The student drew a rectangle and a square with equal perimeters, the length of the rectangle

| September 7, 2021 | education

The student drew a rectangle and a square with equal perimeters, the length of the rectangle is 14cm 1dm more than the width. Find the area of the rectangle and square.

To solve this problem, recall the formula for the area of ​​a rectangle. The area of ​​the rectangle is equal to the product of the length and the width. S = a * b, where a is the length and b is the width. The perimeter of a rectangle is the sum of the lengths of all its sides. Since in a rectangle the opposite sides are equal, then P = 2 * (a + b), where a is the length, b is the width. Knowing that the length is 14 cm, and the width is 1 dm = 10 cm less, we calculate the width.
b = 14 – 10 = 4 cm.
Let’s calculate the perimeter of the rectangle.
P = 2 * (4 + 14) = 2 * 18 = 36 cm.
Let’s calculate what the side of the square is. The perimeter of a square is equal to the sum of the lengths of all its four sides. Since all sides of a square are equal, its perimeter P = 4a, where a is its side. The area of ​​a square is equal to the square of its side. S = a ^ 2.
a = 36/4 = 9 cm.
Let’s calculate the area of ​​the square.
S = 9 * 9 = 81 sq. Cm.
Let’s calculate the area of ​​the rectangle.
S = 4 * 14 = 56 sq. Cm.
Answer: 81 sq. Cm., 56 sq. Cm.



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