How many distinct arrangements are possible using the word CHEER
1 answer:
You might be interested in
Answer:
The perimeter of other rectangles are 50, 28, 22 units
The area of all possible rectangles are same and equals to 24 unit²
Step-by-step explanation:
Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.
To find - What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?
Proof -
We know that
Are of Rectangle = Length × Breadth
And Perimeter of Rectangle = 2 (Length + Breadth )
Now,
Given that,
Orbet used 24 square tiles
And perimeter of the rectangle made = 20
So,
Possible Length of rectangle = 6
Possible breadth of Rectangle = 4
Or Vice-versa.
Total number of Rectangles possible = 4
Possibilities are - 1 × 24, 2 × 12, 3 × 8, 4 × 6
Case I :
Length of rectangle = 1
Breadth of rectangle = 24
∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units
Area of rectangle = 1 × 24 = 24 unit²
Case II :
Length of rectangle = 2
Breadth of rectangle = 12
∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units
Area of rectangle = 2 × 12 = 24 unit²
Case III :
Length of rectangle = 3
Breadth of rectangle = 8
∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units
Area of rectangle = 3 × 8 = 24 unit²
Case IV :
Length of rectangle = 4
Breadth of rectangle = 6
∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units
Area of rectangle = 4 × 6 = 24 unit²
∴ we get
The perimeter of other rectangles are 50, 28, 22 units
The area of all possible rectangles are same and equals to 24 unit²
