Whakaoti i te 20^2=AD^2+12^2

$$20 ^ { 2 } = A D ^ { 2 } + 12 ^ { 2 }$$

Whakaoti mō A

$$A=\frac{256}{D^{2}}$$

$$D\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tātaihia te 20 mā te pū o 2, kia riro ko 400.

$$400=AD^{2}+12^{2}$$

Tātaihia te 12 mā te pū o 2, kia riro ko 144.

$$400=AD^{2}+144$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$AD^{2}+144=400$$

Tangohia te 144 mai i ngā taha e rua.

$$AD^{2}=400-144$$

Tangohia te 144 i te 400, ka 256.

$$AD^{2}=256$$

He hanga arowhānui tō te whārite.

$$D^{2}A=256$$

Whakawehea ngā taha e rua ki te D^{2}.

$$\frac{D^{2}A}{D^{2}}=\frac{256}{D^{2}}$$

Mā te whakawehe ki te D^{2} ka wetekia te whakareanga ki te D^{2}.

$$A=\frac{256}{D^{2}}$$

Whakaoti mō D (complex solution)

$$D=-16A^{-\frac{1}{2}}$$

$$D=16A^{-\frac{1}{2}}\text{, }A\neq 0$$

Whakaoti mō D

$$D=\frac{16}{\sqrt{A}}$$

$$D=-\frac{16}{\sqrt{A}}\text{, }A>0$$