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Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset Vaia Explanations$ Math

Student Edition · 227 pages

ISBN: 9780395977279

Chapter 8: Q3. (page 291)

Complete the simplification.
${(\sqrt{3})}^{2} = \sqrt{3} \cdot ? = ?$

Short Answer

Expert verified

The simplified form is,

${(\sqrt{3})}^{2} = \sqrt{3} . \sqrt{3} = 3$

Step by step solution

01

Step 1. Given information.

The given expression is,

${(\sqrt{3})}^{2} = \sqrt{3} . \underline{?}$

02

Step 2. It can be further written as below.

We have the given expression is ${(\sqrt{3})}^{2}$

Then,

$\begin{matrix} {(\sqrt{3})}^{2} = \sqrt{3} . \sqrt{3} \\ = \sqrt{3.3} \\ = \sqrt{{(3)}^{2}} \\ = 3 \end{matrix}$

03

Step 3. Neglect negative sign.

Since, square root is positive quantity

So, we neglect negative sign.

Therefore, the simplified form is, ${(\sqrt{3})}^{2} = \sqrt{3} . \sqrt{3} = 3$ .

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Most popular questions from this chapter

Simplify each expression.

${(\frac{\sqrt{2}}{2})}^{2}$

The arithmetic mean between two numbers $r$ and $s$ is defined to be $\frac{r + s}{2}$ .

$(a) .$ $\bar{C M}$ is the median and $\bar{C H}$ is the altitude to the hypotenuse of right $Δ A B C$ . Show that $C M$ is the arithmetic mean between $A H$ and $B H$ , and that $C H$ is the geometric mean between $A H$ and $B H$ . Then use the diagram to show that the arithmetic mean is greater than the geometric mean.

$(b) .$ Show algebraically that the arithmetic mean between two different numbers $r$ and $s$ is greater than the geometric mean.

$(a) .$

Find the value of $x$ from the given figure.

Complete the simplification.

${(\sqrt{5})}^{2}$

The early Greeks thought of the Pythagorean Theorem in this form: The area of the square on the hypotenuse of a right triangle equals the sum of the areas of the squares on the legs. Draw a diagram to illustrate that interpretation.

See all solutions

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