Find the lengths of the hypotenuses of all the right triangles in the given figure which is referred to as the square root spiral. Class 9

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· Jul 06, 2026 · Reviewed & updated Sep 17, 2026 · 1 min read

Find the lengths of the hypotenuses of all the right triangles in the given figure which is referred to as the square root spiral. Class 9

Question 1.

Find the lengths of the hypotenuses of all the right triangles in the given figure which is referred to as the square root spiral. Class 9

Solution:

In the square root spiral, each successive right triangle is formed such that:

One leg is always 1 unit

The other leg is the hypotenuse of the previous triangle.

Using Baudhayana - Pythagoras theorem, we find each new hypotenuse.

Triangle 1:

Legs are 1 unit and 1 unit

H1 = [latex]\sqrt{\left(1^2+1^2\right)}[/latex]

= [latex]\sqrt{1+1}[/latex] = [latex]\sqrt{2}[/latex] units.

Triangle 2:

Legs are [latex]\sqrt{2}[/latex] units and 1 unit

H2 = $\sqrt{\left((\sqrt{2})^2+1^2\right)}$

= $\sqrt{2+1}$ = [latex]\sqrt{3}[/latex] units

Triangle 3:

Legs are [latex]\sqrt{3}[/latex] units and 1 unit

H3 = $\sqrt{\left((\sqrt{3})^2+1^2\right)}$

= $\sqrt{3+1}$

= $\sqrt{4}$ = 2 units.

Triangle 4:

Legs are 2 units and 1 unit

H4 = $\sqrt{\left(2^2+1^2\right)}$

= $\sqrt{4+1}$ = $\sqrt{5}$ units.

Triangle 5:

Legs are $\sqrt{5}$ units and 1 unit

H5 = $\sqrt{\left((\sqrt{5})^2+1^2\right)}$

= $\sqrt{5+1}$ = $\sqrt{6}$ units.

Triangle 6:

Legs are $\sqrt{6}$ units and 1 unit

H6 = $\sqrt{\left((\sqrt{6})^2+1^2\right)}$ = $\sqrt{6+1}$ = $\sqrt{7}$ units.

Triangle 7:

Legs are $\sqrt{7}$ units and 1 unit

H7 = $=\sqrt{\left((\sqrt{7})^2+1^2\right)}$

= $\sqrt{7+1}$

= $\sqrt{8}$ = 2$\sqrt{2}$ units.

Triangle 8:

Legs are 2$\sqrt{2}$ units and 1 unit

H8 = $\sqrt{\left((2 \sqrt{2})^2+1^2\right)}$

= $\sqrt{8+1}$

= $\sqrt{9}$ = 3 unts.

Triangle 9:

Legs are 3 units and 1 unit

H9 = $\sqrt{\left(3^2+1^2\right)}$

= $\sqrt{9+1}$ = $\sqrt{10}$ units.

Triangle 10:

Legs are $\sqrt{10}$ units and 1 unit.

H10 = $\sqrt{\left((\sqrt{10})^2+1^2\right)}$

= $\sqrt{10+1}$ = $\sqrt{11}$ units.

The lengths of the hypotenuses are: $\sqrt{2}$ units, $\sqrt{3}$ units, 2 units, $\sqrt{5}$ units, $\sqrt{6}$ units, $\sqrt{7}$ units, 2$\sqrt{2}$ units, 3 units, $\sqrt{10}$ units, $\sqrt{11}$ units.