Here are two squares, A and B. A B The length of each side of square B is 4cm greater than the length of each side of square A. The area of square B is 70cm2 greater than the area of square A. Find the area of square B. Give your answer correct to 3 significant fi

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Answer:

La²+8La+16 = La²+70

Step-by-step explanation:

8La+16 = 70

8La=70-16

La=54/8

La= 6.75

Area b = (6.75+4)*(6.75+4)

Area b = 10.75*10.75

Area b =115.563

The area of a square is the square of its side lengths

The area of square B is 116 centimeter square

Let the side length of square A be (a), and the side length of square B be (b)

So, we have:

[tex]\mathbf{A_a = a^2}[/tex]

[tex]\mathbf{A_b = b^2}[/tex]

From the question, we have:

[tex]\mathbf{A_b = 70 + a^2}[/tex]

[tex]\mathbf{b = 4 + a}[/tex]

Take square of both sides

[tex]\mathbf{b^2 = (4 + a)^2}[/tex]

[tex]\mathbf{b^2 = 16 + 8a+ a^2}[/tex]

Substitute [tex]\mathbf{b^2 = 16 + 8a+ a^2}[/tex] in [tex]\mathbf{A_b = b^2}[/tex]

[tex]\mathbf{A_b = 16 + 8a+ a^2}[/tex]

So, we have:

[tex]\mathbf{16 + 8a+ a^2 = 70 + a^2}[/tex]

Subtract a^2 from both sides

[tex]\mathbf{16 + 8a = 70}\\[/tex]

Subtract 16 from both sides

[tex]\mathbf{8a = 54}[/tex]

Divide both sides by 8

[tex]\mathbf{a = 6.75}[/tex]

Recall that:

[tex]\mathbf{A_b = 70 + a^2}[/tex]

So, we have:

[tex]\mathbf{A_b = 70 + 6.75^2}[/tex]

[tex]\mathbf{A_b = 115.5625}[/tex]

Approximate

[tex]\mathbf{A_b = 116}[/tex]

Hence, the area of square B is 116 centimeter square

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