End-of-Chapter Q4

Given:

Z$\left(5,-6\right)$.

To Find:

One right triangle IZN and its lengths.

Formula:

Baudhāyana–Pythagoras: hypotenuse squared equals the sum of leg squares.

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Solution:

Choose I$\left(0,0\right)$, N$\left(5,0\right)$. IN is horizontal and NZ vertical, so the right angle is at N.

IN = 5; NZ = 6.

${\mathrm{IZ}}^2$ = $5^2$ + $6^2$ = 61.

Results:

IN = 5, NZ = 6, IZ = $\sqrt{61}$ units. Other valid constructions can give different lengths.