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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: Q5. (page 292)

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

Short Answer

Expert verified

The value of $x$ is $10 \sqrt{3}$ .

Step by step solution

01

Step 1. Given information.

The two sides are given 10 and 20.

02

Step 2. Theorem used.

We use Pythagoras Theorem.

Pythagoras Theorem state that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of other sides.

03

Step 3. Now find the value.

Calculation:

$\begin{matrix} 10^{2} + x^{2} = 20^{2} \\ 100 + x^{2} = 400 \\ x^{2} = 300 \end{matrix}$

Taking square root we get.

$\begin{array}{l} x = \pm \sqrt{300} \\ x = \pm 3 \sqrt{10} \end{array}$

Since, distance is a positive quantity. So, we neglect negative sign.

Therefore, the value of $x$ is $3 \sqrt{10}$ .

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Most popular questions from this chapter

State an equation you could use to find the value of $x$ . Then find the value of $x$ in simplest radical form.

Find the geometric mean between the two numbers.

$49$ and $25$

Find the values of $x, y,$ and $z$ .

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 length of the diagonal of the square with perimeter is $16$ .

See all solutions

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