Question

We have seen how we can draw a series of right triangles as in the picture.

(i) What are the lengths of the sides of the tenth triangle drawn like this?

(ii) How much more is the perimeter of the tenth triangle than the perimeter of the ninth triangle?

(iii) How do we write in algebra, the difference in perimeter of the n‘h triangle and that of the triangle just before it?

Answer 1

(i) Sides of the first triangle = 1 m, 1 m, \(\sqrt{2}\) m

Sides of the second triangle = 1 m, \(\sqrt{2}\) m, \(\sqrt{3}\) m

Sides of the third triangle = 1 m, \(\sqrt{3}\) m, \(\sqrt{4}\) m

……………

……………

Sides of the tenth triangle = 1 m, \(\sqrt{10}\) m, \(\sqrt{11}\) m

Hypotenuse of the 10^th triangle is \(\sqrt{11}\)

Perpendicular sides are \(\sqrt{10}\), 1

(ii) Perimeter of tenth triangle

= 1 m + \(\sqrt{10}\) m + \(\sqrt{11}\) m

Perimeter of ninth triangle = 1 m + \(\sqrt{9}\) m + \(\sqrt{10}\) m

More in the perimeter

\(=(\sqrt{10}+\sqrt{11}+1)-(\sqrt{9}+\sqrt{10}+1)\)

\(=\sqrt{11}-\sqrt{9}\) = \((\sqrt{11}-3)\)m

(iii) Sides of the n^th triangle are

\(\sqrt{n}\), \(\sqrt{n}\) + 1, 1

The sides of the (n – 1)^th triangle are

\(\sqrt{n}-1,\) \(\sqrt{n},\) 1

The difference in perimeter

\(=(\sqrt{n}+1)-(\sqrt{n}-1)\)