number of items to be selected (in this case, 4 books).
Plugging in the values, we get:
10C4 = 10! / (4! x (10-4)!)
= 10! / (4! x 6!)
= (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)
= 210
Therefore, there are 210 ways to select a group of 4 books from a list of 10 books.
To learn more about the combination;
#SPJ3
First you have all ten books, but then every time after, you loose one because you can't put the same book in the group twice.
10 * 9 * 8 * 7 = 5040 groups