Two squares have side lengths that are consecutive odd integers. The total area of the squares is 650cm^2. What is their total p

erimeter?

Mathematics

1 answer:

Alenkinab [10]3 years ago

4 0

$the\ consecutive\ odd\ integers\ is:2n+1\ and\ 2n+3\ (n\in\mathbb{N})\\\\Area_1=(2n+1)^2=4n^2+4n+1\\\\Area_2=(2n+3)^2=4n^2+12n+9\\\\Area_1+Area_2=4n^2+4n+1+4n^2+12n+9=8n^2+16n+10\\\\8n^2+16n+10=650\\\\8n^2+16n+10-650=0\\\\8n^2+16n-640=0\ \ /:8$

$n^2+2n-80=0\\\\\Delta=2^2-4\cdot1\cdot(-80)=4+320=324;\ \sqrt\Delta=\sqrt{324}=18\\\\n_1=\frac{-2-18}{2\cdot1} < 0;\ n_2=\frac{-2+18}{2\cdot1}=\frac{16}{2}=8\\\\2n+1=2\cdot8+1=16+1=17\ (m)\\\\2n+3=2\cdot8+3=16+3=19\ (m)\\\\\\the\ total\ perimeter=4\cdot17+4\cdot19=68+76=144\ (m)$

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Use the linear combination method to solve the system of equations. <br> 2a=6−2 <br> 2a+3b=16

PSYCHO15rus [73]

Answer: $a=2$ and $b=4$

Step-by-step explanation:

To solve the system of equations with the linear combination method, you must:

- Arrange the equations in a colum.

- Make one of the set of terms, with equal variables, opposite in signs but equal in coefficients.

- Add both equations to cancel out that set of terms.

- Solve for the the variable left.

- Substitute the value of that variable into one of the original equations and solve for the other variable to find its value.

As you can see, the first equation of the system of equations given in the problem, does not have the variable $b$ , so you can't apply the proccedure shown above. Then, to solve it you must: