End-of-Chapter Q16

Given:

The four vertices in order.

To Prove:

ABCD is a square, then determine its area.

Property / Theorem / Formula:

Equal sides plus a right angle establish a square; use the converse of Baudhāyana–Pythagoras.

Proof:

${\mathrm{AB}}^2$ = (−3)${}^2$ + $1^2$ = 10. ${\mathrm{BC}}^2$ = (−1)${}^2$ + (−3)${}^2$ = 10.

${\mathrm{CD}}^2$ = $3^2$ + (−1)${}^2$ = 10. ${\mathrm{DA}}^2$ = $1^2$ + $3^2$ = 10.

${\mathrm{AC}}^2$ = (−4)${}^2$ + (−2)${}^2$ = 20 = ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$, so ∠ABC is a right angle.

All four sides are $\sqrt{10}$ and the angle is right. Area = ($\sqrt{10}$)${}^2$ = 10.

Results:

ABCD is a square of area 10 square units.