The area of the rectangle is 4225dm squared. Its width is equal to 6m dm.

| January 31, 2021 | education

The area of the rectangle is 4225dm squared. Its width is equal to 6m dm. Find the area of another rectangle, the width of which is thirteenth of the length of the first rectangle.

It can be seen that the condition is not given completely, we will solve the problem provided that the width of this rectangle is 6 m 5 dm.
Let’s imagine the width of the rectangle in decimetres:
6 m 5 dm = 6 * 10 + 5 = 65 (dm).
From the formula for the area of a rectangle, we find the length:
S1 = a1 * b1 → a1 = S1 / b1 = 4225/65 = 65 (dm).
Find the width of the second rectangle:
b2 = 1/13 * a1 = 1/13 * 65 = 5 (dm).
Nothing is said about the length of the second rectangle. Let us find the area of the second on the basis that its length is equal to the length of the first (a2 = a1 = 65 dm).
S2 = a2 * b2 = 65 * 5 = 325 (dm²).
Answer: the area of the second triangle is 325 dm².



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