Question

Hi there!

Construct the "Square root spiral". Take a large sheet of paper and construct the "Square root spiral" in the following fashion.

Start with a point O and draw a line segment $\rm{P_{1} P_{2}}$ perpendicular to $\rm{OP_{1}}$ of unit length. Now draw a line segment $\rm{P_{2} P_{3}}$ perpendicular to $\rm{OP_{2}}$ . Then draw a line segment $\rm{P_{3} P_{4}}$ perpendicular to $\rm{OP_{3}}$. Continuing in this matter, you get line segment of unit length perpendicular to $\rm{OP_{n-1}}$. In this manner, you will have created the points $\rm{P_{2}, P_{3},...... P_{n}....}$, and joined them to create a beautiful spiral depicting $\rm{\sqrt{2}, \sqrt{3}, \sqrt{4},...}$

~ Answers the process step by step~.​

Answer 1

♕$\purple{ \underbrace{ \bm{Explanation}}}$♕

Starting from point O draw a line of 1 cm. Place the protractor at the edge of 1 cm line and make the point at 90°. Now draw a line segment of 1 unit length i.e., 1 cm uptil $\rm{P_1}$ looking at the point 90°. Now join point O and $\rm{P_1}$

Remember :- Perpendicular will always be equal to 1 cm.

$\blue{ \rm { We \: know \: that :-}}$

◑Base = √1 = 1

◑Perpendicular = 1

⛥Hence, by using Pythagoras Theorem :-

H² = B² + P²

➯$\sf{ \small{ OP_1 = \sqrt{ {1}^{2} + {1}^{2} }}}$

$= \sqrt{ {1} + {1} } = \sqrt{2}$

Now, the hypotenuse (√2 ) of first triangle becomes the base of next triangle.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☃Again repeating the same steps :-

Place the protractor at the edge of $\rm{OP_1}$ and in the same way make the point at 90°. Now draw a line segment of 1 unit length (1 cm) upto $\rm{P_2}$ keeping scale straight at the point 90°.

Now join point O to $\rm{P_2}$

Perpendicular is again 1 cm.

$\blue{ \rm{We \: know \: that :-}}$

◑Base = √2

◑Perpendicular = 1

⛥By using Pythagoras Theorem :-

➯$\sf{ \small{OP_2 = \sqrt{ (\sqrt {2})^{2} + {1}^{2} } }}$

$= \sqrt{2 + 1} = \sqrt{3}$

Now, the hypotenuse (√3 ) of second triangle becomes the base of third triangle.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☃Again repeating the same steps :-

Place The protractor at the edge of $\rm{OP_2}$ and again make the point at 90°. Now draw a line segment of 1 unit length (1 cm) upto $\rm{P_3}$ keeping scale straight at the point 90°. Now join point O to $\rm{P_3}$

Perpendicular is again 1 cm.

$\blue{ \rm{We \: know \: that :-}}$

◑Base = √3

◑Perpendicular = 1

⛥By using Pythagoras Theorem :-

➯${ \sf{ \small{ OP_3 = \sqrt {(\sqrt{3})^{2} + {1}^{2} }}} }$

$= \sqrt{3 + 1} = \sqrt{4} = 2$

Here we will not take 2 because we are constructing here square root spiral. Therefore, we will take √4 here.

Now, the hypotenuse of third triangle becomes the base of fourth triangle.

So by repeating same steps you can construct square root spiral of other numbers.

I have construct upto √17 in the attachment above ⬆️.

Answer 2

Answer:

(◍•ᴗ•◍)❤ JAI SIYA RAM (◍•ᴗ•◍)❤

Base = √1 = 1

Prependicular = 1

Hence , by using Pythagoras theorem :-

H² = B² + P²

OP¹ = √ 1² + 1²

= √ 1 + 1 = √2

Now , the hypotenuse ( √2 ) of first triangle becomes the base of next triangle .

_______________________________________

Again , repeating the same steps :-

Place the protractor at the edge of OP¹ and in the same way make the protractor upto P² keeping scale straight at the point 90° .

Now , join point O to P² .

Prependicular is again 1 cm .

We know that :-

Base = √2

Prependicular = 1

By using Pythagoras theorem :-

OP² = √ ( √2 ) ² + 1²

= √ 2 + 1 = √3

Now , the hypotenuse ( √3 ) of second triangle becomes the base of third triangle .

_______________________________________

Again , repeating the same steps :-

Place the protractor at the edge of OP² and again make the points at 90°. Now , make a line segment of 1 unit length ( 1cm ) up to P³ keeping scale straight at the point 90° . Now , join point O to P³ .

Prependicular is again 1 cm .

(◍•ᴗ•◍)❤ Jai Siya Ram (◍•ᴗ•◍)❤