Réitigh {(pir^2)}*2=pi(r+5)^2

$${( \pi r ^ { 2 } )} \times 2 = \pi {( r + 5 )} ^ { 2 }$$

Réitigh do r.

$$r=5\sqrt{2}+5\approx 12.071067812$$

$$r=5-5\sqrt{2}\approx -2.071067812$$

Céimeanna ag Baint Úsáid as an bhFoirmle Chearnach

Cealaigh \pi ar an dá thaobh.

$$r^{2}\times 2=\left(r+5\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+5\right)^{2} a leathnú.

$$r^{2}\times 2=r^{2}+10r+25$$

Bain r^{2} ón dá thaobh.

$$r^{2}\times 2-r^{2}=10r+25$$

Comhcheangail r^{2}\times 2 agus -r^{2} chun r^{2} a fháil.

$$r^{2}=10r+25$$

Bain 10r ón dá thaobh.

$$r^{2}-10r=25$$

Bain 25 ón dá thaobh.

$$r^{2}-10r-25=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir 1 in ionad a, -10 in ionad b, agus -25 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-25\right)}}{2}$$

Cearnóg -10.

$$r=\frac{-\left(-10\right)±\sqrt{100-4\left(-25\right)}}{2}$$

Méadaigh -4 faoi -25.

$$r=\frac{-\left(-10\right)±\sqrt{100+100}}{2}$$

Suimigh 100 le 100?

$$r=\frac{-\left(-10\right)±\sqrt{200}}{2}$$

Tóg fréamh chearnach 200.

$$r=\frac{-\left(-10\right)±10\sqrt{2}}{2}$$

Tá 10 urchomhairleach le -10.

$$r=\frac{10±10\sqrt{2}}{2}$$

Réitigh an chothromóid r=\frac{10±10\sqrt{2}}{2} nuair is ionann ± agus plus. Suimigh 10 le 10\sqrt{2}?

$$r=\frac{10\sqrt{2}+10}{2}$$

Roinn 10+10\sqrt{2} faoi 2.

$$r=5\sqrt{2}+5$$

Réitigh an chothromóid r=\frac{10±10\sqrt{2}}{2} nuair is ionann ± agus míneas. Dealaigh 10\sqrt{2} ó 10.

$$r=\frac{10-10\sqrt{2}}{2}$$

Roinn 10-10\sqrt{2} faoi 2.

$$r=5-5\sqrt{2}$$

Tá an chothromóid réitithe anois.

$$r=5\sqrt{2}+5 r=5-5\sqrt{2}$$

Céimeanna chun an Chearnóg a Chomhlánú

Cealaigh \pi ar an dá thaobh.

$$r^{2}\times 2=\left(r+5\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+5\right)^{2} a leathnú.

$$r^{2}\times 2=r^{2}+10r+25$$

Bain r^{2} ón dá thaobh.

$$r^{2}\times 2-r^{2}=10r+25$$

Comhcheangail r^{2}\times 2 agus -r^{2} chun r^{2} a fháil.

$$r^{2}=10r+25$$

Bain 10r ón dá thaobh.

$$r^{2}-10r=25$$

Roinn -10, comhéifeacht an téarma x, faoi 2 chun -5 a fháil. Ansin suimigh uimhir chearnach -5 leis an dá thaobh den chothromóid. Déanann an chéim seo slánchearnóg de thaobh clé na cothromóide.

$$r^{2}-10r+\left(-5\right)^{2}=25+\left(-5\right)^{2}$$

Cearnóg -5.

$$r^{2}-10r+25=25+25$$

Suimigh 25 le 25?

$$r^{2}-10r+25=50$$

Fachtóirigh r^{2}-10r+25. Go ginearálta, nuair x^{2}+bx+c cearnóg fhoirfe é, is féidir é a fhachtóiriú i gcónaí mar \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-5\right)^{2}=50$$

Tóg fréamh chearnach an dá thaobh den chothromóid.

$$\sqrt{\left(r-5\right)^{2}}=\sqrt{50}$$

Simpligh.

$$r-5=5\sqrt{2} r-5=-5\sqrt{2}$$

Cuir 5 leis an dá thaobh den chothromóid.

$$r=5\sqrt{2}+5 r=5-5\sqrt{2}$$