Answer :
The area of the larger pentagon is 30 m².
If the two pentagons are similar, it means they have the same shape but possibly different sizes. In similar polygons, the ratio of corresponding sides is constant, and the ratio of corresponding areas is the square of that constant ratio.
Let's denote the ratio of corresponding sides of the smaller pentagon to the larger pentagon as r. Since the area is proportional to the square of the side length in similar figures, the ratio of their areas is [tex]\( r^2 \).[/tex]
Given that the area of the smaller pentagon is 30 m², we need to find [tex]\( r \)[/tex] first and then use [tex]\( r^2 \)[/tex] to find the area of the larger pentagon.
Let's say the side length of the smaller pentagon is s meters. Then, the side length of the larger pentagon would be rs meters.
Since the area of the smaller pentagon is given as 30 m²:
[tex]\[ \text{Area of smaller pentagon} = \frac{5}{4} \times s^2 \][/tex]
[tex]\[ 30 = \frac{5}{4} \times s^2 \][/tex]
Solve for s:
[tex]\[ s^2 = \frac{30 \times 4}{5} \]\[ s^2 = 24 \]\[ s = \sqrt{24} \]\[ s = 2\sqrt{6} \][/tex]
Now, since [tex]\( r = \frac{\text{side length of larger pentagon}}{\text{side length of smaller pentagon}} = \frac{rs}{s} = r \)[/tex], we have:
[tex]\[ r = \frac{\text{side length of larger pentagon}}{\text{side length of smaller pentagon}} = \frac{2\sqrt{6}}{s} \][/tex]
[tex]\[ r = \frac{2\sqrt{6}}{2\sqrt{6}} = 1 \][/tex]
Therefore, the area of the larger pentagon is r² times the area of the smaller pentagon:
[tex]\[ \text{Area of larger pentagon} = r^2 \times \text{Area of smaller pentagon} \]\[ \text{Area of larger pentagon} = 1^2 \times 30 \]\[ \text{Area of larger pentagon} = 30 \text{ m}^2 \][/tex]
The complete question is :
The pentagons at the right are similar. The area of the smaller pentagon is 30 m². What is the area of the larger pentagon in m²?
