Question

To represent $\sqrt{6}$ on a number line, which of the following base and perpendicular lengths can help?

• A. $\sqrt{3}\&\sqrt{3}$
• B. $\sqrt{2}\&2$
• C. $1\&\sqrt{5}$

Answer 1

Answer: (A), (B), (C)

$1\&\sqrt{5}$

From pythagorean theorem, ${\text{hypotenuse}}^2={\text{base}}^2+{\text{perpendicular}}^2$
$\Rightarrow (\sqrt{6}{)}^2=6=3+3=(\sqrt{3}{)}^2+(\sqrt{3}{)}^2$

To represent $\sqrt{6}$ on a number line, we perform the following steps:

1. Draw a line of length $\sqrt{3}$ on a number line with 0 on its left end.
2. Draw a perpendicular line of length $\sqrt{3}$ on the number line from the right end of the first line.
3. Connect the other endpoints of both the lines to draw the hypotenuse.
4. Measure the length of the hypotenuse using a compass and mark it on the number line to represent $\sqrt{6}.$

Other possible values of $\text{base}$ and $\text{perpendicular}$ are:

(b) $\sqrt{2}$ and $2$
$(\sqrt{2}{)}^2+(2{)}^2=2+4=6=(\sqrt{6}{)}^2$

1. Draw a line of length $\sqrt{2}$ on a number line with 0 on its left end.
2. Draw a perpendicular line of length $2$ on the number line from the right end of the first line.
3. Connect the other endpoints of both the lines to draw the hypotenuse.
4. Measure the length of the hypotenuse using a compass and mark it on the number line to represent $\sqrt{6}.$

(c) $1$ and $\sqrt{5}$
$(1{)}^2+(\sqrt{5}{)}^2=1+5=6=(\sqrt{6}{)}^2$

1. Draw a line of length $\sqrt{5}$ on a number line with 0 on its left end.
2. Draw a perpendicular line of length $1$ on the number line from the right end of the first line.
3. Connect the other endpoints of both the lines to draw the hypotenuse.
4. Measure the length of the hypotenuse using a compass and mark it on the number line to represent $\sqrt{6}.$