Question

4. The ratio of the length of the corresponding side of two

regular polygons is 3:4. The area of the larger polygon is
320 m². What is the area of the smaller polygon?


A-240 m²
B-427 m²
C-569 m²
D-180 m²

Answer 1

If the ratio of the length of the corresponding side of two regular polygons is 3:4, then the ratio of their areas is (3/4)² = 9/16.

Let A be the area of the smaller polygon. Then we have:

A x (9/16) = 320 m²

Multiplying both sides by (16/9), we get:

A = (320 m²) x (16/9) = 568.89 m²

Rounding to the nearest whole number, we get:

A ≈ 569 m²

Therefore, the area of the smaller polygon is approximately 569 m².

The answer is (C) 569 m².

Answer 2

Let's assume that the smaller polygon has a length of 3x, where x is a positive number representing the common ratio. Similarly, the length of the larger polygon would be 4x.

The ratio of the areas of two similar polygons is equal to the square of the ratio of their corresponding side lengths. Therefore, the ratio of the areas of the smaller and larger polygons would be (3x)^2 : (4x)^2, which simplifies to 9x^2 : 16x^2.

Given that the area of the larger polygon is 320 m², we can set up the following equation:

9x^2 : 16x^2 = Area of smaller polygon : 320

Cross-multiplying, we get:

9x^2 * 320 = 16x^2 * Area of smaller polygon

2880x^2 = 16x^2 * Area of smaller polygon

Cancelling out x^2, we have:

2880 = 16 * Area of smaller polygon

Dividing both sides by 16, we find:

Area of smaller polygon = 2880 / 16 = 180 m²

Therefore, the area of the smaller polygon is 180 m² (option D).