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Sauron [17]
3 years ago
7

Suppose that you had two squares, a small one and a large one. The area of the large square is twice that of the small square. H

ow many times larger is the side length of the large square than the side length of the small square?
Mathematics
1 answer:
WARRIOR [948]3 years ago
7 0

Answer:

The side length of the large square is √2 times larger than the side length of the small square.

Step-by-step explanation:

Suppose we have a small square (square 1) and a large square (square 2). The area of the large square is twice that of the small square, that is,

A₂ = 2 A₁

A₂/A₁ = 2 [1]

The area of a square is equal to the length of the side (l) raised to the second power.

A = l²

l = √A

The ratio of l₂ to l₁ is:

l₂/l₁ = √A₂ / √A₁ = √(A₂/A₁)

We can replace [1] in the previous expression.

l₂/l₁ = √2

The side length of the large square is √2 times larger than the side length of the small square.

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\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
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Hope it helps

Step-by-step explanation:

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