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a. use the diagram. find the areas of square efgh and square ejkl. area…

Question

a. use the diagram. find the areas of square efgh and square ejkl. area of efgh = square units area of ejkl = square units b. what happens to the area when the perimeter of square efgh is doubled? explain the area is doubled because when the side length becomes 2s, the area becomes s² + s² the area is doubled because when the side length becomes 2s, the area becomes 2s² the area is quadrupled because when the side length becomes 2s, the area becomes (2s)² = 4s² the area becomes 8 times greater because when the side length becomes 2s the area becomes 2(2s)² = 8s²

Explanation:

Step1: Find side - length of square EFGH

The coordinates of E(0,0) and F(0,2), G(2,2), H(2,0). The side - length of square EFGH, $s_{1}$, can be found using the distance formula between two adjacent vertices. For example, between E(0,0) and F(0,2), the distance $d=\sqrt{(0 - 0)^2+(2 - 0)^2}=2$. The area of a square is $A = s^2$, so for square EFGH with side - length $s_{1}=2$, the area $A_{1}=s_{1}^2=2^2 = 4$.

Step2: Find side - length of square EJKL

The coordinates of E(0,0), J(0,4), K(4,4), L(4,0). The side - length of square EJKL, $s_{2}$, using the distance formula between two adjacent vertices, say E(0,0) and J(0,4), $d=\sqrt{(0 - 0)^2+(4 - 0)^2}=4$. The area of square EJKL is $A_{2}=s_{2}^2=4^2 = 16$.

Step3: Analyze the effect of doubling the perimeter on the area of square EFGH

The perimeter of a square is $P = 4s$. If the perimeter of square EFGH is doubled, since $P_{1}=4s_{1}$ (where $s_{1}=2$ and $P_{1}=8$), when it is doubled, the new perimeter $P_{new}=16$. Then, from $P_{new}=4s_{new}$, we have $s_{new}=4$. The original area of square EFGH is $A_{1}=s_{1}^2 = 4$, and the new area $A_{new}=s_{new}^2=16$. If the original side - length is $s$ and the new side - length is $2s$, the original area $A = s^2$ and the new area $A'=(2s)^2=4s^2$. So the area is quadrupled.

Answer:

Area of EFGH = 4 square units
Area of EJKL = 16 square units
b. The area is quadrupled because when the side length becomes 2s, the area becomes $(2s)^2=4s^2$